TED-Ed's Frog Riddle Is Wrong
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- Опубліковано 30 лип 2024
- TED-Ed presented a riddle last week based on a classic probability problem. However in the riddle there is a small and seemingly insignificant detail that changes the calculation. In this video I present the pertinent details of the frog riddle, explain its connection to the boy or girl paradox, and then do a detailed calculation of what I believe is the correct probability.
TED-ED frog riddle: • Can you solve the frog...
Blog post (another calculation if the probability a male frog croaks is p): wp.me/p6aMk-4wD
Ron Niles made a video that shows the probability visually and explains an interpretation of a male frog croaking with probability p: • Video Response to "Ted...
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When the two smartest kids in class get different answers
So true
And me getting Antartica: ....
and you still get a different answer
lol
😂😂😂😂
I think what you're uncovering here is the problem with probability thought experiments: the lack of clarity on *where the information comes from.* i.e., the phrase: "Given that we know x."
So in the boy/girl example, "Given that we know one is a boy" - the question of how we know that information is important. If one of the kid's gender is revealed at random, then the other kid's gender distribution is unaffected, and it would be 50/50 that it is a boy or girl. Instead we need to interpret the example as a case where we ask the father "Is one of your kids a boy?" and he answers yes. Which would mean that there is a 2/3 probability that the other is a girl.
What is tripping people up is the interpretation of the second part as us asking "Is one of your kids a boy?" and him answering "Yes and he was born on a Tuesday," as opposed to "Is one of your kids a boy born on a Tuesday?." In other words, using a consistent logic as to how probability problems are usually framed, your sample space is correct.
However its easy to see why its confusing, and if the information becomes known in a different way, the answer would change.
It takes some thinking and requires familiarity with assumptions in probability calculations.
I believe that, in cases like this, we need to specify with our language. We are no longer talking about probabilities in the 'random event' sense, we're talking about *uncertainties* - i.e. the amount of certainty we have in a given conclusion based on the knowledge we have. This is *like* a probability, and we can use the Bayesian method of updating our certainties in the same way that we can use Bayes' Theorem to work with actual randomness. What is being done here is related to that, in that it involves the way our Certainty in a conclusion changes when new information is added.
We're often told to *think* about the idea of Certainty as a sort of 'probability that the statement is true' under the given information, but that is misleading, as the events in question are often not actually random variables - they are facts that we simply don't know at the moment, and perhaps can't directly know in any easy way (which is why we are using secondary information to improve our certainty).
Suppose, in the frog problem, a scientist comes along and tells us: "I examined one of those frogs. It was male." But he does not tell us which frog he examined. We know have the Croak version of the riddle again, because we have to account for, in the MM situation, there are twice as many ways for one of the frogs to be the one examined; our certainty becomes 1/2 that there is a girl frog among the pair. However, if he comes along and tells us "I examined the frog on the left. It is male." that ends up, coincidentally, giving us the SAME certainty of 1/2 that the other frog is female, as now it is the only variable in the situation (MM or MF). Despite seemingly giving us 'more information' (which specific frog he examined), it doesn't end up changing the specific result of this problem.
Consider if the problem had 3 frogs.
"There is at least one male frog" - (MMM, MMF, MFM, FMM, MFF, FMF, FFM) = 6/7 certainty that there is a female frog.
"I heard a croak that can only be a male frog." (MMC, MCM, CMM, MCF, CMF, MFC, CFM, FMC, FCM, CFF, FCF, FFC) = 9/12 = 3/4 certainty that there is a female frog
"The first frog croaked a male croak." (CMM, CMF, CFM, CFF) = 3/4 certainty, exactly as in the nonspecific croak case. Perhaps this is a quirk of binary events, or even 50-50 probabilities. Clearly more exploration is needed to see why the addition of what appears to be extra information doesn't actually improve certainty. There seems to be some limit as to how much knowing things about things other than the things we care about can affect our certainty of them. If we want to know if there are any female frogs in the group, there is only so much help that knowing there is a single male frog in the group can be, and we reach that limit earlier than expected. Fascinating.
This is why I did so poorly in Statistics and Probability in college. Ended up changing my major, and basically my life.
Shouldn't it be 'at least one of your kids' if you ask him? Otherwise he's saying it's 1 for sure and the chance of it being a girl is 100%.
i am glad you wrote this, i was starting to doubt myself. I could see several answers to this problem.
Funny story, I had this actually play out for me.
When my wife and I got pregnant with twins, we didn't know from the first scan if they were boys or girls (too early to tell). However, we had a genetic test that came back with the presence of Y chromosomes in the mothers blood (at least one of the babies is a boy).
I quickly worked out the probability, and none of her family believed me that it was twice as likely that we were going to be having a girl and a boy than twin boys.
Well, in reality the calculations were much more complicated. One in three sets of twins are identical, and identical twins are always the same gender. Also, the stenographer saw separate placentas. Identical twins share a placenta 1/3rd of the time, fraternal twins never do.
I eventually worked out that the odds of us having a girl were 3/5.
My wife is now tree weeks away from giving birth to our son and daughter.
Assuming the probabilities you gave were correct, the chance of having a girl would actually be 6/11, not 3/5.
After eliminating the both girl possibilities from the sample space, it is a 1/2 chance and then after eliminating shared placentas you only eliminate the identical boy boy shared possibility, which was at the time 1/9. Hence the final probability is 6/11
Make an edit about the update dude!
@@manvendrasingh299 They are wonderful kids. Over a year old and walking a bit now.
@@insertphrasehere15 great!
@@insertphrasehere15 take them out for icecream for me dude! i want them to remember you as the coolest dad ever.
The issue with your calculations is that you are counting permutations rather than combinations. It doesn't matter whether or not the boy is older or younger than the other child, so the extra column of possibilities in BB should be removed. Same thing with the frogs, you only care about the number of combinations containing a female, not the number of permutations.
I'll give you a more extreme example on why this is the case that might make more sense (the croaking and order of croak is irrelevant - it is simply used to identify a male frog - you may as well paint one red as in the example below)
I divide 300 frogs up in the following manner (I know which are male because I have tested their chromosomes in the lab - you can't tell them apart any other way)
3 groups
Group A - 100 Males
Groups B & C both 99 Females and 1 Male
I tell you that 1 of the groups contains 100 Males and the other 2 contain 99 females and 1 male. I ask you to pick a group with males only - you have a 1 in 3 chance as the odds are based purely on the basis of selection between 1 male only and 2 mixed groups - and without any other information you are twice as likely to pick a mixed group as a male only group.
I now paint one of the male frogs red identfying it as such - and arrange the same groups and tell you the same information as before and ask you to pick a group with males only. If you chose the group with the red frog in it you will now be correct 98.04% (100/102) of the time - the odds of the red frog being from the all male frog group being A are no longer the same as B or C. I have changed the choice from a random selection of a GROUP to the odds of selecting A KNOWN MALE from A KNOWN DISTRIBUTION and the distribution between the groups is not even.
Same for the scenrio above instead of chosing a random group you are chosing 1 of 4 male frogs - the one which you have identified as definitely male - and 2 belong to group A 1 to B and 1 to C - your odds are now 50/50 of chosing a male only group instead of 1/3. (or 2/3 for female as the question is phrased)
@@TB-up4xi have the 99 female frogs worked out that they all have green eyes?
if the order of the children didn't matter than the BG and GB groups would be the same as well leaving it as 50/50 again. However they do count as separate chances and for the same reason they do in punnett squares
@@AhzaelVoan yes, for boy and girl the chance is 1/2
@@TB-up4xi you paint one of the frogs red - I randomly chose 1 of 3 groups. Chance is not equal to 1/3 only if I would pay attention to the painted frog.
Your example isn't similar to video's example
It was a toad
There are no toads.
+Dylan Macon what is life
What are cows?
Me.
The riddle was actually called Frog Riddle.
do you ever have the feeling that something is not right, but don't know what it is?
That's why we have maths and logic.
I get that more than average people. No, genuinely. I'm autistic, and my brain runs on hyperactive, as do my senses. So, I can tell if something's wrong, even if is that someone is lying, because my brain kinda works faster than my thought... ah. Then you realise that I'm a complete nerd who got offered uni lecture drop-ins at the age of.... 13?
-that does not say that I am 13. It instead clearly states that I am atleast 13, if you have some logic.
(written for the purpose of my privacy)
+Alex Lowe you're lieing, I can tell those things
I know what you mean, Luke. The whole part where he arbitrarily discounts the pair of two boys born on a Tuesday because he wants all the probabilities to be equal seems wrong, but I don't have language to explain why.
Aienbalos Aienbalos If you want, ask cardiff uni maths department, Dr J Gillard is head of admissions, If someone the age of 13 was offered uni lectures. I'm not really called Alex irl, so no point saying that bit.
the error is that you count BG and GB as separate but not BB and BB, as well as GG and GG
this makes the sample space [BB, BB, BG, GB, GG, GG]. given the information that he has a boy, you remove the two impossible events of two girls. the new sample space is [BB, BB, BG, GB]. now if you cancel out one boy for the boy we know he already has, it's [B, B, G, G]. There's a 50% chance it's a boy, and a 50% chance it's a girl
I was thinking about that
The error with your error is that the single BB (without the extra info) has a 25% chance and if you split it into two events you don't change anything about the probability so you get two events with 1/8 chance, so if you do that the 6 options do not have equal probabilities. Take that into account and you are back at the start.
The reason the two BB are not separate is., well for once that we want options with equal probability but how we get that is simple. Take a fixed option for the first like B, then the options are BB and BG both equally likely, take the other fixed option and add GB and GG also both equally likely. It should be obvious that doing it this way gives you options that have the same probability.
there is no "BB and BB" or "GG and GG." There's not a question of ordering. It makes sense to say "Child A is a boy and Child B is a girl, OR Child A is a girl and Child B is a boy," and have those be two separate outcomes.
But it makes NO sense to say "Child A is a boy and Child B is a boy, OR Child A is a boy and Child B is a boy." These are not distinct outcomes, they are logically coidentical. And you could go on doing this infinitely with new cases and say nothing new. "Child A is a boy and Child B is a boy, OR Child A is a boy and Child B is a boy, OR Child A is a boy and Child B is a boy, OR Child A is a boy and Child B is a boy, OR Child A is a boy and Child B is a boy, OR..."
By that logic duplications would always be the single most likely outcome, and we should expect that out of 100 children, the most likely combinations are either 100 boys or 100 girls. But we _don't_ expect that, and for good reason.
It's not that hard. Either order matters or it doesn't.
If order matters, then Child A is known to be Boy. We need to solve for Child B. The only possible sets are BB and BG; GB does not exist because we know Child A is a Boy. So it's 50%.
If order does NOT matter, then BG=GB and Presh is double-counting in his set. The only two possibilities are BB and BG/GB (which are the same) and again it's 50%.
@@RickJaeger Let Bk = the known child (Boy) and (B/G)u represent the unknown child. You have the following possibilities:
Bk Bu
Bu Bk
Bk Gu
Gu Bk
And it's 50%. You say there's no "BB and BB" because they're the same, but they're not the same - they just look it because there's no notation.
The fact is that this is NOT conditional probability. The sex of the known child has no bearing on the unknown child. Trying to frame it any other way is wrong.
In a practical example: Mrs. Smith has a baby in 2020. This is child A and it's a Boy. In 2021 she has another baby. This is child B. What is the probability that it's a girl? It's 50% because child A has no bearing on this outcome.
It works in reverse too: Mrs. Smith has a baby in 2020 (child A) and it's a 50/50 chance on the sex. I miss the birth announcement and don't know what it was. In 2021 she has another baby (child B) and it's a boy. The probability of child A remains 50% for each sex.
So whether the known child is A or B, the odds of the other one being a girl remain 50%.
You may argue that I'm adding order to this problem (similar to how Presh added "Tuesday" to change the probability) and that's why it's 50%. I'll grant you that. But if order does not matter, then BG=GB. In that case there are only 3 initial possibilities:
BB
BG (or GB, order doesn't matter)
GG
We can eliminate GG so that leaves a 50/50 split again.
What I don’t understand is that logically it makes sense to set up these scenarios and permutations, but practically it also makes sense to say that you know one of the two frogs is male, now my sample size is the one I don’t know and that has a 50/50 chance of male or female, as it’s independent of the gender of the other frog. Like, why is that not right?
It's a problem of sampling. Imagine there is a scientist presenting you the problem... He picks up one frog, tests it, and says "this frog is male, what are the chances that frog is female?"
The possibilities for the frogs are these (the first frog being the one tested): MM, MF, FM, FF. Your answer should be 50%, because he eliminated two of the frog combinations: FM and FF, where the test proved that the first frog is not female.
Now imagine the scientist picks up and tests both frogs, and the says "at least one of these frogs is male. What are the chances there is a female?" You have lost information, because you no longer know which frog is male. The scientist has only eliminated one possibility: FF. The other combinations, MM, FM, and MF, all have at least one male, and you don't know if it was the first or second frog which was male. You can then say there are 2 (FM MF) out of 3 (FM MF MM) possibilities that there is a female frog.
@@Greenicegod im confused why MF is different than FM
@@ahmada.1325 there's a difference because only one frog comes first. The possibility space for two frogs is a square (where M1 is the first frog male, F2 is the second frog female, etc.):
____| M1 | F1
M2 | MM | FM
F2 | MF | FF
In the case where the scientist picks up the first frog and finds it to be male, that is, the possibilities have collapsed to the left column, you only have to guess at the second frog.
When the scientist checks two frogs (or doesn't tell you what he did), you can only eliminate the all female case. You don't know if it was the first frog or the second (or both) which was male.
Thus, the FM and MF cases have to be counted separately.
Hope that helps clear it up!
@@Greenicegod but it doesn’t matter which one comes first. There is always a 50% chance that the other is female, whether he shows you or not. Showing you or not showing you, doesn’t actually make a difference logically
@@tavasp Think it this way. If you throw two coins, which are the posible outcomes? It's equally posible to have two tails, two heads and a tail and a head? No, you have 2 in four posibilities of having different results. (HH, TT, TH, HT). If you throw the two coins, I hide the results and say to you that one of them is tail, will you say that the result is two tails or tail and head? The thing here is that you know there is a male frog, but not which, so the probabilities change.
There is no chance that Mr. Jones had a girl.
Because I know those kids aren't his.
Because me and Mrs. Jones, we got a thing going on.
damn
in this case what is the possibility of the kids being yours or his , and what is the possibility of him or you having a girl and what it the possibility of you or him having twins ..... so on so far ...Lol
+Jordan Kelley booooooooooooooooooooooooooooo
Dam the Michael buble reference on point
+Josie Doe Dude, it's a Dr Hook song.
couldnt the frogs croak at the exact same time with half of their regular voice amplitude and sound like one croak
Possible
Maybe but the chance of thes frogs croaking at the exact same time is so small that it barely affects the final outcome
What about the wavelength and amplitude did u consider these in ur assumptions
TheKrazyKid17 thats not how sound works
@@supercool1312 i meeaaannnnn is he wrong thougghhhh
Min. yes.
Just to make point... in the above question add another condition (c) he has a boy born in 1997... I would like to see you come up with an answer for this... 🙂
The probability will go closer to 50% but will still remain higher than it
I think the general formula is (cn)/(((cn/p)-1)):
Where c is the number of outcomes (2 in our case since there are 2 genders) n is the number of possible configurations and p is the probability of the event.
In this case Tuesday is a "day" configuration so you put n=7 and p=0.5.
For the base case, n=1 (they can only be born, no other configuration is allowed) so (2*1)/(((2*1)/0.5)-1)) is 66%. For the frog croaking case the formula changes coz the -1 is no longer relevant (the -1 is done to disregard 2 identical cases but in the frog croaking case the two events aren't identical)
For your case, assuming siblings aren't born more than 41 years apart (Guinness Book of World Records holding siblings are 41 years apart), you consider 41 years before, 41 years later and 1997 so 83 years configurations. So with n=83 you get 50.15%.
You can increase the number of configurations by being more specific e.g. born at 9 o'clock on Tuesday in the second week of February. But of course, at n=∞ you will get exact 50%.
If he was born in 1997, he's not a boy, he's a man.
Yes, this horrifies me too :)
I feel like it is not correct to cancel out the two b2 sets... And when set he had the Tuesday boy first and then the other set he have the Tuesday Boys II they still count equally and they are different sets
Absolutely right
I love watching mind your decision but I have to say that he is wrong on this one not only because he took out one of the b1 pairs but the 2 examples have nothing to do with eachother since the frog puzzle is saying what's the chance that there is a female given that there is a male and mind your decision's problem is saying what's the chance of a girl given that there is a boy born on Tuesday it should be one or the other if he really wanted to calculate the last one he would put all the number of combinations of when a boy was born on Tuesday over all possible combinations
i was also thinking about the same thing
*HE IS TOO DANGEROUS TO BE LEFT ALIVE*
Hi I thought the same thing when I saw him canceling out the B2s, but then I created an Excel sheet and listed out all the possible combinations of having two babies, from B1B1 to B7B7, G1G1 to G7G7, BG to BG, and GB to GB and so on. There are in total 7*7*4=196 possible combinations, among those 27 combinations include a 'B2'. On my sheet the (B2,B2) cell is at the center of a cross meaning it is counted twice that's why He cancelled out when of the B2B2s (otherwise Excel would tell me it finds 28 results). Then, not surprisingly, 14 out of the 27 cells contain the text 'G', so, I guess MindYourDecisions is right on this one😬😬😬
So the more information you have, the more even the probability?
One is a boy = 2/3 chance the other is a girl.
But,
One is a left handed 12 year old boy born on the Fourth of July and likes hamburgers but not pickles and has 2 frogs of questionable gender = ~50.00.......01% chance the other is a girl?
the more details you have, the bigger your sample space with more repeated elements in the sample space that can be removed and so you will approach a limit of equal probabilities for a boy or girl
This is like saying, "I flip a coin. It was heads. If I flip it again, what is the chance it will be tails?" V.S. "I flip a coin on Tuesday. It was heads. If I flip it again, what is the chance it will be tails?"
It's gonna be 50% no matter what.
The general idea is that when you have at least one information, you are separating the 2 elements of your proba. actually, the removing of B2B2 is an error IMO : at the moment you have at least one way of separating the 2 elements, more information don't make it more even ; it already is
No, the removal of the second B2B2 is not a mistake. It is not twice as likely that that particular event will occur just because is showed up twice when creating the sample space. It either happens, or doesn't happen. It can't happen more than that.
+Steven Jacks
I liked that Steven Jacks. It shows that real life isn't as "simple" as statistics theory makes it out to be.
My guess is it depends on the rules the story teller is following.
If it's simply
a) If the father has at least one son I will state that he has a son and the day he was born on
Then the odds remain 2/3 that he has 1 boy and 1 girl. Because nothing has changed.
though in
b) If the father has at least one son who is born on a Tuesday, I will state he has a son born on a Tuesday
Then the odds change dramatically to 52% that he has 1 boy and 1 girl.
Holy shit.. that makes sense right? I'm pretty sure that makes sense!
you roll 2 dice, one falls to the floor.
you see the dice on the table is a 3:
what are the odds that the dice on the floor is a 3?
"well, if you say it could be, uh
31,32,33,34,35,36
13,23,33,43,53,63
you can remove the 33, since it's a mirror, so, it's a uh, carry the 7, divide by 0, and we got a 52% chance of the dice on the floor being a 3"
greaaaaaaaaaaat logic
and theres your fault you dont remove the 33 since its a mirror you remove it because its the same case. so you know the table one is a three then it would first be 1/6 then what would be the possibility for the second be well 1/6 x 1/6 that analogy(i dont know how this is spelled, im german) would have about nothing to do with this problem since you can exactly determine which is on the table and which colour they have . Not knowing that would change the possible outcomes but i dont actually understand why it would change the possibility. Now in his frog case didnt we first establish 1/3 f/m 1/3 m/f 1/3 m/m then wouldnt then follow (c for croak) 1/3 f/mc 1/3 mc/f and 1/6 mc/m and 1/6 m/mc? (assigning the "who croaked" after determining the possibility or am i getting something wrong)
and regarding the thing you wrote about: you wrote down the possible outcomes but! 33 isnt a mirror its the same case if it were like you said rolling three times in a row a 6 would be as likely as randomly having at the end rolled a 1 a 3 and a 4
which is wrong
you have to follow the cases:
6-6-6 is the only case in which you get 3 times a 6 at the end when having rolled three times so-> there is one case
if you have a 1 a 3 and a 4 at the end
1-3-4 1-4-3 3-1-4 3-4-1 4-1-3 4-3-1
there are a total of 3 factorial (3! = 6) cases with equal possibility that is 6* 1/(6*6*6) = 1/32 instead of 1/(6*6*6) = 1/192 which is a huge difference
Oh forget that thing about the croak Hes right it makes the m-m combination more likely and thats why its 50/50%
1/6?
+Toga The fault in your logic is that you know the die on the table has rolled a three. let's say this is the first die. the sample space is:
31,32,33,34,35,36
that are all possible combinations. 13 is not possible because it would mean the die on the table had rolled a 1. we know for a fact it didn't.
However, if both dice had fallen under the table, someone else had looked and told you one of the dice had rolled a 3 your calculation would be correct.
4:53 "And the f-"
Egg and cheese and the With-Fi sucks!
*4:48
i Sableye
L
i Sableye O
i Sableye L
The boy/girl problem is by far less obvious than suggested here (look for the Wikipedia entry 'boy girl paradox'): It depends on how you interpret the statement: a) From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of
1/3. Or b) From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2. Similarly, the rest of the questions discussed here don't contain enough info to decide about probabilities.
EDIT: I provided detailed math behind the statements made in this comment a couple replies down.
This video is flawed at several points.
Part B of the Boy or Girl paradox only has a probability of 52% if we intentionally selected Mr. Jones' family with the precondition that we could examine only families with two children, one of which is a boy born on a Tuesday. If instead Mr. Jones' family was randomly selected from all families with two children, one of which must be a boy, and then asked on which day that boy was born (which turned out to be a Tuesday), then the probability that Mr. Jones' other child is a girl is instead 67% because we didn't start with the precondition that Tuesday was the only day on which the boy could have been born and so the day is still independent of whether or not the other child is a girl. Furthermore, if we start by selecting a random two-child family, examine one child and make the determination that child is a boy, the probability the other child is a girl is 50% as the examined child was a boy just by happenstance and not as a precondition of the selection of the family; asking on what day of the week the boy was born does not change this probability.
TLDR; The statement of the problem must indicate from where we acquired the information about both the gender and the day of the week of the child we know about before we can make an accurate calculation of the probability of the gender of the other child. An easy way to see this is to think about whether we can calculate the probabilities of both children's genders before or after we learn the information about one of the children; if we indeed selected Mr. Jones' family at random, each child's probability of being a girl is 50% and any information we learn about one of the children does not change the probability of the other after the fact. The probability one of the children is a girl being anything other than 50% requires preconditional information about the other child.
The frog part is also flawed (assumptions must be made in either case), but I've already contributed more to this discussion than I should have.
+DirkAlmighty13 This comment is similar to what I was going to post, if the information about the boy's day of birth was acquired after the family was selected, it's not relevant.
It's actually the same as the Monty Hall problem in a sense...did they open the door without a prize at random and it had no prize, or did they intentionally open a door without a prize on purpose.
If you pre-select families with a boy born on Tuesday, it's relevant and has an impact.
If however you selected at random and the boy was born on a Tuesday, it has no impact.
+EnderSword
You 2 are very right. However, I noticed that _if_ the frogs didn't croak, you wouldn't have noticed the 2 frogs in the first place.
Thus, croaking does seem to do something to the probabilities. Personally, I think the rate/frequency/chance of croaking matters a bit. I need to sleep on it.
+pierrecurie The reason the frog portion is flawed occurs at 8:00. "We're going to assume these are all equally likely events". How can we? Maybe the frogs croak more frequently around female frogs? Maybe when two male frogs croak, we can only hear one. The situation is presented in the context of a mathematical problem. We are mathematicians, not biologists. We can't take simple facts presented in the problem and extrapolate them. If a question on a math test asked me about how many oranges I can deliver in a truck, do I assume some of them are going to rot? Of course not.
+DirkAlmighty13
Remember that P(A given B) = P(A and B) / P(B).
So for situation B in the Boy / Girl Paradox:
Let there be two children, C and D.
Let P(A) be the probability that C or D is a girl.
Let P(B) be the probability that C or D is a boy born on a Tuesday. (Or in both cases is inclusive)
Assuming it is equally likely for a boy to be born on any day of the week, and the it is equally likely for a child to be either gender,
Since the events in P(B) are NOT mutually exclusive, P(B) = P(C is a boy born on a Tuesday) + P(D is a boy born on a Tuesday) - P(C and D are both boys born on a Tuesday)
= P(C is a boy and C is born on a Tuesday) + P(D is a boy and D is born on a Tuesday) - P(C is a boy and D is a boy and C is born on a Tuesday and D is born on a Tuesday)
= P(C is a boy) * P(C is born on a Tuesday) + P(D is a boy) * P(D is born on a Tuesday) - P(C is a boy) * P(D is a boy) * P(C is born on a Tuesday) * P(D is born on a Tuesday)
= (1 / 2) * (1 / 7) + (1 / 2) * (1 / 7) - (1 / 2) * (1 / 2) * (1 / 7) * (1 / 7) = 27 / 196
Since event A is _DEPENDENT_ on event B, P(A and B) =/= P(A) * P(B). Instead, P(A and B) = P(C is a girl and D is a boy born on Tuesday) + P(D is a girl and C is a boy born on a Tuesday)
= P(C is a girl) * P(D is a boy born on Tuesday) + P(D is a girl) * P(C is a boy born on Tuesday)
= (1 / 2) * (1 / 2) * (1 / 7) + (1 / 2) * (1 / 2) * (1 / 7) = 1 / 14
Therefore P(A given B) = (1 / 14) / (27 / 196) ~ 52%
And so the answer of 52% is correct.
If you would like to dispute the math, feel free, but please, keep it civil.
+KWG
Your description of the event when examining the children is not necessarily accurate. I would actually argue that according to the wording used in the video, it is NOT accurate.
Indeed your calculations are accurate IF we are assuming the condition that Mr. Jones' family has a boy born on a Tuesday is determined prior to the selection of the family (i.e. Mr. Jones' family would not have been chosen had he not had a boy born on Tuesday).
To restate your calculation more concisely:
P(G.Bt|Bt)=P(Bt|G.Bt)*P(G.Bt)/P(Bt)=(1)*(1/14)/(27/196)=14/27 ~~ 52%
where P(G.Bt) is the probability a girl is in the family AND a boy born on Tuesday is in the family
P(Bt) is the probability a boy born on Tuesday is in the family (regardless of the other child)
| indicates dependence; P(X|Y) = P(X given Y)
This is Bayes' theorem; for which you provided a simplified version. In your case, P(G and Bt)=P(Bt|G.Bt)*P(G.Bt)
Put simply, this is the probability of one of the children being a girl GIVEN that one is a boy born on Tuesday. The boy being born on a Tuesday is a GIVEN piece of information (it is assumed true at the start).
However, at 3:41 in the video, the narrator states "...the fact we LEARNED that the boy is born on a Tuesday...." The word LEARNED indicates that the information about on which day the boy is born was gathered after the family was chosen (the family wasn't chosen out of a pool of families only with a boy born on Tuesday). This seems like a more typical situation.
This information wasn't GIVEN, it was LEARNED.
This means that P(G.Bt|Bt) is NOT the correct probability.
Instead, it is P(G.Bt|bt) where P(bt) is the probability that any one particular child we examine will be a boy born on Tuesday after we select the family at random
P(bt)=P(b)*P(t)=(1/2*1/7)=1/14
P(bt|G.Bt) is the probabilty of a particular child being a boy born on Tuesday in a pair of children (one a girl, one a boy born on Tuesday)
In this case P(G.Bt|bt)=P(bt|G.Bt)*P(G.Bt)/P(bt)=(1/2)*(1/14)/(1/14)=1/2 == 50%
Your case is equivalent to "if we take a family that has a boy born on a Tuesday, what is the probability that the other child (in a pair) is a girl?"
My case is equivalent to "what is the probability that a child (in a pair of children) is a girl when you find out that one is a boy born on Tuesday?"
What's important is how we know there is a boy born on Tuesday.
A very easy intuitive way to see this is considering a similar case with coins.
I have a jar full of coins. I flip two coins and cover up the results.
In my case, I look at one and notice that it is heads and it is a coin minted in Philadelphia. What is the probability that the other is tails? Still 50%.
In your case, I ignore the coins and keep flipping (more coins) until I know (somehow) that one of them is heads and was minted in Philadelphia. What is the chance the other is tails? P(T.Hp|Hp) =/= 50%
Doesn't my case seem more typical? Why are we ignoring sets of coins that don't have ones from Philadelphia?
For more information, look at the Wikipedia article on the "Boy or Girl paradox", particularly the "Second question" section and the following section on ambiguity.
Because you have not demonstrated any indication that you are hostile, I will not assume in advance that you are; I trust that you will provide a response that is civil and that you have the faculty to do so.
You are counting sequence probabilities not results.
BoyGirl = GirlBoy
Ya that's what I thought
Michael MAnville ah, but in that case, a boy and a girl has a higher chance that 2 boys, as there are twice as many ways of it occurring.
Here’s an explanation: say Mrs. Hull has a baby in January . 50% chance she has a girl and likewise for a boy. Then in November, she has another baby. Again 50% chance she has a girl and likewise for a boy. So our sample space would be:
B(j) B(n)
B(j) G(n)
G(j) B(n)
G(j) G(n)
As you can see, BG and GB both count because they had switched positions
It should be given as 2*boygirl, say if you flipped a coin twice, the probability of both events giving the heads is 1/4, likewise for tails giving 1/2 probability of those outcomes and a heads tails mix is a will have the other 1/2 of the probability.
@@kent2670 i thought that too
you could also say that hearing a croak identifies the frogs, as you now have "frog that croaked" and "frog that didn't croak". You only have to determine the gender of the "frog that didn't croak" which has obviously a probability of 1/2 to be female
I disagree if you follow his reasoning; you are assuming that the frog which croaked was identified, but the problem as presented does not say that, so 2/3 is the correct answer. Unless you _see_ which frog croaked you only know one of the frogs is male, which eliminates the FF pair, leaving the MM MF FM space.
That said I disagree with the OPs analysis, and agree with yours, just not for the reason you stated. See my comment on the video.
But why would he count FM and MF as different variations, they are the Same. It doesn't matter whether the male one Sits on the left or the right. This would also reduce the possibilities to FM/MF and MM which means ½
@@lawrence-dol Consider the situation in which you hear 2 croaks from the pair of frogs, without knowing if the same frog made the two or different ones. In such situation you'll end up in a probability of ⅖ for finding a female frog. So this video is right
@@joelthoss8255
The chance the first frog is a male is 50%.
The chance the fsecond frog is a male is 50%.
The chance both frog are male is 25%.
The chance both frog are female is 25%.
The chance that one frog is a male and the other is a female is 50%.
The probability is 1/2 but not because of this. This isn’t true because you don’t know which frog croaked. Let’s say that, instead of the frogs croaking, there is a big sign that just says “at least one of these frogs is male”.
Since you don’t know which frog is male, you have to lick both of them, which gives a probability of 2/3. If you did know which frog is male, then you would just lick the one of unknown sex, which gives a probability of 1/2.
I think it is a mistake to remove the one of the B2B2 sets, these are two separate sets which you are taking the results from to provide the calculation in a top set. They are not the same data point, they just look the same. Imagine you have two sets of lizards with lizards of different colours. On your left are three cages one set contained a cage with two red lizards, a second a green and red and a third a green and black. On your right you have three more cages a set contained a cage with two red lizards, a cage with a blue and green lizard and a cage with a black and green lizard. It would not be correct to remove one of those cages with two red lizards just because it looks the same as the data point in the other set. Similarly the two B2B2 sets are not the same data point they are two separate data points.
Yeah he shouldn’t have removed it in the boy-girl paradox but I guess he just made a mistake trying to prove his point. However imo he did do it correct the second time with the frogs and get’s the right answer at the end.
Actually, if you think of the data set presented, it's like a data table, where you can lock first child as line and second child as colomn. In that table, the B2B2 only shows once. That's the same data point in the sample, because it's the same pair of child in the table.
@@rafaelsobrozavieira1530 No I think you made the same mistake he did, just because two data points are named the same doesnt make them the same. Think like this if couple a have two children Ted and Jane and couple b have two children called Ted and Jane it doesnt make them the same data set. They are two separate children altogether they are just named the same. In the case of this video he used the names b2 for two different data sets. Each b2 was a separate data set.
@@rafaelsobrozavieira1530 Yes, the video is correct, as you say. An anology is that you are twice as likely to roll a 1 and a 2 with a pair of dice (because this can be done in two ways) than two 2s. In the same way it's twice as likely that you will have two children born on Monday and Tuesday (two different ways that can happen) than two children both born on Tuesday.
Riddle me this, Paul.
I have two boys, Abe and Ben. At least one of them was born on Tuesday.
What is the probability that both boys were born on a Tuesday?
To all those claiming he's "overcomplicating" the issue and it should be a simple answer, there's a guy with some doors and some goats and a brand new car who would love to make a bet with you.
+martijn van weele It isn't the same formula. They are not different versions of the same riddle. Big deal if they have one similarity.
+Evan Leeds They are ALL conditional probability problems. The classic Monty Hall example (it's not a paradox) is similar.
+Frample Tromwibbler He isn't say "this is not the same riddle". He said "this is unrelated to the Monty Hall Paradox". Which is false. In both cases, the key to the solution is to enumerate your options, carefully making sure they have equal probability, then see what happens when options are removed. In both cases, in fact, you get a good answer by using a decision tree. They're pretty closely related.
John Aldis
The Monty Hall problem is not about eliminating impossible solutions to find the new probabilities. In the Monty Hall problem, the information about the probability comes from the fact that the door was not opened at random, but rather opened on the basis of revealing a goat. They are both probability riddles but the thought behind solving them is not similar.
+Frample Tromwibbler It's actually a really close problem. In both, you have a first case probability, then you give one extra information (One male, or goat behind one door) and you have to carefully calculate the new sample probability.
I think you are wrong with the croaking frog. The probability should be around 0% that 1 is female. Male frogs croak to attract a mate, once one is present they stop croaking.
The Point
You
yeah, apparently this is only a math problem and shouldn't include biology or logic outside the problem... this is why I hate these math only problems.
This is a very valid point and made me laugh out loud.
How do you know the mating habits of a hypothetical animal? Did you just imagine going on a expedition to a fictional forest?
What if the mate adjacent to the male frog is somehow an undesirable female? Would it continue croaking?
1:51 "CHILDRENS", hear carefully.
It's a common (and understandable) mistake for non-native speakers (like me). Almost all plurals in English end with an "s". When one doesn't, there is an opportunity for mistake.
Problem: just because only one frog croaked doesn't mean they can not croak or did not croak and you missed it.
You only know that a male croaked.
Therefore, you can take the information that there is a male, and ignore the rest.
When are you more likely to hear a male croak, when there is 1 male frog or 100?
@@Miscio94 basically if you hear only 1 male croak then it's a more likely scenario than with 100 croak. But i get your point
Yes cuz after all we are counting probability of the frog being male. But in this video he's considered a third set of frog not croaking which increases the probability of the frog being male but in reality it doesn't .uh this is hard to explain
whydoyou needmyname it’s increase the chance in reality if you been given a random pairs of frog
@@fos1451 i meant which increases the probability of male and decreases the probability of female
You are calculating the probability that (he has a girl) AND (has a boy born on Tuesday) given he has (at least one boy) AND (at least one boy was born on Tuesday). This is a totally different problem... You can't combine sample spaces for two events and only apply them to one event.
+Opine o' mine I'll be honest, I thought like hell about why it didn't make sense to come to that conclusion. It was pretty obvious that the Tuesday event was irrelevant, but I couldn't find any flaw with how he did his sample space. I eventually realized the sample space is right, only it's answering a different problem.
+mizuhonova I was stuck on the fact he removed B2B2 thinking it should be equivalent to the other B2B2. You're right, his sample space should be half the size because the order the children were born has no meaning in this context.
+Xeridanus
The two children are different children. You have to separate their outcomes somehow, he just did it with order of birth.
His sample space was correct. He listed all possible outcomes given there was at least on boy and at least on boy was born on Tuesday.
+Brooks Ogborn But it doesn't matter if they are different children, as long as one of them is a boy. Order of birth isn't correlated with gender.
My teacher showed us the ted-ed video a couple months ago in class and I tried to explain to her how actually both options would be 50% but she wouldn't listen, she just kept saying "well the video says it's 2/3 so you must not understand properly"
You were 100% right. Your teacher needs to go to summer school, along with Presh and Ted-Ed.
Why did she choose that one? Why not choose one with a less controversial answer
@@alex2005z It's bold of you to assume the teacher realized the answer was controversial.
@@zombieperson3695 true
Bet your teacher wasn't in science major. People who study science always question the source and always looking for possibilities that established truth can be proven wrong with new data.
I usually enjoy these videos but find this one to be confusing, and incorrect in the case of the Boy or Girl paradox.
At 5:06, the author assumes that all combinations have equal probabilities, which is not the case. The overall probability for having a girl (BG + GB) must be 2/3 since we know that one child is a boy (see (A)). The B2B0 and B0B2 are two-fold less probable than the B2G0 and G0B2 possibilities.
The removal of one B2B2 at 5:10 leading to 52% does not make sense either. If you do that, you need to weigh the probability for that instance accordingly (double it!)
The wording of the question is not specific enough to change anything about the probability anyway.
Are they born in the same week? On which day does the count of 7 days start?
I just disagree with the boy-girl paradox sample space in general. In the BB case, the fact that one boy or the other is born on Tuesday is irrelevant, so the entirety of the second column should be removed because it should not be weighed twice. If you have a boy born on Tuesday, there are only 21 different possible combinations for the other sibling, 14 of which involve a girl, so it's still 2/3.
@@Harmonic14 No probility doesnt work like that. Its weighted twice? Yes exactly, it must be weighted twice because there are 2 ways to achieve the same final result so its more likely that other result but he present it by ways to make it not the final result.
"The B2B0 and B0B2 are two-fold less probable than the B2G0 and G0B2 possibilities."
and why is that?
But isn't having boy girl and girl boy the same thing?
Considering that they are two individuals, being a boy or a girl makes quite a difference for each of them. :)
No. If you flip a coin twice, HT and TH are different.
Looks good but your production value is lower, so Ted must be right.
+George Cakiades You have managed to make a sarcastic comment on UA-cam that can be distinguished as such. Congratulations! :)
Let's play a game: find the flaws in MindYourDecis videos
@@lovelindqvist4765 LOL I lost
Why are you getting rid of the "B2, B2"? They relate to different possibilities so they have to be included.
The first - is where the elder son is the one being defined as "being born on Tuesday" in the given information.
The second - is where the younger son is the on being defined as "being born on Tuesday" in the given information.
They refer to two different eventualities.
I am still thinking about this, but I tend to agree. In the frog story it would relate to the case that both frogs croak... The story does not say how many times the frog croaks or whether you can differentiate their voices
+omyyer: I totally agree with you that this is one possible way to look at the problem, yet it's not the only one. The reason why many people in the comments are confused about this is that +MindYourDecisions failed to include one tiny but important Detail: The probability of Mr. Jones having a girl actually depends on the process of randomization used to select Mr. Jones.
Let's take a look at possibility number one (which would lead to the outcome discussed in the video): Imagine we created a list of all parents who have two children, at least one of which being a boy born on Tuesday. If we selected one parent at random from that list, it wouldn't matter if he had one or two sons born on Tuesday. The probability stays the same, since each parent appears on the list exactly once. Therefore B2B2 must only be included one time, leading to a probability of 52%, as discussed in the video.
If we did, however, create the list by looking at all children who have exactly one sibling and created a list entry each time we found a boy born on Tuesday, a parent with two sons born on Tuesday would be mentioned in the list twice, doubling the probability of that parent being selected. In that case we would have to include B2B2 twice, resulting in a probability of 50% for the second kid to be a boy/girl.
P.S.: Constructive criticism is always welcome. This does not only include the content of my comments, but also my language skills since I'm not an English native speaker. ;)
It's also true to remember that if there are an equal number of male frogs and female frogs, the fact that you heard a male frog means that either of the other two frogs have an added probability to be female, just because there are more left after the male one was removed.
Lol Indeed!
+omyyer I agree. It is not clear why some are treated as combinations and others as permutations in calculating the probability. We also have unspoken assumptions about M/F distributions, sample sizes, and total populations and NO information about behavior (frogs travel in mated pairs, etc...)
It's interesting to consider a group of three people walking in the woods. Alice, Bob, and Cindy. Bob is wearing headphones, and Cindy is deaf.
All three eat a poison mushroom and are searching for frogs. All three stumble across the frog on the stump.
Alice hears a frog croaking behind her, and sees the two frogs in the clearing. Alice then gets the attention of Bob and Cindy. Bob takes off his headphones to hear Alice say, "one of those frogs is male." Cindy, being deaf, doesn't hear anything.
Cindy is only aware that there are two frogs in the clearing. She calculates that she has a 3/4 probability of survival if she goes to the frogs in the clearing and a 1/2 probability survival of going to the frog on the stump. So she goes for the clearing.
Bob is only aware that there are two frogs in the clearing and that one of them is male. He calculates that he has a 2/3 probability of survival if he goes to the frogs in the clearing and a 1/2 probability of survival of going to the frog on the stump. So he goes for the clearing.
Alice is aware that there are two frogs in the clearing and that one of them is male *because she heard one of them croak* . She calculates that she has the same probability of surviving going either way*. Who can say where she goes?
The resolution of this seeming paradox is that the probabilities calculated here do not mean that Cindy is more likely to survive than Bob and Alice. If all go for the clearing, they will either all survive or all die. It's impossible for Cindy to survive but both Bob and Alice to die. The probabilities they calculate are their *estimates* for survival (by going to the clearing) based on the information they have. If an omniscient being were there too, the omniscient being would conclude that there is either a probability of 0 or a probability of 1 for survival by heading to the clearing (since the omniscient being knows the sexes of the frogs). The omniscient being has more information than anyone, and everyone else is either overestimating or underestimating based on the information they have. The degree to which their estimate is wrong depends on how much (or little) information they have.
*When it comes to Alice's calculation, there are various arguments here, depending on the probability that a male frog croaks. No matter what the probability it is that a male frog croaks, you would calculate that you are equally like to have a female frog in the clearing as you are to have a female frog on the stump - both occur with probability 1/(2−p) where p is the probability that a male frog croaks - thanks to HumptyDumptyOakland for pointing this out in the comments of this video and Presh Talwalkar for doing a computation like this in his blog post (see the video description)! Regardless of what p is, however, both the stump and the clearing have the same probability of having a female, so regardless of the value of p, Alice would not decide one way was better than the other.
Wow best comment ever
@MuffinsAPlenty I can't concentrate on the solution you suggested for an example of the frog video.
Well, Alice knows for sure that there is at least one male in the clearing, so wouldn't that affect her decision? Wouldn't grabbing both frogs give her 100% chance of survival because she knows for sure she has a male frog?
@@benjaminmorris4962 They want a female frog.
doesn't having heard the male croaking in the clearing make it _more_ likely that the lone frog is female, as it hasn't croaked yet?
The main gripe I always had with probabilities being portrayed like this is that it is based on information, like someone said here already. It is not based on known or unknown chance factors that can be calculated, like sides on a dice. This is the difference between true randomness and randomness based on information given.
It happened to me.
When my wife and I got pregnant with twins, we didn't know from the first scan if they were boys or girls (too early to tell). However, we had a genetic test that came back with the presence of Y chromosomes in the mothers blood (at least one of the babies is a boy).
I quickly worked out the probability, and none of her family believed me that it was twice as likely that we were going to be having a girl and a boy than twin boys.
Well, in reality the calculations were much more complicated. One in three sets of twins are identical, and identical twins are always the same gender. Also, the stenographer saw separate placentas. Identical twins share a placenta 1/3rd of the time, fraternal twins never do.
I eventually worked out that the odds of us having a girl were 3/5.
My wife is now tree weeks away from giving birth to our son and daughter.
@@insertphrasehere15 Nice story & good thinking! How are your wife and children now, are they all healthy and happy? Life must be busy for you both.
Same applies to the Tuesday case, it is an i.i.d. event. The probability of the day the second child is born is independent of the day of the first. This is similar to rolling a dice, the probability of getting any given number in the second roll is independent of the first roll. I don't think you should mix the sample spaces.
I think so too. I was confused why the two sample spaces was converged into one.
Looking at it once again, I think the video is correct.
Let's look at this simpler problem:
"Among the two children, what is the sample space if he has a boy?"
By applying the same logic, it could be the first-born is a male or the second-born is a male.
If the first-born is a male, then we have either MM or MF.
If the second-born is a male, we have either MM or FM.
If we'll not converge the two, our sample space would be MM, MF, MM, FM, which is wrong since our sample space are supposedly only MM, MF, FM.
They should have just said that two females can't stand each other.
***** Humans can mroe or less determine from where the sound comes and iirc therw was a big clearing or so. If we consider that we could've heard an unseen frog then the whole calculation is wrong anyway.
It's one of those riddles that looks good while it's numbers on paper, but falls short when we apply rational logic to it.
Actually, it is Presh Talwalkar's reasoning that makes no sense, since he considers P(F sub 0)=P(F sub 1), when the problem SPECIFICALLY STATES THAT ONLY MALES CAN CROAK, which means that his sample space in 8:10 is NOT one of likely events, since a frog can be 0.25 a male that croaks, 0.25 a male that does not croak, and 0.5 a female that does not croak, as well as 0.0 female that croaks. If a frog is female, it HAS TO BE one that does not croak.
@@ErmisSouldatosyes, exactly. It's driving me insane people don't notice this. Also the fact that he just deleted one of the B2B2 in his example for no reason despite the fact it was flawed to begin with.
And now, i have one more posibility: both male frogs croak at the same time 😂
Or B2B2 and B2B2 can also coexist if we consider the time of day where they were born (one after another or at the same time)
B2b2 is an event. B2b2 either exist or it doesn't you don't have two b2b2. The probability of getting b2b2 is the same as any other combination in the sample space
It says “you heard A croak” so M1M1 is not possible. If you heard 2 croaks, without any calculation, I can tell you for sure that probability is 0%.
@@brianmak5406 by the same logic, mr jones has A boy, so the BB event is not possible
It would not be 2/3. if BG and GB are unique, then B1B2 and B2B1 are also unique. Therefore there are FOUR remaining options, two of which include a girl. Therefore the chance is 2/4 or 50%
It doesn't work like that, though. If I flip two coins C1-head and C2-tails is different from C1-tails and C2-heads. However, C1-heads and C2-heads cannot be uniques from C2-heads and C1-heads, because nothing changed, just the way you structured the sentence
+Antony Di Placido yeah but it's still 50% chance...
+Lillian Theuma True!!!
maria mlakic there's an odd amount of remaining possibilities, how could it be 50-50?
Antony Di Placido We know ONE is male. Let's.. change the example (for explanatory purposes, does not change chance) You have two coins. You hold one coin in each hand, and flip both (but keep them hidden).
You are told "One of them is definately heads"
This leaves two possibilities.
Possibility 1:
The one in your left hand is heads
Possibility 2:
The one in your right hand is heads
They both have equal chances.
So let's assume 1 is true.
There are two possibilities.
HH
HT
Let's assume 2 is true.
There are two other possibilities.
HH
TH
Adding them together, we have our list.
HH
HH
HT
TH
As you can see, 2 of the 4 possibilities contain a tails, ergo the chance is still 1/2.
Wait, with the boy/girl paradox, how come the BB probability isn't there twice based upon the fact one might've been born before the other?
there is no destinction between the two boys
it doesn't matter what order you get two "heads" or two "tails" there is no difference.
+The mario fan (also a regular guy) You're right. If he has two kids and one is the boy, the chance of the other being a girl is 50/50. The same as if he had 5 kids of which 4 were boys. This isn't a great video overall tbh.
exactly! this guy is an idiot. If the man has 2 kids of different genders, it doesn't matter which was born first. Either he has 2 boys, 2 girls, or 1 of each regardless of age. i think he studies powersets to much.
+Charles M. Matos
Hold on a sec dude. This guys knows what he is talking about. With two children, the chance of getting two boys is 1/4. Same with two girls: 1/4. The chance of getting a boy and a girl is 1/2 because you combine the 1/4 chance of getting a girl then a boy with the 1/4 chance of getting a boy then a girl.
Oh wait, but we saw the frog twitch it’s leg, now we have to factor in the probability of all of the other frogs twitching their legs.
Here is a mistake with the video’s solution for the frog problem:
If you assume that the probability of croaking/not croaking is 50%, then the probability of the 4 scenarios you present WONT have the same probability.
This is a list of all possible scenarios:
M0M0 = 50%^4 = 6.25% (Since there are 4 variables of 50% chance now)
M0M1 = 6.25%
M1M0 = 6.25%
M1M1 = 6.25%
M0F0 = 50% x 50% x 50% x 100% = 12.5% (Since a female frog has a 100% chance of not letting out a male croak)
M1F0 = 12.5%
F0M0 = 12.5%
F0M1 = 12.5%
F0F0 = 50% x 50% = 25%
Total: 100%
So let’s take the sample presented in the video:
M0M1 - 6.25%
M1M0 - 6.25%
F0M1 - 12.5%
M1F0 - 12.5%
In this case, the probability of a female appearing in this sample is actually (12.5+12.5)/(12.5+12.5+6.25+6.25) = 2/3, coincidentally the same answer Ted-ed gave us.
The biggest issue is you are assuming a particular probability of hearing a male croak in the timeframe of this problem, which is like turning this into a different problem. And even when you assume the probability to be 50%, you still made a mistake calclating the actual probability of a female.
Ok, but on the problem with the boy and the girl, is it true what he says?
@@santiagofederico8259 About the boy/girl problem, in order to treat all members of the sample space as equally likely to happen, these 3 assumptions are required:
1. The probability of a boy or a girl is the same.
2. The probability of being born on any day of the week is the same for both boys and girls.
3. The probability of being born is the same on any day of the week.
Under normal circumstances, it is reasonable to assume that either gender is 50%, and that there is a 1/7 chance of being born on Tuesday, just like for any other days of the week and for both genders. Therefore, it becomes reasonable to do addition and division with these equally likely members, thus making the "14/27 ~ 52%" calculation valid.
Similarly, we can only treat all members of the sample space in the male/female frog problem as equal and make the "2/4 = 50%" calculation valid if similar assumptions are satisfied:
1. The probability of a male or female frog is the same
2. The probability of croaking is the same for both male and female frogs.
3. The probability of croaking is the same as not croaking.
For the first assumption, sure. For the second assumption, the word "distinctive" in the problem already defines that the probability of croaking is different for male and female frog (0% for female and >0% for male). For the third assumption, one has to assume the probability of hearing a croak in the given time frame. It may or may not be 50%, but strictly speaking it needs to be 50% in order for the members to be considered equally weighted. Since at least one assumption (the 2nd one) cannot be met, one cannot treat all members of the sample space as equal and thus cannot add them up and then divide by the total size, making the "2/4 = 50%" calculation invalid.
Yes, he got the boy-girl problem right, but not the frogs
3iX - TrashGaming - no, he is also wrong on the boy/girl problem.
100% agree to this
Both answers to the riddle are incorrect because the order does not matter, the variables are independent of each other, and one of the two variables is known (or eliminated as a variable). The answer is still that there is a 50% chance that one of the two frogs is female in a group of two frogs where you absolutely know that one of the frogs is male (even though you don't know which frog is male).
If frog 1 is the male, we have eliminated 2 possibilities - FM and FF, leaving MF and MM. If frog 2 is the male, we have still eliminated 2 possibilities - MF and FF. It doesn't matter which frog is the male, as either way, there are still only two possibilities left. Thus there are only two possibilities left - MM and EITHER MF OR FM. The fact that
we don't know which of the possibilities of MF or FM has been eliminated does not mean that they are still viable possibilities - one of them has still definitely been eliminated, thus still leaving us with a 50% probability of having a female frog in the group. Just because you don't know which two possibilities have been eliminated doesn't mean that they haven't both been eliminated.
Don't believe me? Let me give you a demonstration. I am running a betting game where we bet on two flips of a coin. You win the game if one of the two flips turn out to be a tails. But one of the two flips is guaranteed to be heads. We will count the foregone heads flip as occurring in whichever order you choose - either prior to or subsequent to the actual flip. So, since one of the two flips is guaranteed to be heads, I will only flip the coin once - and if that flip has a tails you win and if it has a heads, I win.
Well now, since I have only removed one possibility out of four - the possibility that both flips wind up tails - and the remaining three possibilities are still out there (Heads-Heads, Heads-Tails, and Tails-Heads), and you win in two of those three possibilities
(Heads-Tails or Tails-Heads), It is only fair that I only give you 2/3 odds. In other words, you have to lay down 2 dollars to my 1 dollar in order to play the game. If I win, I get all 3 dollars and if you win, you get all 3 dollars.
If you take that bet, I would seriously advise you to never, ever go to Vegas.
The video explained that in the second case (you heard a croak), your logic holds. There are two possible ways you could have MM and only one each for MF and FM, so there are 3 possibilities, but MM is twice as likely to occur as MF or FM so the probability is 2/4 = 1/2. In the first case there is only the knowledge that one frog is Male, not that a frog croaked, so MM, MF, and FM are all equally likely, for a probability of 2/3.
Your example isn't the same as the frog scenario at all, however - here's why: The order *does* matter and I still get to flip two coins. I flip the coin once; if it's tails I win (the other flip is automatically heads). if it's heads, we still have to flip to see if the other coin might be tails. Thus the probability that I win is Pr(first flip is tails) (1/2) + Pr(first flip is heads and second flip is tails) (1/4) = 3/4. The information that one of the two coinflips is heads is irrelevant. With 2:1 odds for you my advantage is slim, but if we play long enough the House always wins. See you in Vegas!
No Zach. The fact is that by virtue of one of the frogs being known to be male, at least two possible choices have been removed. One of the four possible choices that has been removed is FF. The chooser KNOWS that that possibility has been removed. The other possibility that has been removed is EITHER MF or FM. The chooser just doesn't know which of those possibilities has been removed. But the fact that the chooser doesn't know which of those possibilities has been removed doesn't mean that the removed possibility is still available.
In the coin flip scenario, one of the coins has already been predetermined to be heads - in other words it is a coin with heads on both sides (a two headed coin, if you will). Now whether you want flip the two headed coin first or second is irrelevant, as you are now effectively just betting on ONE coin flip. In the scenario you describe, the possibility of two tails flips has not been removed (since you are claiming that both coins you are flipping have the possibility of coming up tails) and is not the same scenario. On the other hand, if you are playing that game with a two headed coin (which is an identical scenario to the frog riddle), if the first coin you flip is the two headed coin, and it obviously turns up heads, you have not eliminated the Tails-Heads combination, as that combination was never really a possibility if you flipped the two headed coin first. On the other hand, if the first coin you flip is the regular coin, and it comes up heads, then you don't really have the combination of Heads-Tails left, do you? Even though you don't KNOW that that combination has been removed by virtue of the second coin to be flipped being a two headed coin.
So - still want to play my coin flip game with one of the coins being a two headed coin and you only get 2:3 odds? I don't care the order you flip them.
Imagine you have two frogs in a bag. They both have a 1:1 chance of being male or female.
The probability for having two male frogs is then 1/2 * 1/2 = 1/4.
The same goes for having two female frogs: 1/2 * 1/2 = 1/4.
The only option left for the two frogs is to be one of each, and since all the probabilities must add up to 1, we are left with a probability of 1 - (1/4 + 1/4) = 1/2.
Now one of the frogs in the bag croaks, giving away that it is a male.
This doesn't change the fact that you are twice as likely to have one frog of each sex rather than two male frogs.
Then you stick your hand into the bag and randomly pick up one of the frogs. The probability for this frog to be male is _not_ 1/2, but 2/3.
If the frog you pick up is male, you have a probability of 1/2 that the other frog is female.
If you on the other hand pick up a female frog, there's a probability of 1 that the other frog is male.
This gives us three possibilities:
MM = 2/3 * 1/2 = 1/3
MF = 2/3 * 1/2 = 1/3
FM = 1/3 * 1 = 1/3
Jerry Nilsson
"This doesn't change the fact that you are twice as likely to have one frog of each sex rather than two male frogs."
Absolutely it changes that calculation. The probability for having two male frogs has gone from:
1/2 * 1/2 = 1/4
to
1/1 * 1/2 = 1/2
And the possibility of having one male frog and one female frog has gone from:
(1/2 * 1/2) + (1/2 * 1/2) = 1/2
to
1/1 * 1/2
And the possibility of having two female frogs has gone from:
1/2 * 1/2 = 1/4
to
0/1 * 1/2 = 0
When you reach your hand into the bag the first time, the odds of getting a male frog are NOT 2/3 - they are 3/4. Your odds of getting the frog that croaked (the frog that is DEFINITELY MALE) are 1/2. Of the remaining possibilities (the possibilities stemming from choosing the frog of unknown gender first), there is a 1/2 chance that it is male and a 1/2 chance that it is female. So you add the possibility of choosing a known male first (50%) to the possibility of choosing the unknown frog first (50%) MULTIPLIED by the possibility that that frog is male (50%), and the result is that you have a 75% chance of choosing a male frog on the first pick. The same calculation holds true for the second pick - with their being a 75% chance of getting a male and a 25% chance of getting a female.
Further you stated: "This [the fact that you know one frog is male] doesn't change the fact that you are twice as likely to have one frog of each sex rather than two male frogs."
Absolutely it does. You have removed half of the possible combinations (you have removed all of the combinations that would include the frog that croaked being a female). You are now equally as likely to have the combination of one frog female and one frog male as you are the combination of both frogs being male.
+cazinger I'm sorry. I spoke prematurely and was wrong.
I believe you made a mistake in your Bayesian calculation. Let's assume the probability of a male frog croaking is p. We have the following possible scenarios:
M0,M1 and M1,M0, both with a probability of p*(1-p)/4
M1,F and F,M1 both with a probability of p/4
So, given that we heard a croak, we have the Bayesian probability of having one female of:
P = p/2/(p - p^2/2) = p/(2p-p^2)
Let's test this out for some limit values. If p->1, that is, male frogs croak all the time, then P->1. Which is expected, because if we had two male frogs we were almost certain to hear two crocks. As we have only heard one, the most probably scenario is that we have at least one female.
On the other hand if we had p->0 (as we can't have p = 0), the limit of the probability is 1/2. Again, kind of expected, as the fact that is so rare for a male frog to croak that hear a single croak almost gives us no new information. It is almost equally likely for them to be either (male,male) or (female,male) pair.
Finally, if we assume p = 1/2, we get the 2/3 answers, as stated by TED. So we FINALLY get to the result the video expected. You are right about how the method of obtaining information changes the result, but you did the calculations wrong. Your main mistake is around 8:07, when you say the scenarios are equally probable. They are not! FM1 is twice as probably as M1M0, because the female NEVER croaks! So even though there are two ways you can write M1M0 and M0M1, the probability is halved, so you would still get 2/3 when doing the whole calculation
I'm glad we have someone like you to put it in plain words, without video.
sometimes I believe that teaching via video is bad, because of dimensionality. one has to scroll back and forth and cannot jump with the eyeballs back and forth between paragraphs and sentences in text.
I like entertainment. I like learning. but when combining them, it is skewed towards entertainment, and less towards improving critical thinking.
@@vaneakatok thanks. I appreciate the feedback, and hope that the comment was useful. Bayesian probability is really cool, you should look into that if you are interested in learning more
You are correct, assuming that females never croak! If females also have 1/2 chance to croak, (but you can tell the difference,) then the scenarios are correctly computed as equally probable. (M0M1, M1M0, M1F0, F0M1.) I think that was this video's assumption.
Guys can you explain this because I am having troubles to understand.
So you guys saying when we heard single croak then turn back to see possibilities we have 2/3 to get a female frog.
but if I had a friend which saw which frog is croaking he must have 1/2 to get a female frog.(right)
We assume he can't present which is the male frog because of extraordinary things.
Am I still have 2/3 to get a female frog? Definitely not because it is impossible.We can't have a frog that has both genders.
Can you explain this please?
Sorry for my bad english but I am sure you guys understand it.
I think you have also mitaken, why the probability of having M1;F0 or F0;M1 is p/4.
The probability of that is also (1-p) [probability of not croaking from female) times p [probability of croaking of men] divides by 4 [possible options]
So, (if p is probability of croaking)
P (M1;M0) = p (1-p)/4
P (M0;M1) = (1-p) p/4
P (F0; M1) = (1-p) x p/4
P (M1; F0) = p x (1-p)/4
You can Say "The probability of a Female not croaking is not 1-p, it's 1 (if there is a female), because the exercice says that the female didn't croak" but if You use that, by the same lógic, You can Say "The probability of a male not croaking is not 1-p, it's 1 (if there is 2 males), because the exercice says that the male croak was heared only once, so if one croaked, the other can't croak"
Probability is sometimes counterintuitive and that's why I love it even more.
In the girl/boy paradox, why did you assume that the day of the birth of the other child is relevant just because the day of the birth of one child is mentioned?
The reason is because since one is relevant, the other must be relevant as well or the sample size won’t make sense. For example: If a boy is born on Tuesday and we make a sample size considering all days of the week, the other person would be underrepresented if we also didn’t make a sample size for each day of the week.
Because this is how sample spaces work ig
Those are different ways it COULD be a girl, but the question asked what are the odds of it BEING A GIRL!
Nope, those two questions (mathematically) still mean the same thing
+CRISPYrice At least the guy figured out it was wrong. That's better than randomly believing something you were told in a video. Here's a detailed explanation of why it's wrong.
If you are told the month but not the year the child is born on it becomes 30 x 4 instead of 7 x 4 so 120 possibilities (pretending all months have 30 days). Two are still cancelled so because only two of the months will be the same so add one and its 60/119
Now you are told the minute of the day the child is born. There's 1440 minutes in a day so now it's 720/1439.
Wait, now you know the second the child was born on a given day. That's 43200/86399 that it's a girl.
Now let's go the other way. There are two options: the child was born during the day or during the night. 2 x 4 is 8 and one of those possibilities cancels. So now it's a 4/7 chance that it's a girl.
We already know 1 option gives a result of 2/3
So are we to believe depending on how many choices you are given the chance of it being a girl fluctuates from 66.67% with 1 choice to 50% with infinite choices (that's the downward limit if you set it up as a function of choices versus probability). This becomes a paradox because the original riddle is no longer two thirds since there's an infinite amount of choices about the boy birth not specified. On what planck second was he born? In what hospital? What was his first word spoken? In fact, no matter how much you specify an infinite amount of choices will not be specified. Therefore, the probability is always one half assuming this explanation its valid since there are an infinite amount of choices for each boy of girl with only one ever cancelling.
Fortunately, having to consider an infinite amount of arbitrary sets doesn't affect probability or else the whole concept of probability would be fucked. The problem here is irrelevant information. Sure, you can line all the children up from Monday to Sunday when it has nothing to do with the problem. You were never asked to find the day the child was born on, but go for it. Why not do a sample size with 69 choices or 1337 choices while you are at it?
So yeah, the video is convincing, but its complete crap because the sample size of seven isn't with reference to anything.
In most simple terms you aren't allowed to increase a sample size by adding extraneous information. That would be like having the question "What is 2 apples +2 apples" then saying its 7 because you ate 2 apples for lunch three days in a row. Just doesn't make sense.
+Aaron Harkleheimer I'm still baffled as to why the odds of both aren't 50% .Since we're finding the presence of a single unit, a girl, why do we assume order matters? If order doesn't matter then it goes down to two choices, one boy exists and one girl exists or two boys exist. Since adding extra data scales up the two evenly, that'd mean that the probability is always 50% like most people assume with this problem. (unless an infinite amount of data is added, at which point I drop out of calculus)
+Aaron Harkleheimer
You are correct that the solution given is wrong, and your explanation is useful, but the actual probability in all of the cases you propose is 2/3, same as the original problem. It doesn't matter day of the week, month, hour, minute, nanosecond. The only thing that would change the calculation is if they told you that the first child was a boy, or that the second child was a boy. Then the answer would be 50%.
+liger04
Google "The Monty Hall Problem." That will help you understand the concept.
shouldn't BG and GB be the same thing in sample space. if not what difference is in their representation.
+Vishal Devgire I have the same question!
+Vishal Devgire In the solution, the two children are sorted in an arbitrary way. This is because the chance of there being a boy and a girl is twice as high as the chance that they are both boys. If you consider BG and GB the same, all outcomes that involve a BG pair are twice as likely, because that pair is twice as likely to occur. By making a distinction between BG and GB, all outcomes are equally likely.
The order should not matter
Patrick Polcuch
The problem with saying that the order doesnt matter is that you give the same amount of weight to the BG outcome as the BB outcome. However, they are not equally likely to happen.If you take a group of 2 children with random genders, there is a change of 50% to get BG/GB, but only a 25% to get BB.
+Patrick Polcuch na you're right Patrick, they should'nt remove the b2b2 from the equation this is wrong
Oh! Now i get it! It is more posible to hear a croak if you have 2 male frogs from having only one!
Yea I didn’t fully wrap my head around it until the final part where he said it’s twice as likely for 2 males to croak, which obviously will decrease the chance of a female
Wouldn’t having two b sub 2s just make that event twice as likely as any of the others? Why should it make it more likely for him to have a girl?
The first set of result is from a column where the first boy is tuesday and the second is from the row where the second boy is tuesday. B2b2 is the point they intersect which is why it is written down twice. But should only be counted once. In the same way that probability of getting HH is the same as any other possible combination when flipping two coins
No. There is one way to have a boy born on Monday and another one on Tuesday. The older one can be born on Monday and the youngest on Tuesday, or the older on Tuesday and the younger on Monday. However, there is only one way that both are born on a Tuesday which is... well... that both are born on a Tuesday.
One could be born on last Tuesday, and one on the Tuesday before that. Or vice versa. I still don't think you can remove one case, because "Tuesday" is not one event - it is many events in the calendar.
@@amcconnell6730 but it doesn’t specify which Tuesday, so with the information given, all Tuesdays are equivalent.
3:20 Absolutely not.
BG and GB are the same thing. They're one result. The chance is still 50/50.
so if your older kid is boy, young is girl is same as older being girl and younger being boy?
Yes, but what if one of the two brothers is born in a Friday morning at two thirty and twelve seconds? What are the odds of the other sibling being a girl then?
***** Or, rather, the odds are not to calculate on the information you posses but on the probability that would make a difference. This kind of thinking is wrong, and if I wouldn't be able to point out precisely why, lacking a proper five-year course in mathematics, I know that the results of the thought experiment are inconsistent with what would happen in reality.
+Batrax If one of the two SIBLINGS were born at that exact time, and the probability of being born at a certain time is uniform for all times, then the probability would approach 50% from above as your time gets more and more specific. The only discrepancy comes in when you have twins of the same gender, which would only be counted once in the sample space.
+KWG The probability doesn't approach 50% since it's always 50% assuming that any child born is equally likely to be a boy or a girl, which we are told originally. The maths he does after assuming that day of birth is a factor is wrong, I and several other people have posted answers as to why.
KWG: Even when there are twins one is older than the other, maybe by a minute or less but still older.
Still 14/27 because of the Friday morning part. Not unless you cant have both kids being born at different days at 2.30 12 sec
Oh man, I am late to this party.
So let me ask this, if we reword the problem
"Suppose there are two frogs. If you know the gender of one frog, what are odds the other frog is the opposite gender."
does it change the outcome?
1/2
It could be
BB
*BG*
*GB*
GG
And I Think I am later than you XD
Also I dont think knowing the gender of one of the frogs will change the result.
You could try to use the example in the video, but because it isnt clearly said if the Frog we know the Gender from is female or male, it wont affect the 1/2.
B1 B2
B2 B1
G1 B2
B2 G1
G2 B1
B1 G2
G1 G2
G2 G1
Due not telling the exact gender, not only the MM possibility will double but also the FF and with both possibilities Doubled MF and FM will double as well.
"The key in variation A is that you hear this information by random chance."
This is not quite true. It is important to KNOW that FF is not part of the sample space (as opposed to not being the case in one observed random event). In the boy or girl paradox, it is important that Mr Smith remains silent if he has two daughters. Only if this is assumed, his chance of having two sons is 1/3 (see wikipedia). If on the other hand he always gives a random but true information about any pair of children, FF is part of the sample space.
This is why TED-Ed designed their riddle the way they did. FF is not part of the sample space because we would never notice any pair of female frogs. In the described situation, we only looked because one frog was croaking.
Still, your conclusion seems right. The unambiguous assumption of the paradox goes: "From all families with two children, at least one of whom is a boy, a family is chosen at random." While croaking is indeed a random event, it is twice as likely to occur in MM pairs of frogs.
I might change my mind again, though.
EDIT: I just have. Two frogs do not have twice the chance to produce at least 1 croaking. The ratio varies from more than 1 to less than 2. Therefore, TED-Ed is correct that your chances are always at least slightly better with the pair. However it cannot completely reach 2/3.
I agree with most of what you think but consider this:
If a male frog has a specific, non 0 probability of croaking (in a specific time interval), that would make a non-croaking frog more likely to be female. If the croaking chance, let's call it P, is 50%, a non-croaking frog has a 2/3 chance to be female. If P=10%, it's 10/19. Considering this, I did some calculations, and my conclusion is that there is an equal chance of the two frogs having a female frog, as the one alone frog being female, which is always more than 50%, and the less likely a frog is to croak, in other words, the smaller P gets, the closer this probability gets to 50%.
And here is the interesting part: in mathematics, 0% doesn't always mean "impossible". For example, tossing a coin infinite times, and always getting heads is 0%, but it could theoretically happen. So if the riddle says that a non-croaking frog has 50% chance to be female, that would mean that P=0, but you still hear the croak, so it's exactly twice as likely to come from MM as from MF, so equally likely, as MF or FM, which would make the result exactly 50%.
Surely the genders are independent of each other? This problem reminds me of the Monty Hall problem, except in that case the events are dependent. A dependent version of this problem would be something like:
A man has 4 children, 2 girls and 2 boys. What is the probability that the youngest child is a boy if the oldest child is a boy.
Without knowing the oldest is a boy, the probability that the youngest is a boy is 50%. But once we learn that the oldest is a boy, we know that only 1/3 of the remaining children is a boy, so the probability that the youngest is a boy is 33%.
But in the frog problem, the genders of the frogs are independent. Saying that the probability of the second frog being female goes up to 2/3 because we know that the other frog is male is the same as saying that the probability of getting tails on a coin flip is 2/3 because I got heads on the previous flip.
This is further convoluted with the addition of more variables which are also independent. Day of the week has no effect on gender, nor does the means by which one frog was gendered effect the gender of another frog. This whole thing stinks.
The frog gender problem is "You grab two frogs. At least one of them is male. What is the probability that one of them is female." NOT "You grab a frog and it is male. What is the probability that you grab a second frog and it is female." These are not the same problem.
The coin flip equivalent is "You flip two coins. At least one of them landed heads. What's the probability that one of them landed tails." NOT "You flip a coin and get heads. What's the probability that you flip a second coin and get tails." These are not the same problem
Okay. So it's not that one is affecting the other. It's just that the order is unknown, so we have twice as many possible M-F arrangements as we do M-M arrangements.
While I agree with your coins explanation, how does day of the week affect anything. It is independent variable. It is like saying I floipped two coins this week and on Thursday i got a head. What is the probability of getting one tail?
Tracy Coxon the fact that the order is unknown is irrelevant as the order does not matter. therefore, the possibilities are mm/mf OR mm/fm. either way it's 50%
The probability isn't going up it's going down. Before you know the gender of one of the frogs there is a 75% probability that at least one frog is female, and a 75% probability that at least one frog is male. The options are MM, FM, MF, and FF. If you know that one frog is male, you've lowered the probability of having a female frog to 67% because you lost one of the options (FF).
As I see it when you want to know the probability of having a girl when there is a boy in the couple of kids is 1/2 since the rank as no consequences on the result. The permutation BG/GB as no weight for the end result being only to know the probability of having a girl in the couple of kids.
+Ultack The trick here is that the chance of two children being boys is only half as likely as two children being one boy and one girl. To clarify, imagine a couple having a child. Regardless of the gender of that child, they have a chance of 50% that their second child is a different gender. So in 50% of the cases the children have a different gender. However, the chance of both being boys is only 25%.
Now imagine a collection of groups of two children. 50% are BG pairs, 25% are BB and 25% are GG, equal to how likely they are to happen. Now remove all the groups with no girls. This leaves the BG and BB groups. However, there are twice as many BG groups. So if you pick a random group from the remaining groups, you have a 67% chance of picking a BG group.
+IVIasterIVIind Summing up:
BB BG GB GG
0.25 0.25 0.25 0.25
Therefore:
BB (BG/GB) GG
0.25 0.5 0.25
I have two frogs and one is a male, what is the chance of my second frog be a female? 50/50 is the only answer.
@@IVIasterIVIind This would be correct if the sequence matters. Which it does not. The question was is there on girl/one female frog. This means that we have a combination not a permutation which you are using. This in turn means that BG/GB is one "group".
You know I've seen this a few times and one thing I dont get is why the order matters, wouldn't it be more intuitive to condense pairs that are practically the same (i.e. B2G0 and G0B2; B2B0 and B0B2). If someone would be willing to explain that, it would be helpful
Those are different senarios tho, just add a sub strcit that means older or younger
There is no difference, and the author is also incorrect.
To make order important, you need to change the problem/question.
@@F_A_F123 The order matters because its a different scenario, no matter what the question says. Yes you could technically swap the BG around to make GB, but that doesn't change the fact that they are 2 possible scenarios. Hopefully that helps.
I believe it's because both of those are distinct equally likely events. When we discuss the sample space of 2 coins being simultaneously tossed, we take HT and TH as different events for the same reason
If you don't get why order matters, try an experiment. Take two six-sided dice, roll them, record the sum, and repeat a bunch of times. (Probably you'd want a computer automating this and rolling thousands or millions of times haha). Under the "order doesn't matter" theory of probability, you'd expect that rolling a sum of 12 is just as likely as rolling of sum of 11 since you only get a 12 by rolling 6+6, and the 11 by rolling 6+5, and you're just as likely to roll a 6 as a 5. But roll a ton of times and you'll find that on average you're rolling twice as many 11s as 12s. This is because you can only roll a 12 if both dice roll 6s, but a roll of either 5+6 or 6+5 will get you 11.
Why would you diffrentiate m1m0 and m0m1 ?
For the same reason you differentiate between f0m1 and m1f0
that is, you shouldn't, because in no way nor place were you asked about the order of the frog kids.
@@lorefox201 Yes even I was wondering why the order mattered so much when you are going to lick both anyways
I'm confused here. My statistics may be a bit rusty, but it seems like this video is treating this as a permutation and not a combination. In a permutation, the order matters, and in a combination, the order does not matter. It seems to me that in a case of having two children, the order that you have them in doesn't really matter, and it should be treated as a combination. i.e., the two possibilities GB and BG are the same as far as this question is concerned.
Yup, this is exactly what I was thinking.
+bungalo71 Nevertheless, the chance to have a certain combination might not be equal for all possible combinations.
Go toss a coin twice :-). Repeat that a lot of times. Then you'll see that having a combination of Head and Tails in two tosses is more likely than just Tails, or just Head.
+bungalo71 yes. the two events are completely separated
Let's freeze the video at 5:52 and look at all the wrong.
1) We've decided that we no longer just want to compare BB, BG, and GB; we also want to break BB into two columns to represent whether the first boy or the second boy is known to be a Tuesday boy. Well okay we can do that, BUT we also have to halve the probabilities of each column. We started with a simple BB arrangement; if we want to subdivide the BBs we are allowed to but the total of the subdivisions has to equal the original BB probability.
2) We also can't eliminate one of the two "both Tuesday" cases, because they are not exactly the same: one of them has the first boy as the known Tuesday, and the second has the second boy as the known Tuesday. That makes a difference because WE decided it makes a difference. If which boy is the known Tuesday boy matters, we need to stick with that proposition all the way through.
But the thing is, I ran a few tests through a big randomized sample set, and... he's right. It's super weird, but the first boy-girl problem had a ~66.7% chance of a pair containing a girl, and the second boy-girl problem had ~51.9% change of containing a girl. Keep in mind this was about 100 million randomized samples too.
I don't doubt for a second that, if you program according to his logic, you will get numeric outcomes as he described. I'm saying that his logic itself is wrong.
I agree: there is no justification for removing the perfectly valid second route through the probability space to the 'both Tuesday' outcome. This error is catastrophic to the structure of the problem.
+kingbeauregard you are correct in my opinion
+Joseph Masuda
Sorry, but this is pure logic, opinion is not a factor here.
'i think you are correct', is a perfectly fine statement (one i intend to take a look at)
'in my opinion, you ARE correct' is not.
+kingbeaurega
"BUT we also have to halve the probabilities of each column."
No we do not.
The way he got to that sample-space was worded terribly, i'll give you that.
But he is accurate.
The first case had: the older one is: Boy or Girl: 2 options: equally likely, same (but completely independent) for the younger sibling.
results in 2x2 = 4 (equally likely) options, looking at those options afterwards, we find 1 of them not meeting the requirements of (after the fact) observations: resulting in 2 out of 3 equally likely combinations contain a girl.
The second case had (for either sibbling, independent of the other) B0,B1,B2,B3,B4,B5,B6,G0,G1,G2,G3,G4,G5,G6 as equally likely outcomes
which results in 14*14=196 equally likely combinations, of which only 27 are possible (given the after the fact observation).
Yes he did word it somewhat clumsily, probably because listing all 196, in the same way he listed all 4 at 2:35, would have become an unreadable mess
"2: We also can't eliminate one of the two "both Tuesday" cases, because they are not exactly the same"
While it is true they aren't exactly the same, this is one where we can argue, what it is we're talking about here.
The moment we declare 'which one' we have the info on as relevant, we're back to the whole 'the variables are known to be independent'-point where (by definition) a 50/50 would be in effect.
Or to compare (again, the whole, 'the variables are independent, you might want to break out the square of options' aspect applies here), your point would boil down to you roll 2 dice[standard d6]: a red and a blue one.
given the red one rolled a 2: the blue one has a 2 in 7 (instead of the regular 1 in 6) chance of also rolling a 2.
Or to bring out the example the video refers to: there are 3 frogs, 2 on the field: one on a stump, you know (with absolute certainty) the left one on the field is male, this somehow makes the odds the right one on the field is as well, different from the odds the one on the stump is.
-> no it does not.
It seems that the problem is in the wording. In the first scenario, we are asking for the probability of one of the children being a girl if we know that one of them is a boy. The differentiation comes from HOW you obtained this information. Either you know it outright as a matter of fact -or- you obtained it through observation.
In the initial condition of knowing it as a fact, that means that you are indeed modifying the initial conditions to exclude the possibility of there being two girls and declaring that one of the children must be a boy. In this scenario the probability that one is a girl is absolutely 2/3.
In the second condition of knowing there is a boy through observation, that means that the initial conditions are THE SAME and all outcomes are still just as likely. You aren't actually introducing any new information about the initial sample space (contrary to the Monty Hall problem where the 'all-knowing host' introduces pre-determined information about the conditions of the initial sample space). If our observation doesn't affect the initial conditions, the sample space still includes (G,G) even if we KNOW that it didn't come to fruition. So the probability is simply 1/2. Assuming that the occurrence of the observing one result can change the probability of a second, independent, result in this way is the gambler's fallacy.
So the problem is in the ambiguity of the conditional sets. Knowing this, it even helps to clarify what's happening in condition 2 (He has a boy born on Tuesday). Again, this depends on how we come about this knowledge and if it affects the initial sample space. Ultimately the inclusion of the boy's birthday does not affect the sample spaces and the resultant probabilities coincide respectively with the first condition.
I love watching mind your decision but I have to say that he is wrong on this one not only because he took out one of the b1 pairs but the 2 examples have nothing to do with each other since the frog puzzle is saying what's the chance that there is a female given that there is a male and mind your decision's problem is saying what's the chance of a girl given that there is a boy born on Tuesday by having 2 variables in the given area he is reducing the chance that there is a girl with a sibling that matches the given
if he really wanted to calculate the last one he would put all the number of combinations of when a boy was born on Tuesday over all possible combinations for example at the ones pairs in the video over (B0 B1)(B1 B3).... all the ones that don't have B2
Well
Atleast we know he too watches the TED-ED videos😊😊
"At" and "least" are two different words by the way
Man's just corrected someone 2 years later
@@AntonVladimirovic atleast not 3 years later
@@joevincent1987 Atleast...
@@joevincent1987 nice
A very interesting video, but unfortunately, you’ve made a critical mistake. When you count up items in a sample-space to compute a probability, you implicitly assume that all items being counted are equally probable. In your example with the croaking vs. non-croaking frogs, this assumption is not only unfounded, it's actually impossible. Allow me to demonstrate.
p(M) = p(F), given by the constraints of the problem
p(M) = p(M0) + p(M1), because croaking and not croaking are mutually exclusive and exhaustive cases
therefore,
p(M1M0) = p(M0M1) = p(M1) * p(M0), because we’re looking at two independent events
p(M1F) = p(FM1) = p(M1) * (p(M0) + p(M1)), by substituting out p(F) for its equivalent p(M0) + p(M1)
Therefore, we see that p(M1M0) = p(M1F) if and only if p(M0) + p(M1) = p(M0), or in other words, if and only if p(M1) = 0. But p(M1) is the probability of a male croaking, and we cannot have heard a male frog croak if the probability of a male croaking is 0. Putting that in terms of numbers, if p(M1) is 0, every item in the sample space has probability 0, which means our sample space doesn’t sum to 1 and is therefore not a sample space. So the events you’re counting aren’t equally probable, so counting them is useless for determining the probability of a given outcome.
Interestingly, when you play the math out a bit further, you find that the TED-Ex answer can be correct, but only for the case where a male frog has a .5 probability of croaking. In general, the more likely a male is the croak, the more likely the pair is to include a female frog.
Your including the probability of an event occurring even though it has already occurred. We already know that the frog croaked.
True, but that doesn't change the relative probabilities of the cases where it does croak. It only removes the cases where no frog croaks, and normalizes the remaining cases so that they sum to 1.
That's... not true. Order does matter. That's an intrinsic property of probability. No amount of "but we don't care which one is which" will get you away from the fact that they're two different events and must be considered as such. You may not agree with that statement at face value, but let me propose a demonstration that you can do in your own home. Take two coins and start flipping them. Do it enough times, and you'll see that you get a heads/tails combination more often than two heads or two tails. That's because heads + tails is different from tails + heads, so having some combination is more likely than flipping two of the same face.
(Technically, this is only a demonstration if you accept that all unique combinations have equal probability. They do, but if you don't believe that, I can walk through the math.)
Actually, the equation p(M1M0) = p(M1F) isn't representative, since we know that the female didn't croak. More exact would be: p(M1M0) = p(M1F0) p(M1)=p(F1).
The reason why in your calculations that equation holds if and only if p(M1) = 0 is because in your calculations you implicitly assume that p(F1)=0. That also means that all of your results are only valid under that condition.
If we assume that the chances to croak are the same for both sexes, then the actual croak chance doesn't matter, the chance to have a female will always be 0.5(the actual croak chance will cancel out).
Good point, although the original problem says nothing about females croaking. Importantly, it doesn't say that if a female frog is present, it hasn't croaked. That is simply not given by the constraints of the problem (and in case you're thinking "it's obvious", frogs vary a lot in volume and can even drown out each other's croaks). So I maintain that the correct equation to use is p(m1F) rather than p(m1f0), given that the problem does not provide any information about whether or not the female frog has croaked.
"We're going to assume that [croaking vs. not croaking] are equally likely events."
How about not? Lets assume ALL male frogs croak instead and not assume that we can tell whether one frog or two frogs are croaking. Or if you don't like that, assume that only one frog in a group croaks at a time. Or, while we're assuming things, lets assume that male frogs only croak in the presence of females. The problem with this question isn't the answer, it's the vague wording of the question itself that forces people to make assumptions before answering.
Exactly.
The vague wording isn´t great but the answer is wrong. So is this video. MF/FM are actually one and the same as nothing in the riddle absolutely nothing makes sequence matter. In fact the way it is stated sequence clearly does not matter as both frogs are licked.
@@XMysticHeroxyeah but M/F is twice as likely as M/M which is the point. Sequence doesn't matter sure, but it is more probable nonetheless since there are two cases in which a female appears compared to one case where it doesn't (2/3) chance of female.
I really like how you added the stutters to the subtitles. Loved the video. Thank you.
i thought the MM event would split 1:1 to M1M0 and M0M1, each having a probability of 1/6 instead of M1M0 being as likely as MF
The whole problem with these types of probability riddles (and riddles in general) is that there is not enough information to come to any conclusions in any definitive mathematical sense. If you want to solve a problem you have to first know what problem you are trying to solve. In general, this is the "trick" behind most word problems and riddles. They do not supply enough information and they expect you to make assumptions based on little more than socially accepted norms. So, these types of problems are really a test of social conformity more than they are any kind of intellectual test.
+ThinkTank255 Agreed. For example, I'm not sure it's at all proper to double the chances of the other child being a boy, simply because the one known was born on Tuesday. In the original sample space, there was no enumeration of older and younger brother, yet the solution provided here enumerates those possibilities once we add the day of birth. This seems an improper differentiation between the situations.
trying to convince my self im not crazy for thinking this video is wrong, your answer put my thoughts into words perfectly
But in this problem, in (A) the answer is 1/2, in the (B) - also 1/2.
I think it is right to think about 2 frogs as 2 unordered elements, because there is 2 unordered frogs.
But if we think about 2 frogs as 2 ordered elements, there is no a difference between MM and MM, MM is just 2 times more probable thing than MG (or GM) . . . and the answer is also 1/2.
Since the frog croaked, he has not attracted a female yet; therefore the second frog is also male.
+Panos Triantaphillou That's exactly what I said when I saw the Ted-Ed video. It seems very obvious, but I'm amazed out how few people pointed it out in the comments.
+Panos Triantaphillou This is the most correct answer I have read here.
Haha. That's actually a good point.. Haven't thought about that yet. :D
This reminds me much more of Foshee's two boys question. He just hoped to provide a rather simple conditional probability test - but he put the question wrong🙂
You should make a video of that!
6:35 No, it is not equally likely to happen. The propability for (MM) is the same as the sum of (MF) and (FM) because the order is not relevant. Otherwise u would have to count MM twice
While order isn’t relevant to the question, it is relevant to the probability.
Generate random pairs of frogs by flipping two coins, heads is male and tails is female. The more pairs you generate, the closer you'll converge on 1:2:1 for MM:MF:FF. You can also do this in a spreadsheet very easily, thousands of pairs in under a minute including the time to set it up. So out of the pairs with a male, the pairs with a female represent 2/3 of the total since the ratio is 1:2.
Now, have one random frog in each pair croak, and rule out the pairs where a female croaks. None of the MM pairs will be ruled out. Half of the MF pairs will. All of the FF pairs will.
Reducing the pool of MF by 1/2 takes the ratio of MM:MF to 1:1. That is, in the remaining pool of frog pairs, there's a 1/2 chance that a random pair has a female frog.
why can't the day of birth be extra irrelevant information?
+Timothy Heimbach That's the rules :D
The major flaw in this logic is the failure to realize that it is inconsequential WHICH frog croaked. In your example, M0M1 and M1M0 are actually repeat events. It's equivalent to arguing that their position on the rock would somehow affect the probability. "Two male frogs" is a single possible outcome regardless of who croaked. The croaking itself isn't the relevant information, it was simply used to determine the relevant information(that one frog is male). Therefore it shouldn't be used in the calculation. At least not in that way.
PROOF
Simply replace the statement "frog croaked" with the relevant extraction "is male" and it becomes clear.
M0M1: Two male frogs of which the right one "is male."
M1M0: Two male frogs of which the left one "is male."
These are very clearly redundant statements describing the same condition.
The initial scenario has 4 possible outcomes. When you acquire the new information it drops to 3. How you acquire that information does not matter. I could point to a frog and tell you its male, I could whisper in your ear, write you a note, pick one up and show you its penis, or you could hear one croak. The outcome is always the same. ALL you have learned is that one of them is male, making the probability that the other is female 2/3.
+str8dominican However the probability is calculated with the knowledge you do not know which one croaked. You only know one of them did. Because you do not know, and treat each frog as an individual, you take the probability of encountering the relevant options out of the relevant sample space.
That is how probability works, assuming each option is equally likely.
Since a different thing happened from M1M0 to M0M1, depending on who croaked, they are different scenarios that happen to have the same outcome, so they should be counted separately.
+DarkBabyIon but the question is not asking which frog was the one that croaked. That information is inconsequential. If you argue that the frog croaking has any effect on the probability that the other frog is a girl you are just wrong. In a similar fashion I can then bend probability to my whim by creating a different scenario based on the boy girl paradox and come up with an outcome that suits my design by saying "what is the probability that mr jones has a girl if he has a boy and that boys name starts with the letter A" then by the same logic your sample space must include a sample space for both possible genders with every letter of the alphabet. If it is possible to change the probability of an outcome based on such senseless logic then there is absolutely no point in even going through the effort of finding out the potential probability of an outcome.
Alex Motor Did you even went through probability, which to my knowledge is HIGH SCHOOL material?
It is also taught in physics and math in ALL universities with science subjects. That exact material is what is taught in ALL probabilities.
By taking extra information you are able to reduce the sample size sometimes dramatically.
["what is the probability that mr jones has a girl if he has a boy and that boys name starts with the letter A" then by the same logic your sample space must include a sample space for both possible genders with every letter of the alphabet]
You are right. We DO take into account all the possible letters, but notice the sample size has changed DRAMATICALLY because we no longer have families with two boys with whatever name, one of them must have a boy with a name starting with the letter A.
We also don't know if either of the two kids is first born, so we take into account the fact the girl can be a first born or born later.
The change of sample size changes probabilities.
Tell you what, same probability logic is applied to quantum mechanics, *which act as though based on probability and not deterministic* and we managed to find almost all particles as described by the standard model.
Truth is, all probability theory was made by professional gamblers, which are not strangers to probability.
+DarkBabyIon you guys forgot to take into account the odds of the frogs being equal distances from you. ie. the odds that the one on the stone and the pair are equi-distant from you AND the odds the two frogs were at the same or near zero arc radius from each other.
Please add these odds to the calculation.
PS. Thank goodness this guy who ate the mushrooms was not an excellent mathematician, only a good one.
having little mathematical experience (1 semester of calculus), but having touched on probability in a genetics class, it seems to me the question could be said as this:
you give mr. jones two 14 sided dice (b0-6+g0-6). he rolls both and says he got a b2 (boy on tuedsay). what is the probability he got g0-6 on the other die? 7/14=50%
Does anyone have a good and clear mathematical explanation for why it's permutations and not combination for the 1st one?
What was done in the video was to say gb and bg are not the same. This is incorrect as he would be using permutations which means to arrange in order. Order does not matter in this scenario, as just because the first child is a boy doesn't effect the parents chances of the 2nd child's gender. it's like dice.
Permutations should be used when you are picking marbles out of a bag, you are removing a choice from the sample space for your next pick.
@@someguy86 Order DOES matter in this scenario. If you flipped a coin twice in a row, you would count HT and TH as being two different results, because they are both equally likely as HH and TT. Similarly, two boys born on Tuesday is half as likely as a boy born on Tuesday and a boy born on Friday, because the boy born on Tuesday could be the younger or older sibling.
If it doesn't make sense to you, consider this: It's easier to have a boy *born on tuesday* if you have two boys, because you've given yourself and extra chance to have a boy *on tuesday*. It's easier to hear a male frog croak if there are more male frogs to do the croaking.
+Davy Ker Well, that does make sense
It all depends on how Mr Jones was selected. If he's been selected because he has a boy born on a Tuesday then yes, it does increase the probability that he has two boys. But if it's just additional information about the boy then it makes absolutely no difference to the probability. We are not told which of those is the case.
Simon Coward We should take all the information provided as given. Why wouldn't we?
+Davy Ker Whether the information is obtained via random sampling matters. Re: Monty Hall Paradox
Adam Hoelscher It's not 'the Monty Hall Paradox' it's the Monty Hall Problem. Please explain what you mean by this. We're talking about conditional probability in a mathematical way, we're not concerned with the physical methods being used to obtain information.
The Boy or Girl answers depend on how you learned that there is a boy, or a boy born on a Tuesday. If you specifically asked "Do you have a boy," or "Do you have a boy who was born on a Tuesday," then the video gets the correct answer. In my opinion, it got there by incorrect logic. Sample spaces don't change when you gain information. A correct solution should use conditional probability: the sample spaces still contain GG, and B6G4, but the probability that Mr. Jones will answer "yes" to either question is 0% for those events.
But if, say, Mr. Jones volunteered the information "I have a boy," or "I have a boy who was born on a Tuesday," then you have to consider the equally-likely possibilities that he would have said "I have a girl," "I have a boy who was born on a Saturday," or "I have a girl who was born on a Thursday." If this is how you learned it, the importance of using conditional probability becomes more obvious.
The probability that Mr. Jones will volunteer the information "I have a boy" is 100% fro BB, 50% for BG and GB, and 0% for GG. With similar probabilities for the Tuesday variation.This makes the answer 1/2 for both questions. So as expected, there is no change.
The reason "Tuesday" seems to affect the answer to the first problem, is because a man with two boys is almost twice as likely to answer "yes" to that very unusual question, than a man with only one. Since there is no evidence that the question was asked.
And this is not an idle criticism, or a new one. Look at the Mr. Jones riddle this way: Mr. Jones asks "I have two children, what are the chances I have a boy and a girl?" The obvious answer is 1/2. Then Mr. Smith adds: "I know the gender of at least one of them." Regardless of what that gender is, the problem is now identical to the video's first variation. If K is the known gender, and O the other one, Your "sample space" {KK, KO, OK}. And your answer has changed to 2/3 without you having gained any information! This paradox was first described in 1889 by Joseph Bertrand, as his "Box Paradox."
The first part of your very good comment seems to refer to a priori beliefs, which is a Bayesian approach to the problem.
I disagree with you on the last part, you gain information by knowing the father knows the gender of one of his kids, which affects the probabilities.
I think the 2/3 answer is the right one without considering a priori beliefs.
So the question here is : what is a probability?
My answer : A probability is a measure of the frequency of realization of an event depending on the information available about this event and the a priori beliefs held by the observer of the event.
Guys, stop posting such long comments and replies
It took me 1/2 hour to understand the conversation
@@xavierplatiau4635 You must assume the father knows the genders of both. This means he deliberately choose to conceal information that would allow you to identify both children's genders. If he actually has two boys, he had no choice about what gender to reveal. If he has one of each, he had a choice. He could have told you that at least one is a girl.
The point is that the events A="Mr. Jones has a boy and a girl" and B="Mr. Jones has a boy and a girl, and told you he has at last one boy" are not the same. The probability of event A is 50%. The probability of event B is somewhere between 0% and 50%, but we can only assume it is 25%. (Otherwise, the probability he has two girls when he tells you "at least one is a girl" is 0%.)
@@jeffjo8732 When thinking of that paradox, I've always assumed the father only knows that "at least one of them is a boy", as if he just discovered by a third party he had 2 children and at least one of them is a boy, I never assumed he concealed information.
But I agree, if the father knows more, then in a bayesian approach you must consider that the father could have said "at least one of them is a girl" etc.
@@xavierplatiau4635 Then to get 1/3, you have to assume that whoever told him about the family was the one concealing information.There are six possible events leading to this kind of knowledge, represented by (First)(Second)->(Known). Two are trivial: BB->B and GG->G have prior probability 1/4. Two more are part of the condition for this problem, BG->B and GB->B. Say they each have probability P/4. The the last two, BG->G and GB->G, have probability (1-P)/4. There has to be some mechanism in transmission of information that makes P=1.
The gender probability each frog (or child) for each individual selection of a frog (or child) are individual events. The probability of the unknown frog (or child) being male is 1/2.
I thought that at first too, but now I think it's similar to the monty hall problem. In monty hall you pick 1 of 3 closed doors (only one is a winner) and one of the doors you didn't pick is opened to be a loser. Now you can choose to switch to the other closed door which has a 66% chance to be the winner vs the 33% chance of your initial pick.
For the frog problem let's say there are 4 closed doors, 2 have male frogs and 2 have female frogs. You pick 2 doors at random and 1 of the doors you picked is opened to reveal a male frog. Now the second door you picked is 33% chance male and 66% chance female.
@@runescapefan0001 The Monty Hall problem is fundementally different because Monty will only open one of the doors you didn't pick AFTER you've picked a door already. Not only that, you can only pick ONE of the doors. In the frog riddle you have 3 frogs, one of which is irrelevant because you know it's male. You can choose to lick ONE frog which has a 50% chance of being female, or you can lick TWO frogs where one frog is irrelevant and the other frog has a 50% chance of being female.
You didn't "pick" a frog already. You KNOW one of the two frogs is male. You're picking between one frog or another.
isnt MF and FM the same? because you can put a plus there so you have M+F and F+M and because you can switch these numbers it is the same thing. if i am wrong pls tell me what is wrong with this thinking
Because when you think about it, if they were the same, then you would be saying that with two children you can only have 3 combinations instead of 4. We have to separate them because by combining them the sample space would not have equally likely events (MF and FM together make 2/4 = 1/2). Order does in fact matter in sample spaces, unless you specifically put in it that one is more likely than the other, which is possible but annoying for doing math with it.
That sounded so counter intuitive untill you said that it's twice as likely to hear the crook in a m/m pairing.
+akakalliba It's not twice as likely to hear a male croak in a MM pairing because it doesn't say anywhere in the riddle that all males croak at same rates. The croak's parameters are never defined. Also, if a male frog croaks only once every 5 hours, we can repeat this scenario 100 times in 3 hours, and maybe not hear a single croak after the 1st time, even if both frogs are male. Also if a male's croak is distinctive, wouldn't that by default make a female croak distinctive. Yet we're not told if males are as likely to croak as females. Each new assumption changes the outcome. Ted's riddle was indeed presented poorly, but I still disagree that 50% is the correct answer. The correct answer can't be found using the given information.
+Aleksey Z. We're operating under the assumption that both genders croak with equal probability. If you were watching this video, you would have heard him say that.
+codeMurp :3 that helps, and is better than ted riddle approach. it's still ambiguous because it forces you to assume something. the flaw is in the riddle for failing to provide enough info for a sample space without assumptions.
Why are we operating under that assumption? The TED-Ed video gives us that male frogs croak and female frogs do not croak. Croaking is not a new category, it is simply the defining characteristic that makes a frog male.The timing of the croak is actually the random event that the person who made this video claims is not practical. TED-Ed is actually right.
for the boy or girl paradox why is BG different from GB? the question simply asks if he has a girl in this case order shouldn't matter
my question exactly.
It doesn’t matter he just needs content
It does not. Both videos are wrong in the same way. And actually this guy is more wrong in a sense. If you falsely assume they are separate the original video is totally correct. Here he just adds arbitrary variables for no reason that are not in the original riddle.
Indeed, it seems very odd, but if you think about it chronologically, it's not that weird. Say Mr Jones has his first child. It's a boy of a girl with equal probability 1/2. Les us denote B1 and G1 these two scenarios.
Now Mr Jones has his second child, still with equal probability 1/2 of either gender. In scenario B1, there are two possibilities: scenario B1-B2, where the second child is a also a boy, and conversely with scenario B1-G2. Both are equally likely and since scenario B1 bore a probability of 1/2, it follows that both scenarios B1-B2 and B1-G2 have a probability of 1/4. This also applies to scenarios G1-B1 and G1-B2.
Therefore there are four scenarios with probability 1/4: B1-B2, B1-G1, G1-B1 and G1-G2. Now say order doesn't matte. It's kind of as if we fused scenarios B1-G2 and G1-B2 into one scenario B-G = G-B (along with scenarios B-B and G-G). Then, as you can see, the probability of having two boys (B-B) is 1/4, same for two girls (G-G), but the probability of having a boy and a girl (B-G or G-B) is 1/2.
I know it seems counter intuitive, but hey, probability theory is full of "paradoxes".
A quick remark: I don't like the boy-girl analogy though. Sex and gender are far more complex and subtle than the boy only / girl only dichotomy. I prefer to consider coin flips.
Now it can help to understand why it would be so absurd to assume that scenarios B-B, G-G and B-G = G-B have probability 1/3. Let's make that assumption. It still holds that, for the first child, scenarios B1 and G1 have equal probability 1/2. Consider scenario B1. Then when the second child arrives, it must be the case that it's a boy with probability 2/3. Because we know that, in the end, there's a 1/3 chance to get scenario B-B. Indeed, 1/2*2/3 = 1/3. But that's absurd. When the second child comes, it's a boy with probability 1/2. (Kind of.)
It seems like just taking all the possible scenarios and then assigning each an equal probability is a mistake though.
It's always a bit iffy to do stats in calculations like this, but let's assume that any frog pairing and croaking behaviour would have been possible, and you just happened to hear one frog croak. I'm also going to take your 50% chance of croaking at any given point. The way a random frog pair is generated is that each pairing MM, MF, FM, FF has a 25% chance of occuring and each M has a 50% chance of croaking.
In that case, an MM pair has a 50% chance of containing one frog that croaks (25% of no croak, 25% of 2 croaks), so you have an overall 1/8 chance of randomly finding exactly an MM pair with exactly one croaking frog from a random sample (1/4 chance of finding that pairing and 1/2 chance of finding one croak). By contrast, both the MF and FM pairs also have a 50% chance of a frog that croaks, since the male might or might not croak with equal probability. Since theres two pairs, the chance that you may find a croaking frog in a pair with one female and male each is 1/4 (2 pairs times the 1/4 chance for each pair times the 1/2 chance that the male in the pair croaks).
You can see that in the full sample space, the possibility of exactly one croaking frog is twice as likely for the pairs that contain one F as they are for the ones with two Ms. Once the sample space collapses because we hear one frog croak, we can see that the probability of a female in the pair is indeed 2/3. The mistake you make, I think, is that you assign M1M0/M0M1 an equal probability as FM1 and M1F. This is incorrect as we have seen, you cannot just look at all possible outcomes and declare them to be equally likely.
Interestingly, this does depend on the probability of a male croaking at any point in time. If males almost never croak, the probability does become 1/2. However, if male frogs (almost) always croak, you can be (almost) certain that you have at least one female. This is intuitive, since assuming males always croak and you find a pair with only one frog croaking, you know that the other is very very likely to be female.
Yeah i fed up with people mistakenly assume that each scenario has equal chance of occurrence😂
And those that cancel out mf and fm pair too 😂what the h
@@theeraphatsunthornwit6266it's like saying there is a 50% chance of winning the lottery. Either you do or you don't but for some reason people are serious about it when it comes to the frog riddle.
why do we need to consider FM and MF as diffrent cases? does it make difference when we just need to know if second frog is female or not?
its also more likely for a male and female frog to be together because male frogs usually fight between each other for territory ;)
(you're mad because i didn't use math)
Well if you want to go down that far, then we have to know if its mating season or not, since that increases the chances of a male frog looking for a female. And to go a step further, we should determine the size and strength of the male frog, for if he is the alpha frog, the the probability that he has attracted a mate is increased even more. That's the whole point of this video: Showing how the probability of something can change and be more complex the more information we have.
male frogs dont hang around so there is no alpha frog
But male frogs don't croak while near a mate so that reduces the chance of the second frog being female greatly. #JustSaying
No math, so no validity
So if we say we know he has a son born on Jan 13th. Making it 730/1459, or 50.03%
The word that makes the probability change is actually when you say "a" which means only 1. You change it to "at least" and probability wont change.
It depends on the likelihood that a male frog will croak. The less likely that a male frog will croak the closer it is to 50% and the more likely it is that it will croak the closer it is to 2/3. So if you know that a male frog croaks every second (but you can't know if it is one that's croaking or both) then it is 2/3 chance that there is a female there.
Why are the options: BB, GG, BG, GB? why include both BG and GB but then not BB flip flopped? So wouldn't it be B1B2, B2B1, G1G2, G2,G1, BG, GB?
Almost correct. Remove the two GG's. We know one is a boy. But yes, your point is completely valid otherwise. It is a 50% chance. Presh was completely wrong here.
You're assuming that a frog is equally likely to croak as it is to not croak
He is not. In this scenario, you know that exactly one out of the two frogs croaked. You could have sat there for a hundred years before one of them croaked.
No he isn't. In each scenario there is one frog croaking, so it doesn't matter what the odds are, they are still equally likely.
+professorgaming except in the riddle, he has to make a split second decision
+trayshawn ugubu
not to croak*
Edit your comment.
It's not just counter-intuitive that knowing the boy was born on a Tuesday alters the probability, it also appears to be wrong.
Wdy mean?
These questions are generally just nonsensical as they don't define a repeated experiment. Only a repeated experiment defines a probability.
@@StefanReich Indeed in classical probability only a repeated experiment defines a probability. There are other theories of probability that are "epistemic", such as Bays theorem is based on...but these theories describe different kinds of probability and this video tries to mix them and ends up with a mess. There are indeed ways to defined a repeatable experiment where the probability of a girl is 2/3rds when we find out one of the two children is a boy...but HOW we find out has to be specified in a way that could in theory be repeated. For example if somebody took a database of all the families in America and filtered out only those that had two children where at least one was a boy, and then randomly selected one from the results.
I my opinion Presh made the mistake of counting BB twice by pretending to add information. Both children are born on certain days of the week, so you can just pretend to know the days for both children and be back at 2/3.
Lets take middle way to solve this:
Just forget it, and who the heck will eat mushroom if he even dont know its?
What’s the name of that feeling you get when a bunch of people think a TED video is wrong because they don’t understand the logic, and then it turns out they’re right because the puzzle made a mistake, but they’re not right for the reasons they think they’re right?
Sums up my thoughts precisely
No, the Ted Ed video is correct
From the top: Why is he including permutations? I think this should be done strictly on combinations. But who am I to comment?
Yeah it's 50% not 2/3.
That’s correct
I thought the same but since he is showing possible cases, he needed to have all of those be of equal probability. If he had done combinations and then possible cases, he would've needed to then somehow factor in the higher probability of the certain cases. By keeping the equal chances between them all, it allows for an easier setup.
However, since he is including permutations, he should've made it two separate ways of getting BB as then it would be B1B2 and vice versa. Then it would be the proper answer which is one half.
@@noahchristman2077 1/2 is not a valid answer in both cases. In case (A), there is only one BB since time is not a variable in this situation. In case (B), there are 13 BBs since B2B2 = B2B2 and the answer is thus slightly higher than 50% (14/27).
What if it was a male frog that identifies itself as female? Triggered.
+Robin Powell [SJW INTENSIFIES]
What if the female frog is actually a trap and the male frog is actually a reverse trap?
You didn't remove G2B2 as you did B2B2. G2B2 is duplicated in B2G2 - so 13/26=50% chance, which makes sense as the probability of having a girl is independent from the day on which the boy was born.
With the frog riddle
Conditional probability
If A is the only male, probability A croaked is 100%
If B is the only male, probability B croaked is 100%
If A and B are male, probability A croaked is 50%
If A and B are male, probability B croaked is 50%
If A and B are female, probability either croaked is 0% (unnecessary but helps lower down)
Unconditional gender distribution probability
MF - 25%
FM - 25%
MM - 25%
FF - 25%
Multiplying out the two events
A is the only male and croaked - 25%
B is the only male and croaked - 25%
Both are male, A croaked - 12.5%
Both are male, B croaked - 12.5%
Both are female, one of them croaked - 0%
A male in an MF pair croaked: a male in an MM pair croaked 50:25 = 2:1 (2/3 against 1/3)
There is a reason this is considered a paradox. The probability is a false probability. Even in the first cases when it doesn't specifically say wat day they where born, you can assign a day because they had to be born on a day of the week. the can assign all kinds of numbers to manipulate the probality like saying the month they where born, the hour, or even the minute.
i dont exactly see the difference between MF and FM. in both cases there is a female and a male so what is the difference?
Same pls explain
+Mohtada Jokhio I agree completely. Seems to me you can have 2 boys, a boy and a girl or 2 girls. Why does the order in which you say "boy and a girl" matter?
+Ravinian Because we're looking at equally likely events.
Echaen Gaming
After reading many of the responses I found one that made sense to me. If you take a population of 50% male and 50% female and just make random pairings...you'll get 25% all male, 25% all female and 50% male/female. This makes sense to me that if you have a pair, and you know one is male then odds are good the other is female.
+Mohtada Jokhio Because one is called Tom and Tom thinks these cases are not the same at all.
Seriously, once we distinguish between two actual frogs, we have to carry that all the way to the end. These cases look identical (to you, because you don't know their names), but they have equal probability to any other case so you can't just mix them together.
Considering you're not asked to identify whether or not a specific frog of the two is male or female, in practice the listings of "MF" and "FM" would be the same situation. You are asked if there is a female in general, so more accurately the possible results would be "there is two males", "there is two females", or "there is a male and a female", as which is which matters not. Thus "is there a female" becomes initially a 2/3 chance. Upon discovering there is a male among them, that chance would become 1/2, as, again, the order of the frogs does not matter. If, however, the question asked "what are the odds that frog 1 is female?" or "what are the odds of either frog being female?" (as opposed to "what are the odds that there is a female?"), then "FM" and "MF" would actually be different results. When it comes to a lack of order in labelling, you can't really consider differently ordered results as the same, because whether you say "there is one male and one female" or "one female and one male", there is still only one of each sex. If labelling isn't necessary, then neither is order. That's like saying there's a difference between 2+1 and 1+2. Without labels, they are the same result, not different.
+tubefan90000 someone already said it... TL;DR: the question doesn't reference an order so BG and GB are the same thing.
"What is the probability if there is a female in the pair..." =/= "what is the probability the [a specific frog] in the pair is female..."
Daonitre Ok, so? First off, I never saw that comment, so of course I wouldn't know that. Second, why does it matter if someone says something in the comments section that someone else already said? It happens all the time, so why is this any different?
tubefan90000 Defensive much? I was going to say what you did; so instead I replied to your comment and made it less of a wall of text. If you get irritated by people replying to you then you're on the wrong site. Enjoy your drama though - I don't need it.
Daonitre Sorry, just very used to people getting mad at me for making a comment that others have made, or commenting in a way others have, while ignoring the others doing the same. The way they do it was the same as yours, however part of that problem is that it is text online, thus removing changes in tone and attitude
Thank you! Finally someone that gets it. I can sleep well today knowing not everyone around me is retarded .
By saying that one of them croaks is just explaining h9w we know 1 is male and not that we know that by random chance
Very subtle. So, if a male frog ALWAYS made a croak sound, with a probability of 1, then the two formulations of the problem would be equivalent. However, as this video points out, if male frogs only make the croak noise SOMETIMES, with probability less than 1, then the two problems are different. It helps to consider a case where the male frogs croak with a very low probability, such as 1/1000. Now, the sound of a croak makes us think, intuitively, that there are likely two male frogs, rather than just one, for such an unlikely event to have occurred. This way of thinking about this Frog problem is somewhat similar to how the classic "Monte Hall" problem, with 3 doors, makes more sense intuitively if you ask the same question with 1000 doors, only one of which hides the prize.
I actually agree with you on your analysis of the TEDed video, but I find the Boy Girl Tuesday paradox ambiguous and uninteresting.
...and wrong, the real sample space is BB, BB, BG and GB. Or short BB and BG.
Why do you differentiate between MF and FM? What is the purpose of this?
Let me simplify. Ignoring that you know one's male, what is the probability that they're both male. It's 1/2 for each. 1/2 times 1/2 = 1/4 Now let's look at why it's that. One possible combination is that frog one is male and frog two is male. Another is frog one female and frog two female. Or frog one male and frog two female. Or frog one female and frog one male. So basically, there's a 50% chance that frog one is male, and if he is, frog two could be either female or male. There's also a 50% chance that frog two is male. If frog two is male, frog one could be either male or female. That gives the possibility of MF or FM
Does anyone have a better way of explaining this? I'm only twelve I haven't learned this yet.
+Ava Nicole I understand this, but if the question is "What is the probability that there is a female in a pair of frogs if one is male?", then what is the purpose of differentiating between MF and FM? They both say the same thing: There is a Male and a Female in this pair. The order of the words does not change the meaning of the sentence. If I can say, "There is a Male and a Female in this pair.", then I can say, "There is a Female and a Male in this pair." I can find no justification to warrant the differentiation of the two. It seems to me to be an arbitrary decision that messes with the results.
+Phoenix Fire There is a double chance because the frogs are distinct entities, even if you can't distinguish them. One of them can be male or female, and the other one can be male or female, so both FM and MF combinations can occur.
+Phoenix Fire
because that idea is the proof that the male/female pairing is twice as likely as the male/male pairing.
+Phoenix Fire there are two frogs and we can label them 1 and 2 (they each stand in a different positions) so the first letter in the sequence represents the frog number 1 and the second number represents the frog numbered 2 so in the sequence FM the frog we labeled as 1 is female and the frog we labeled 2 is male, however in MF the frog we labeled 1 is male and the frog we labeled 2 is female, I hope that explains it
The original video states something not considered in this video. "You don't know which frog made the sound." This is important because if we hear the croak but don't know which frog croaked, then the probability one is female is 2/3. On the other hand, knowing which frog croaked only leaves a probability for the other frog being a girl, which is 1/2. Presh would be correct if we knew which frog croaked.
But the problem is that if there are two males, they are twice as likely to croak, so it ultimately cancels out.
Couldn’t both frogs croak
They 'Could', but it specifically stated you hear A frog croak, not frogs.