Can you solve the frog riddle? - Derek Abbott
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- Опубліковано 26 вер 2024
- View full lesson: ed.ted.com/less...
You’re stranded in a rainforest, and you’ve eaten a poisonous mushroom. To save your life, you need an antidote excreted by a certain species of frog. Unfortunately, only the female frog produces the antidote. The male and female look identical, but the male frog has a distinctive croak. Derek Abbott shows how to use conditional probability to make sure you lick the right frog and get out alive.
Lesson by Derek Abbott, animation by Artrake Studio.
*got poisoned by a mushroom*
"And it may just be your lucky day"
......
🤣
TED loves us.
Llololololol
Buy lotto in dying
My question is, how did he know all this about the antidote when he didn't know the mushroom was poisonous?!?
xp
Isabella Rose Nikle so true
Yeah, but if he didn't know the mushroom was poisonous in the first place, how did he know about the antidote???????? My question still stands
LoL, true question here. Maybe he was in doubt if it was Frognous, the ultra super poisonous mushroom, or Mushtart, the most delicious mushroom in the world.
Isabella Rose Nikle i know right
Guy: *Gets poisoned and has a few seconds to live.*
Also guy: hang on, lemme go over the probability first.
Lol
Ikr
Well you cant just risk it and go yolo
@@senbonkzakura7991 trust me at the verge of death that IS a possibility
@@MrSasukeSusanoo hmm ok
me: smart enough to deduct which frog is a female, and which is not
also me: eats a random mushroom, expecting to be okay
Yeah
To be fair, there are many safe to eat, edible species which look similar to deadly species. For example, "false morels" are mushrooms that are poisonous and resemble edible "true morels."
It’s for science!
Deduce*
I just went with "Oh, males make sounds to attract mates, so the other frog heard it and came there, therefore is a female"
Guess I'm not a math person lol
Oh Deer - Sabrina same
Oh Deer - Sabrina I also am bad at math
Me tii
Me too😆
What if they're both male and trying to attract the female on the tree stump
I wish I had a 67% chance of getting a female.
Lol😂😂
Haha u made my day xD 😂
+Zenn Exile dude....lmfao this is so true. It really crushes that "there's someone for everyone" bullshit and instead replaces it with truth.
***** There are Times, where I feel alone with my German sense of Humor - very alone.....
I was referring to a common german mistake mixing up "to get" and "become"....
lol
I've played enough fire emblem to know a 67% chance ain't goin to save anyone's ass
XD true
Basically like 50% except one side has like 17% more. Not much difference, might as wel pick the smaller percentage.
SOOOO RELATABLE
R.I.P. Cherche
except with criticals
better than 50/ chances
“You only have enough time to go in one direction”
*spends 10 minutes calculating*
“Ok time to go to the cleari-“ *dies*
Lol 🤣
Wow just wow
Yeah true
I was gonna comment this but found this one
well technically during the situation you wont really spend so much time in calculating the odds since its quite obvious which is right, they've simply expanded the explanation to make it easier to understand for other people. its like saying you need to sit down and do written calculation to find 18/3, when the answer for most people is direct and neednt be calculated
real solution: dont eat random mushrooms
lol
Yes! Someone agrees with me. You also beat me to the punchline XD
Or don't go into a random rainforest...
+Tay Unicorn Confess, most of us would be Dead before being done with the riddle
I said look at their areas where the sun doesn't shine....
Plot twist! The frogs were just hallucinations from the mushrooms he ate and were actually just more poisonous mushrooms
Patmike2 ded
Patmike2 XDDDDD
Patmike2 lol
There is actually a hallucinogenic frog that contains dmt and can be licked.
My idea was to run to the clearing because the other frog was actually an hallucination.
I think you'd die by the time you finish calculating...
Yah
thats what i said
I thought the same
Well you aren't meant to calculate before you go, you're meant to already know this type of probability business so you just kinda go the right way from the start through intuition. This video is here to save your life ahead of time. :P
MrServantRider They're both actually even chances, but ok...
I love these kinds of riddles, where the puzzle necessitates that you don't have time to go after all three frogs, but you do have time to calculate conditional probability in your head.
The calculation is super quick if you actually know the math though...
@@CerberusPlusOne If you know the math, you will understand that both options are equally as likely to save you, ted-ed got this one wrong.
Here is how I see this problem. Let's assume the one on the left is the male we heard. Let's also put a (c) next to it when notating the possible combinations. The one on the right has a 50% chance of being a male and 50% chance of being a female. So for this scenario, the possible combinations are M (c) - M and M (c) - F.
If we assume the male we heard is on the right, the one on the left has a 50% chance of being a male and a 50% chance of being a female. So for this scenario, the possible combinations are M - M (c) and F - M (c).
Thus, with no assumptions, we have these 4 combinations, with an equal (25%) chance of occuring:
M (c) - M -> no antidote recieved.
M (c) - F -> antidote recieved.
M - M (c) -> no antidote recieved.
F - M (c) -> antidote recieved.
In conclusion, we can see there are 2 instances in which the antidote is recieved, each with a 25% chance of occuring, and 2 instances in which the antidote is not recieved, again each with a 25% chance of occuring.
For both options we calculate the odds like this: 25% x 2 = 50% chance of antidote recieved.
25% x 2 = 50% chance of no antidote recieved.
This means that the option in which we lick the frogs in the clearing has a 50% chance of saving us, which is equal to the 50% chance of being saved by the frog on the tree stump. Thus, both options are equally correct.
@@rorangecpps1421 But we don't know who is the male we heard and without this information M(c)-M and M-M(c) are the same. Info changes probability.
@@rorangecpps1421 They are not equally likely. Member that probability that event A happens if event B happens is probability that both A and B happen divided by probability that B happens. Here A is at least one of two frogs is female and B is at least one of two frogs is male. Both A and B happening means that two frogs must be male and female. Probability for that is 1/2 and and B happening means that not both of them are female so that would be 3/4 chance. 1/2/(3/4) = 2/3.
@@yary2343 by that same token, MF and FM are the same, since the only way for them to be different is if the position of which frog croaked matters, which would mean that M(c)-M and M-M(c) would have to be different, since the position of the croaking frog is different. removing one requires removing the other and leaves you with two options, MF and MM.
"you only have time to go in 1 direction"
*Does math that takes over a minute*
Logic: this is fine.
Don't worry, the answer is wrong and the actual math can actually be solved in an instant.
@@Chraan it was due to exhaustion, doy
How time be moving in anime
And the math question in a 1 hour exam
Plot twist: the guy watches Ted-Ed and knew about this riddle and the answer
(Before seeing answer)
Odds are equal, unless mating is implied to have an affect.
Clearing: 1 male frog, 1 unknown frog. (0% + 50%)
Stump: 1 unknown frog. (50%)
(After seeing answer)
There _aren't_ four possible combinations in the clearing, because the order doesn't matter at all.
Having
F M
F F
M M
M F
isn't logical because two of them are functionally identical.
Yeah.
It doesn't apply to each of the two frogs separately. It doesn't matter which one croaked; one of them is guaranteed to be male. We can forget that one exists, as far as calculations go. The only reason it might not be 50% is if the social patterns of the frogs played a part in the calculations, which was not specified.
This is what I thought too.
Then you were right
You are correct. Trust me i'm an engineer! Also, if you go at the clearing, you only have to lick one frog instead of two, one of which is definitely male frog!
You take 10 mins to do math and die of the poison mushroom
CONGRATS! YOU SOLVED THE RIDDLE!!!
i love you knukels
gogomen101 change your pp
Actually once you know how to do conditional probability, it kinda takes seconds for that case since the sample space is that little
Vincent William Rodriguez what's pp?
"So youre stranded in a huge rainforest, and youve eaten a poisonous mushroom."
Very relatable situation, Ted-Ed.
Well, i do live in Brazil, so it could happen one day
You should see their other ones. Seriously, this has got to be the most normal riddle out of all of them.
But its just a riddle
Lmao
Not to mention the fact that you have to lick frogs to survive...
BUT
The male frog croaked loudly, and so depending on what time of year, he could be looking for a mate, and he wouldn't need a mate if he already has one and so that means that both frogs in the clearing are most likely male, and so the tree stump frog is more less likely to be male.
Thats what i thought
But
This isn't real life and you have to act on the facts given in question
I thought that too!
EXACTLY
I thought of that too. Would be nice if the riddles are labeled Probability and Logic so you have a better shot and more choices
I'd be dead before I figure this out
same
+GugulynnPenguin True That
+GugulynnPenguin what's not to get?
me to
doesn't matter, I'd automatically go to the two
I guessed the clearing because one of the frogs was male and I was thinking, "hey, what if they're like mating or something?"
😂
Yea same
same here to
Me too
The too frogs were looking at the tree trunk. Like stalkers ready to speak up and try and get a date with the frog at the tree trunk. What if they are both males try to attract that female.
If you think about it in a more realistic setting, two frogs next to each other are very likely to be mates, as they are not fighting over the one on the stump. This means that it is instantly way more likely for one of two frogs to be female.
You can make the odds say whatever you like if you inject arbitrary variables.
Your argument is easily destroyed if the frog on the stump is male or the frogs in the clearing aren't aware of it. But of course, the original question does not provide any of this information, so it is meaningless to make hypothetical assumptions.
yes this is the exact problem with these videos, they're not logical and also their math is just straight up wrong anyways
Even then both could be males and are lekking. Sometimes male frogs will group up especially males with weaker croaks will stay near males with louder croaks to imcrease their odds of finding a mate. Plus it could just not be the breeding season so they aren’t reproducing or just chilling. But yeah the two frogs are the better option.
But I don't really get the odds they calculate. I kean I get the calculation, but also:there's a 100% chance one of them is male, so therefor I would think there still is a 50% chance tou get the wrong one... Is this not true???
@@DaniqueEmiliaSteinfeld no, you're right, the video gets it wrong.
but both ways are bad because the frogs would run away after seeing u
basically dodo birds didn't recognize humans as a predator and by the time they started to adapt they all got hunted to extinction because of the new, deadly predator that it encountered. so if the frog was like the dodo bird it could have gotten hunted however it is in the rain forest so chances of hunters frequenting the area drop a little, also the frog should be rare and hard to find so it's unlikely it would get hunted to extinction before the destruction of its natural habitat kills it.
In reality, all of them are male. Life.
Your picture used to be my Bebo profiler about 10 years ago.
I solved it, really simple math
Step 1: Don't go into a forest alone
Step 2: Don't randomly eat mushrooms, idiot.
Step 3: What kind of weirdo licks frogs?
Drink Bleach Please... XDD ikr
exactly, who would go in the forest and eat random mushrooms
+Amy agwumezie Mario of course
bjgeantil True tho XD
Amy agwumezie XD
So if you see which frog croaked the other one has a 50% chance of being female, but if you don't it magically goes up?
Moral of the story: don't pay attention and you'll be luckier
That reasoning appears consistent with the flawed logic used in this video, similar to common wrong approach to Bertrand's box paradox and red/blue card problem. The questions are phrased in a way suggesting real-world interactions, not quantum-mechanics in which merely observing something can totally alters things.
Part of the confusion people have is that the above examples as well as this one do technically involve conditional probability, but does not affect the sample space in the way incorrectly used to solve it. Based on just the given probabilities and simple facts given, without making any erroneous assumptions (based either on implied verbiage like in the original wording of Monty Hall problem, or completely imagined, or inserting outside knowledge flawed or correct), this case is analagous to Bertrand's box: the probability of **_the_ other** (i.e. at least one of the two, but one is guaranteed not, so one of the one) frog from the pair being female is 1 in 2, identical to the probability that the lone one is.
Imagine the same scenario but with the numbers higher. Suppose the total number of frogs is multiplied by 8: eight on the log silent, eight in the clearing. You perceive at least 2 separate frogs croaking from the latter group. You require the lickings of at least 3 female frogs instead of just 1.
Yes yes, ignorance increases your odds of survival
Actually you could of seen the frog and the odds would have been the same this is shown in the much more Famous version of this riddle.
You are on a game show. Three doors. behind one is a car, you pick it you get it. You pick one of the 2 wrong doors you get nothing. You pick door 3. The host who knows what is behind the doors reveals that behind door number 1 is nothing, it was a losing door. He asks you if you want to switch and get what is behind door 2 or stay with your original pick door 3. What do you do?
That one tends to make more sense to people because you are basically taking the 2/3 odds you were wrong when you picked. instead of the the 1/3 odds you were right originally . If you have switched you have essentially picked both doors 1 and 2 instead of just door 3.
@@dragonlogos1
This riddle is not a clone of the Monte Hall problem.
MasterChief0522 name one Functional difference
Step 1:confirm you have green eyes
Step 2: ask the poison to leave
TRUIEEEEEEEEEEEEEEEEEEEEEEEEEUEUEUEUEUEUEEJLUFSAGDFIYLSGDI:
Me: **quickly runs and licks all of them**
Frogs: We are all male you were doomed from the start
Me: **slowly dies**
It’s a probability
Very small percent of that to happen
@BlazePlayz YT huh
xX Samara Xx I mean that’s a 12.5% probability (I think)
No it’s 1/6 the chance of having a male on the log is ½ but the ground one only has ⅓ ,½ x ⅓ =1/6
It still feels weird... I know that on one side there is one frog that is either male or female, and I know that on the other side there is one frog that is either male or female. The fact that there is an extra male on one side seems like it shouldn't matter since I'm gonna lick both anyway, so I'll be licking a male + another frog that is either male or female. O_o
That's what I thought too.
its like one is going to be male, but the other one you dont know. however the chance is not 50% because both frogs can be the male one, and that gives you a higher chance because you dont know which one is male. since both are not guaranteed male, both also can potentially be female, though not simultaneously
+Haran Yakir . Yeah, I think their answer is wrong, I think I've heard this before. In the case of two males, the croak could have come from male1 or male2 so that case must be counted twice, giving 50/50 % again. A similar problem has been discussed a lot, and depending on how you interpret it you get different conclusions...
ikr
+ZyTelevan
Video doesnt take in the fact that only a male frog can croak.
If you heard one frog croak, that gives you these options
Frog 1 male frog2 male, frog one croaks
Frog1 male frog2 male, frog two croaks
Frog 1 female frog 2 male, frog two croaks
Frog 1 male frog two female, frog one croaks
Now if it was impossible to find two female frogs but both male and female could croak, the video would be right because then you are getting these other possibilities:
Frog1 female frog2 male, frog 1 croaks
Frog1 male frog2 female, frog2 croaks
My take on this is that in the animal kingdom, most of the male species would produce noise/scent to attract the females. It's more likely that the single frog would be a female and the other 2 frogs are males trying to get her attention?
I've an IQ of a potato
but you don't often see two males together as they would fight each other for territory, so the group of two would be more likely to contain a female as a mate to the male that we heard.
My take on this is you probably don't understand what a logic puzzle is.
still lost on this one. I see it as the stump has a 50% chance to have a female which has the antidote right. the cleared path has at least one male but we don't know which one. so one for sure is a male 100% that can't help and the other has a 50 chance to b a male as well or female. so 2 out of three combinations are gonna have a female but that's the theory of probability. I see it as only 33% chance roughly that the pair has a female Cuz the gender of the male frog in the scenario has no power of the gender of the other frog. someone please help explain this to me. I would like to either get it or see if I'm correct
+Alex
You're right on the workings, but eh.. 33%..?.. Anyhow; Here is my now copy+paste'd message which I made after giving up writing a new one each time, though eh, then again I'll just make telling you the last stop:
If the Male was frog A, the possible outcomes were MF and MM, if the male was frog B, the possible outcomes were FM and MM, FM and MF do not cross, regardless of whether you do not know which it is, since they are different situations, they still add up to 50%, with 1/2 representing each of them, while MM also represents 50% and 1/2.
Not knowing which one is male does not make there more chance that one could be female, it simply means there are more possible outcomes than if you did know which it was, while less than if you didn't. More positive outcomes likewise does not mean more chance at a positive result when the positive outcomes have a 50% chance as to which is possible alongside the negative one, to then roll the dice again between the select positive and the negative.
Having a male and a female on different positions dosent actually add to probability it would still be 50% their is one frog and that frog has a 50/50 chance of being a male or a female same with the frog on the trunk both answers are 50/50 we can just ignore the male frog as the male frog isn’t there for a specific reason no reason to account it to our decision
I thought that too. if you're gonna lick both of them why would it matter which side each was on ?
a lot of people are trying to argue that the solution posed in the video is incorrect. I just wanted to point out that this riddle is just another version of a well known and researched paradox called the "monty hall' paradox. It's structured differenty (you must choose between one DOOR vs Two Doors instead of frogs, but the actual paradox is the same). I'm not sure why these guys decided to make up their own monty hall paradox instead of using the original, but if you have doubts, please do some research on the original paradox and you will see that it has been proven that the probabilities in this video are correct. Mythbusters even did an episode on it.
@@MrCruztm There is very little in common between this and the monty hall problem. You have a confirmed losing choice and two remaining choices.. that's it. Every key aspect of the monty hall problem is missing here- they aren't comparable.
@@MrCruztm this isn't even close to the monty hall dilemma because in the riddle ted-ed presented the poisoned man licks both frogs, which in the monty hall dilemma is like picking BOTH doors you didn't select
@@k473r No, in the monty hall problem you select one door, one you haven't opened is revealed, and then you can switch to the second, unopened door.
If I'm dumb enough to eat a PURPLE mushroom in the rainforest then how the hell am I smart enough to know how conditional probability works..
none of these riddles are realstic
Hollyhart19 ya
Hollyhart19 You're probably not.
Or maybe don't be an idiot to eat a wild mushroom.
That doesnt make any sense
it does
Except this video is wrong because the order of the frogs doesn't matter, so one of the two combinations of a male and female frog is also eliminated giving us a 50% chance there's a female in there... combinations vs permutations are also important. Although if it's mating season then definitely go to the pair.
But hey, let's give it a twist. Let's say you were thinking for too long, and now you are so nauseous that if you do go to the clearing, there's only enough time to lick one of them due to them being a few feet apart. Where do you go?
Exactly lol
+Fledhyris Proudhon Yes you're correct, if you could only lick one at the clearing, you should go to the stump.
But no, the order doesn't matter for the math in the original version. It wasn't because I didn't care, but because of the type of probability the question is asking. In the original context, you can lick both at the same time, so the question is of combination, not of permutation, so a M/F and a F/M pairing must be treated as one and the same and so the clearing still has a 50% chance of having a female just like the stump.
+lenno 15697 Deciding whether order matters or not isn't of personal interest, it's of the type of probability the question is asking. You can't treat this question as that of a permutation because it's very nature is that of a combination and therefore gives a 50% chance of having a female in the clearing, not a two-thirds chance.
Daniel Choi
That's like saying when you flip a coin twice, getting a heads and a tails is equally probable to getting two heads.
Not all combinations are equally likely (although all the permutations are). Just as you are twice as likely to get a heads and a tails than getting two heads (even though they are both one combination), you are also twice as likely to get the MF combination than you are getting the MM combination.
In this case, the probability of the combinations form a binomial distribution.
Answer: Confirm that you have green eyes
Tell the poison to leave
That makes no sense, but it is a reference to the green eyes prisoner riddle.
Ozo
@@GayahakJ ulu, man. Ulu.
Makes 0 sense
@@xpearl_heartx no
I didn't distinguish between male, female and female, male
Ikr
+AZ N
and that is why you are correct and the video is flawed.
it claims that permutation matters. but it doesn't. the two frogs can change places all they want, it has no influence on your choice.
the question is not whether the female is sitting on the right, the left or is not there.
the question is whether it is there or not there. there are really only 2 possible scenarios because two of the 3 scenarios are identical and do not influence the outcome as they can be easily turned into one another by the two frogs swapping places...which would not change anything at all.
+CaptainObvious0000 It may not matter to your choice (as if you are choosing which of the two to lick). But it is vital to determining the probability. We start with 4 possible outcomes: MM, MF, FM and FF. All are equally likely (1 in 4 each). A mixed sex pair is twice as likely as a male-male pair. When we hear a croak we can eliminate FF as an option. The remaining 3 are still equally likely (now 1 in 3).
The video is right. And so are the 3 earlier commenters.
They aren't different. Not in a way that will affect us here. You'll be licking them both anyway, in the hope that one of them will be female.
But they are different possibilities which have to be accounted for in determining the probability.
You can choose to think of the variations as both-male 1 in 4, both-female 1 in 4, one-of-each 2 in 4 (or 1 in 2). That's fine because, as you say, we don't care which is which.
The problem is that some people aren't watching the video properly and are saying /it's either male or female, must be equal chance of either/. (That's what I thought at first. Then I watched the rest of the video.)
Paul Kennedy
what your thinking implies is that:
if there is 1 frog sitting to the left and 1 to the right of you, your chances of survival are 50% no matter what you choose.
now suddenly a male frog appears. it walks to the left and joins the other frog. male frogs are useless to you and do not provide any information about the frogs they choose to join.
just because that male frog chose to join the frog on the left, you are implying that the chances of survival when choosing this side are now randomly going up from 50% to 66%.
you have to realize that you can tell the same story but leave all male frogs out. or make all male frogs invisible and unhearable.
it wouldn't make any difference. male frogs provide no cure and information to you in this scenario and you can't treat them as if they did.
your scenarios are MM, FM, MF, (FF)
the actual scenarios are:
1 F, 0 F
your math may be valid for other problems of probability, but not for this one.
If I was in that situation I would be dead by now cause I don't have the brain to figure that out
You wouldn't be interested in watching these kinds of videos if you weren't smart.
+TALKINGtac0 not true. You can just do it for fun if you want.
Omfgg same im so dumb i would never be able to solve this i would be dead by then
Ctystalgaming right!😂
Same my brain wouldn't be able to figure this maths out I'm only 10 XD
But think of it this way, you have 2 coins and your friend has 1. One of your coins has heads on both sides so no matter how you flip it, it always lands on heads. If you both flip your coins, what are the odds that each person will have at least one coin that lands on tails? You each have a 50% chance as the double heads coin is obsolete. They both have a 50% chance and its the same here.
Exactly! I agree with you.
Wrong. You’re eliminating the probability of one of the coins. To do the experiment correctly, flip both your coins, but discard any flip that both land on tails, since one coin must be heads. When doing this, you’ll find that the two-coin flip will have a tails coin 67% of the time, simply because getting a heads-tails is the most likely result when flipping two coins.
@@chininckenwang6004 If you flip a coin and it hits heads, what is the probability that the next time you flip the coin it will be tails?
@@chininckenwang6004 If you have to discard 25% of the results, you've designed the experiment incorrectly
The difference is there is no frog that is guaranteed to be male. You only know that at least one of them must be male.
Idea: dont eat mushrooms in the first place
Lol
Lilac Pastry makes perfect sense though
He said that at the end
Lilac Pastry it's a riddle
I know that. But seriously who would eat any strange mushroom?
Obviously go to the tree stump. Chances are that was a mating call for a _female_ frog at a distance. And Guess what. That frog at the stump is at a distance.
That's what I thought!
+Daniel Nelson Thank you!
EXACTLY!
Except that frogs mate by the female laying eggs in the water and the male fertilizing the eggs after they are laid so there is no direct mating. All of the responses indicating that the croaking has anything to do with mating are fundamentally flawed.
at first i thought like that too :D
Plot twist: all three frogs jumps away and you die no matter what
Omg lol!
jump*
I still stand by this one being 50% either way. You know at least one is a male, it doesn't matter which one. So male female and female male at this point is the same. So it is 50% either way.
a lot of people are trying to argue that the solution posed in the video is incorrect. I just wanted to point out that this riddle is just another version of a well known and researched paradox called the "monty hall' paradox. It's structured differenty (you must choose between one DOOR vs Two Doors instead of frogs, but the actual paradox is the same). I'm not sure why these guys decided to make up their own monty hall paradox instead of using the original, but if you have doubts, please do some research on the original paradox and you will see that it has been proven that the probabilities in this video are correct. Mythbusters even did an episode on it.
@@MrCruztm the monty hall problem is considerably different from this one. it's just that this problem is super difficult to wrap your head around
Agreed, the sample is actually incorrect in this video. I’m no mathematician, but there’s only 2 distinct possibilities given what we know, not 3. Knowing that one frog is undoubtedly male (we’ll call him frog #1) the only unknown is the other frog (frog #2). The two possibilities are
Frog #1 is male & frog #2 is male.
Or
Frog #1 is male & frog #2 is female.
It’s like if you decide to flip a coin twice, you’ll have a 75% chance of getting tails at least once. However, if you get heads after the first flip, your odds of getting tails does not remain at 75% from there, it drops down to 50%.
Initially your options were
Head- head
Head- tails
Tails- tails
Tails- head
But after landing heads it’s just
Head- tails or
Head- head
Because that first flip is known.
Same with the frogs.
@@MrCruztm this isn't the same as the Monty hall problem
@@slipshinobi4749 Your example is flawed. You are specifying that your first flip is heads. Similar to if in the video, they specified the frog on the left was the male. This changes things significantly. A better analogy was if you were told one of the flips was heads. Then there is a 2/3 chance one was tails as well
I would not eat the mushroom in the first place because I'm not a dumbass.
Your name and picture fits so well with your comment
IKR ( ͡◉ ͜ʖ ͡◉）
EGGSACTLY.
Same, I'm not even a fan of mushrooms so I wouldn't eat it
I'd smoke it instead.
My logic- More frog = More chance. ...... that was it.
Kiramki Lilo haha same 😂
But when you hear that there is male it lowers your chances.
The question is how much.
mai logic since the other didint croak it is female dont they croak literally every second
But one of the two frogs is 100% useless, so there is one possible female on the left and one possible female on the right. In my opinion it doesnt matter where would we go; its 50 to 50
Yes, they didn't considerated a thing:
There is a male and a female if the first is male and the second a female or the opposite there're two options.
But if the first is male there is only one the second is female, so it doesn't change where he will go
I'd argue that at least one male frog from the clearing was trying to attract a female, the frog on the tree.
Exactly!!
That's true...
And it found it ?
Following this logic, one can assume that a female would be found on both sides. Since he heard ONE croak coming from the side of the two frogs, the other frog next to the male one is a female since she isn't calling to the one on the stump. So theoretically he had a 100% chance of survival
To be consistent with the video's logic, there are actually two variations of male-male, one where it was the left frog who croaked, and another where it was the right frog. With this, you have two variations of male-male and two variations of male-female, resulting in 50% each.
Since you lick both frogs anyways, only the total matters, so I wouldn't put any distinction between male-female and female-male.
One way you can test this is with two coins, with one of them being double-sided heads. One of them will always come up heads, and you won't always know which one's which, but that doesn't matter, the other one still has only a 50% chance of landing tails.
For dramatic effect, if you saw 100 and heard the 99 of them croaked, you would still have 50% chance of one of the 100 being female, whereas this video would suggest you have a 99% chance.
I get what the video is trying to teach, and I've thought about how the riddle could be modified to get the 67% it's looking for, but I can't think of anything.
I thought I was the only one thinking this. It really is annoying when they're just going to ignore these things..
It's not the chance of one frog being female, though. It's the chance of two. The more frogs the merrier, really.
this comment is really well written, illustrates the flaws in this puzzle perfectly
@@novelyst it is still the chance of one frog because you know for sure that the other one is male and you need to find a female. The chance is 50% in both scenarios.
@@andreapizzichini it's not. The probability issue in the question is based on a simple evaluation of the frog population, not the gender of an individual frog: about 50% male, about 50% female. had you 100 frogs and 99 croaks, because the frog population is about 50% male and 50% female, it is far more likely that *a* frog is female (assuming that this happened by chance).
Think of it like this: if you tossed two coins, the possibilities are HH, TH, HT, and TT, right? Having a combination of H and T is more likely than the individual possibilities of H and H or T and T. Now, if you can guarantee that it's *not* TT, it is now more likely that you have a combination of two different faces than only heads. A 75% chance of at least one T goes down to a 66.6 . . .% chance, not to a 50/50. The same works for three coins, and so on.
If you get just one question wrong on a test, no matter where, you lose a 100% score. You can see how with an accuracy rate of 50%, 1/2 is more likely than 2/2, 2/3 is more likely than 3/3, and so on and so forth for (x − 1)/x.
I have a riddle for you: how can the guy actually know the antidote to the poisonous mushroom if he didn’t even know what it looked like?
Because he's hallucinating
Stolen comment
Brennan Kretzinger Maybe he ate the mushroom, started to feel sick and pulled out some reference book (or even his cell phone and googled it.)
You don't need to know what the cure is in order for the cure to work, therefore you automatically know the cure. I think that's how maths work...
MinishMoosen No that doesn’t make any sense at all, like literally that makes no sense in any scenario unless you know underlying circumstances
IT'S FIFTY FIFTY
MATHS IS LIES
Sup Brewis!
It is 50/50, but math isn't a lie, the creator of the video just isn't good at math.
It's a reworded Monty hall problem, and it's pretty funny you say he's bad at math when mathematically you are incorrect. en.m.wikipedia.org/wiki/Monty_Hall_problem
This is not a reworded version of the Monty Hall problem, though it is similar in many ways.
Yeah, the one thing it definitely isn't is the monty hall problem.
In MH, only one door can have a prize, when in this one theoretically either direction could save your life. You're choosing between a set of one door and a set of two doors (but one of which definitely doesn't contain the prize).
real answer 0%.... why would frogs allow him to lick them like that? wouldn't they run away? 1:06
Kevin Hsieh exactly
I've been defending everyone with the 50/50 logic, I now found this. I need to rethink my life.
Kevin Hsieh ГMAO
They're nice frogs lol :3
Kevin Hsieh true
Ted Ed puzzles in a nutshell :
All disasters will stop themselves to give you time to think
Yup
After you've calculated your odds, you collapse and die.
Oh well. You've lived a good life.
Too bad your last moments were full of math and brain pain...
The odds are 50%. If you are making MF and FM two separate odds, you also have to recognize M(croak)M(silent) and M(silent)M(croak) separately.
EXACTLY
You're not getting the point here.
No matter which male is croaking, there are always 4 variants.
And it's just the 50% M-M variation of 25% wrong answer
That was what i was finna comment. like did noone realise that FM is the same as MF
@@minhphanle3978 actually not. FM and MF are two variants of the same result, if we need at least ONE to be female then that means MM and MF are the possibilites so 1 in 4. It's irrelevant whether or not the first or second frog is female in this scenario.
Thank god, Im not the one who thinks the same!
Wouldn't it just be 50/50? Because you can ignore the male frog right? It's just whether the other two are female
Ignoring the male frog changes the entire problem. Ignoring information fundamentally changes any probability problem.
If I roll a die, then tell you the outcome is greater than or equal to 4, and you ignore the fact that I told you that, you would calculate the odds of a “2” showing up as 1/6.
@@BizVlogs this would be like rolling 2 dice and someone telling you that one die has a “4” in every face, what are the odds that at least one die is a “2”. You’re not ignoring information if you ignore the “4” die, you’re using it.
Ignoring information would be if we said “having a die with a 4 on every face just means that at least one die is a 4”… or “hearing a croaking male just means at least one frog is male”
No you cant ignore a male like that
@@theeraphatsunthornwit6266 you really can. technically the underlying process has a few more steps, but in this case it works just fine to ignore the male.
@@thejackscraft3472 in some set of assumption you cant. The same way you cant ignore opened door in goat door problem.
Just dont eat the fucking mushroom.
Right if I was stranded in a rainforest I wouldn't eat anything I see in there.
Rebzyy Well if you have really good knowledge about that kinda stuff you would know whats deadly and whats edible
ugh
2:48 Your initial probability space is flawed. F/M and M/F are functionally the same. You're counting the same combination twice while F/F and M/M are only counted once.
When you remove the repeated combination, the probability is 50% in either direction.
Seriously, how could you have missed this?
I think they got this wrong. They drew up the sample space as if there is a "left frog" and a "right frog," at least one of which is male, and came up with three possible scenarios.
But if instead of having a "left frog" and a "right frog," you draw up a sample space with a "croaking frog" and a "silent frog" you only get two possible scenarios.
It's tempting to think that they way they did it in the video is correct because you're going to lick both frogs, and they each have an chance of being female. But in actuality, you know that one of them has no possibility of being female, and the only reason you lick both is because you can't tell them apart.
This is what I thought too, thus you have a 50% chance going in either direction. Either i'm getting whoosed big time, or they presented this one wrong.
I think where you might be getting confused is the point where you say "one of them has no possibility of being female"
Let's change what we're looking for to make it easier to understand. Instead of looking for the female, we try and find the male. If you hear a croak, you know that one of them has to be a male. Frog 1 has a 50% chance of being a male, and so does frog 2. But if both have a 50% chance of being male, that means the other 50% must be the possibility that they are female. So therefore both frogs, individually, have a chance of being a female. You said "one of them has no possibility of being female". Once you consider this, you realise that it makes sense splitting the frogs into the left and right frog.
@@Matthew-rl3zf OP is correct in their thinking, you have 2 possibilities- case 1: frog 1 croaked and can't be female. case 2: frog 2 croaked and can't be female. In either case your probability of survival is only dependent on the remaining silent frog.
Not sure what you're trying to point out in the second paragraph. Frog 1 has 25% of being female, frog 2 has 25% of being female. There is a 50% survival rate according to your logic.
@@kfbr3923 By your logic, it’s the same likelihood of getting 2 heads in a double coin flip as 1 heads and 1 tails. Try it.
@@Owen_loves_ButtersNo that's not the same thing. Flip 2 coins and look at 1, if it's tails, re-flip. If it's heads, mix them up (if you insist) so that you don't know which you looked at. You'll end up with the same likelihood of getting 2 heads as 1 heads and 1 tails despite the possible combinations of HH, HT, TH. Try it.
The odds are still 1 in 2, after all you knew at least one of the 2 frogs in the clearing was a male before you even acted so it is redundant to add it into the probability, hence there is still only a 1/2 chance that you will get a female frog in either direction.
If the scenario in this video was to be likened to the Monty Hall problem it'd be like having three doors with one of the doors already being opened before the game started. The whole point of conditional probability is that you update your old/used probability based upon new information, the problem with this scenario is that the new information was given before the original probability was acted upon. In the Monty hall problem each of the three doors present a variable, afterwards after you pick one the host picks one of the wrong ones which rules out one of the existing variables meaning you recalculate your probability using this new information. The scenario in this video can't be thought of like that is because the variables don't change, the stump and the opening are 2 different events that don't influence each other. The other reasons why the scenario in the video isn't like the Monty Hall problem is because in this scenario all the frogs could be male, whereas in the Monty Hall Problem at least one of the remaining 2 doors is a winner, also because you do not get to pick a second time.
What people seem to have a hard time understanding is that the fact that switching makes your chance of winning more likely has everything to do with the host, your chance of picking the right door the first time is 1/3, well the host knows for sure which one is the winner and makes sure not to pick it. What happens next is the part where most people get confused, the host basically choose 2 doors, the one he opened and the one he didn't, so the chance the other door the host chose but didn't open is the winner is 2/3.
This is why this video is nonsense.
Exactlyyyyy. The problem is that they’re making out like female-male is different than male-female, but if you lick both it doesn’t matter.
You can't just remove one of the frogs from the probability pool, because you don't know which one is the male, and you didn't know this before the 2 frogs were in the clearing. The video is not clear, but let's try it in a coin fashion.
Your goal is to pick a circle that has a Tails coin in it. In one circle, I flip a coin and put a cup over it. In the the other circle, I flip two coins and put two cups over them. I reveal one of those two cups to show that it's a Heads coin. Now, which circle do you pick?
People are getting confused because they think a Heads coin being revealed is a guaranteed presumption of the question, but it's not. Revealing that one is a heads drops the chance of at least 1 Tail from 75% to 66%.
What this video did is known as gambler's falacy.
@@Dubaikiwi Ah yes, when the heads is revealed for one of the coins for the 2 coin option, you have a 66% chance of getting tails from the other cup, of a 2 sided coin. I get you
not you thinking you're a smartass and ate it up💀 embarrass yourself
Sorry this might be dumb but isn't "male and female" same as "female and male"?
I mean you know that one of the frogs are a male so really there are only 2 frogs in question both of which have a 50% of being female.
That's wat I was thinking
Imagine if you flip a coin twice, what's the odds of getting 2 heads, 2 tails or one of each?
To get two tails you need to flip tails twice, so 0.5X0.5=0.25 (25%)
To get two heads you need to flip heads twice, so 0.5X0.5=0.25 (25%)
To get one of each, you need to *either* flips heads then tails 0.5X0.5=0.25 (25%)
*Or* flip tails then heads: 0.5x0.5=0.25 (25%)
As such there is two combinations giving you one of each, making it twice as likely to occur.
If you made a table a spliced both combinations that give one of each together, you'd end up with skewed odds (33% to get either option).
Thanks man
both the frogs are different. They have to be taken as separate cases
TED-Ed You are wrong on this one.
Imagine you can see the frog that croaked. Then you know that frog is a male and the other is 50%. Since you are going to lick both frogs, it makes no difference the one that croaked was number 1 or number 2 from the pair.
True
@Everstruggling a pair of frogs, where you know for sure that at least one is useless. Coin flipping where both sides are tails.
Vojta Vojta it’s 2/3 because you don’t know which frog did the call
There is no easier way for explain things than the easiest way.
There is no bigger blind than the one who doesn't want to see.
@@anonymousclown3872 It doesn't matter which frog croaked, just that one is male, because you're going to lick them both anyway.
It blows my mind that 6 years later people are still debating if this video got it right or not. You'd think they would give this video an annotation by now
ikr, I just learnt this conditional probability this month in class 12
Its because the problem itself is a lot of a "haha gotcha" while still being wrong
@@hunterpeterson1495 No it's correct. Everyone just doesn't get that.
@@samuelsoliday4381 The video in the context of frogs is not correct. assuming that a croak is a rare occurrence it means that both have an equal chance. If you want I can explain further but I have to know you'll respond
@@hunterpeterson1495 It is correct. You're just picturing it wrong. If the frog didn't croak there would be a 75% chance that at least one would be female. That's because there's a 50% chance that one of them is female while the other is male, a 25% chance of them both being female, and a 25% chance of them both being male. Hearing the male croak, only invalidates the 1/4 chance of them both being girls. Even though there are three possibilities, one of them naturally has a greater chance of occurring than the others, and the elimination of one of the possibilities doesn't change that.
you only have time to run not to think get a notepad pull out a pencil and do math
right
If you know conditional probability before hand, this takes two seconds to do.
+Awesome Gameplays less
EXACTLY
+Awesome Gameplays If you know conditional probability before hand, you will realise that it doesn't matter whether you go for the clearing or the treestump (and you will realize that TED-Ed is wrong).
So if I knew which of the two frogs made the noise, would I have a 50% chance in surviving?
Yes
+erifetim No
+Yacine Benkirane Yes
+erifetim as amazing as it sounds, if you knew that one of the frogs is male and the other one female, your chance of surviving would rise to 100%
but the chance of surviving or probability in general has no effect on the actual result, which is determined by the laws of physics
+erifetim No,
If you know which frog did the noise, and if you don't- It still leaves you clueless about the 2nd frog.
The location of them doesn't actually matter sense you lick both of them.
The tree stump has a 100% chance of having a female frog because the frog croaking was looking for a mate so it croaked to the female on the stump
RubyHamster EXACTLY
That's what I thought!!
Rubyhamster ted ed might know math but he sucks at biology
That's what I thought! Although he didn't gave us an exact example of how they mate or attract other frogs :/ but other than that you are correct!
Eliza Roll AND PEGGY
If you examine the 4 possibilities, you notice that 2 of them are identical: male-female, female-male, therefore making the chance still 50-50. Another way to approach this is, if you number the frogs and say frog number one made the croak, then you can exclude the possibility of the frogs being female-male, meaning that there is still only a 50% chance of surviving
Yes exactly, this was my logic, not sure why the video is different because simply having a male confirmed means you can rule out one frog. Meaning it’s really just “okay do you want a 50/50 chance on a log or in a clearing”
The situation where 1 frog is male and the other female is twice as likely to happen.
@@_sparrow0 Why?
@@foopy7677 There are 4 outcomes: 1. 2 male frogs, 2. a single male and female frog, 3. a single female and male frog, 4. 2 female frogs. Each has a 25% chance of happening. Because we know that there is at least 1 male frog the first outcome is impossible. So situations 2, 3 and 4 have 33% chance of happening and because 2 and 3 are the same we can add up their percentages. So there is a 66% chance that there is a F and M frog and 33% that there are 2 M frogs.
@@_sparrow0Let's number the frogs: 1 and 2 and say that frog 1 is male, now let's examine the 4 outcomes again. 1: frog 1 is male and frog 2 is male - possible, 2. frog 1 is female, frog 2 is female - impossible, 3. frog 1 is male, frog 2 is female - possible, 4. frog 1 is female, frog 2 is male - impossible, because we labelled frog 1 as the male one and the fact that one of the frogs is male doesn't matter, what matters is that the correct frog is male. I hope i made it clearer
Concerning the pair of frogs, either you heard the frog on the left, in which case the possibilities would be MF and MM
or you heard the frog on the right, in which case the possibilities would be FM or MM
so the possibilities are FM, MM, MM, MF
But, you say, MM is listed twice, and it's the same configuration. Now let me use a Capital letter for a noisy frog, and a small letter for a silent one then we get:
fM, mM, Mm, Mf
as the possibilities. It would make sense that double male is listed twice, since with 2 males, it's twice as likely that one of them would make a sound
So, bottom line: 50% chance either way...
if all frogs are male, youre dead and your decision doesnt matter.
if only the croaking frog is male, you live and your decision doesnt matter.
so the only time your decision matters, is when there are exactly two male frogs and one female.
which means this is the monthy hall problem in a disguise, so the answer is 2/3.
if youre not familiar with it, 3 frogs in total, 1/3 any one of them is the female. so you choose one, but before you lick it one male identifies itself. if you stick with your original choice, youre sticking with 1/3, but if you switch, youre improving your odds to 2/3, because at that point only two out of the three frogs are unidentified and youre choosing one of them.
TheGundeck the reason we were able to break our options down into MM, MF, FM, FF is because we knew the probability of getting any one of those combos was equal. We don't have any information about how often a male frog croaks, or how likely they are to croak in a certain amount of time. So we can't say Mm and mM are equally as probable as MF or FM. The video assumes that P(mM) + P(Mm) = P(FM) = P(MF). Not P(Mm) = P(mM) = P(FM) = P(MF) like you suggest.
@@mitch9237 in order to make the probabilities like in the video, male croak rate would have to be 50%. That would make the probability of the single silent frog being female 67%.. survival rate is the same in both directions. If you change croak rate to approach 0%, survival rate is 50% in both directions.
I don’t think ted Ed is trying to assume anything here, they just don’t care if they’re wrong.
I haven’t seen that many flamewars as crazy as this one
It's because the video is wrong
Petr Novák Yes it it wrong
@Sophie Toma think again
Sophie Toma the possession of the frog doesn’t change things. It’s still 3 probable outcomes. Saying that the frogs possession is a different outcome isn’t valid because the position can change without the outcome changing. The odds of 1 frog being male are 100% and the odds of the other being female are 50%. Doesnt matter which frog is which
@Sophie Toma Don't forget the "male frogs may croak" part.
The problem I feel with the 67% answer is that it is taking into account the fact that Male : Female and Female : Male are both probable outcome. While this is technically correct from the view of finding a cure the order of male and female frogs is not a factor and thus the two outcomes are the same, this means that separating them into two probable outcomes is incorrect as they are not actually separate outcomes.
Another problem here is that they say that information is giving you a higher chance of survival but it is actually a lack of information that is skewing the conditional probability as if you knew which of the two frogs was male there would be a 50:50 chance of a female being the other frog (This is also of course not taking into account the fact that in there being a male frog this raises the chances of either of the other two frogs being female by a slight fraction as there is now 1 less male frog in the pool thus altering the 50:50 chance)
The answer is correct from a mathematical standpoint but the question is wrong which is why the answer doesn't sit right.
+Quiczor ^ this. :S
Agreed
+Quiczor Thank you . They are trying to use the Monty Hall solution here, but the set up doesn't allow for it. Male:Female and Female:Male is a different set of data in theory, but not in the probability of survival in this situation.
+Quiczor Each frog has it's own independent odds of being male or female. It's not a matter or order but of each frog being unique.
+Vnxnymxus Machiavelli but this isn't the monty hall problem. This problem is simply deducing which choice is more favorable and why. Thats it. The monty hall problem is different because at no point did you ever divide doors into two groups and also you are given exact information about one door, rather than a condition that effects only two.
They got this one wrong, it's a 50/50 either way. This riddle is inherently different from the monty hall problem because the frogs have no connection to eachother, and there is no all-knowing being that eliminates chances for you. all of the frogs are an individual coin-flip
Exactly.
Or you can run to both
Nope! Search Bayes Theorem or the Monty Hall game and you'll get the same answer
So 1 of them is 100% male
And 2 of them have 50/50% chance of being female
@@silentofthewind The Monty Hall paradox is completely different, that is conditional probability and has everything to do with the host. i.e. You pick door one(1/3), host picks between door 2 and 3, host opens door 3, chance of door 2 being the winner is 2/3. Why? because the host had to pick between door 2 and 3 and cannot pick the winning door, so if you conclude the host picked 'both' doors than the one the host didn't open has a 2/3 chance of being correct.
The situation in this video is very different.
Guy: doesn't know mushroom is poisonous
Also guy: knows the antidote
What if he is study about frog, and he knows if a female blue frog cured all of the poison no matter what poison it is
Difference between Zoology and Botany
@@accidentallyaj5138 Actually, herpetology (study of amphibians and reptiles) and mycology (the study of fungi). Botany refers to plants, and fungi are not plants.
@@CerberusPlusOne An error on my part , apologies because I know better, that was a hasty reply which I didn't think through as in the moment I was thinking about plant based antidotes.
I think I am the only one who noticed a logical error. The croaking sound from the clearing not only tells us that there must be at least one male, but it also tells us that one of the frogs on the clearing can't be female, so either [male, female] or [female, male] must be removed from the sample space for the different combinations of frogs in the clearing, leaving the survival rate of going for the clearing to 50%.
Edit: Since my comment has apparently caused so much controversy, let me solve it once and for all.
We are not told the probability that a frog will croak before we pick a side, given it is a male, so I will use 𝑝 to represent that probability.
Now, we list all the possible combinations of frogs without the knowledge of the croak (a means the frog on the left is female, and b and c means the frogs on the right are female respectively):
{}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
We then give each outcome above a "weight" representing the probability that only one male croak comes from the right and no croaks come from the left (this step converts the sample space of the possible genders of the frogs from before the croak to after the croak):
{}: (1−𝑝)(1−(1−𝑝)²−𝑝²)
{a}: (1−(1−𝑝)²−𝑝²)
{b}: 𝑝(1−𝑝)
{c}: 𝑝(1−𝑝)
{a, b}: 𝑝
{a, c}: 𝑝
{b, c}: 0
{a, b, c}: 0
Using these "weights," we can calculate the probability of there being a female frog on the left:
𝑃ₗ=((1−(1−𝑝)²−𝑝²)+𝑝+𝑝+0)/((1−𝑝)(1−(1−𝑝)²−𝑝²)+(1−(1−𝑝)²−𝑝²)+𝑝(1−𝑝)+𝑝(1−𝑝)+𝑝+𝑝+0+0)
=1/(2-𝑝)
and on the right:
𝑃ᵣ=(𝑝(1−𝑝)+𝑝(1−𝑝)+𝑝+𝑝+0+0)/((1−𝑝)(1−(1−𝑝²)+(1−(1−𝑝)²−𝑝²)+𝑝(1−𝑝)+𝑝(1−𝑝)+𝑝+𝑝+0+0)
=1/(2-𝑝)
I know it's very messy.
From these results, the survival rate of going either direction is the same and is greater than 50%, but the survival rate still depends on 𝑝.
Edit 2: Just realized that my calculations in the first edit did not account for female croaks and successive croaks.
You are by far not the only one noticing flaws in this video.
Yeah this is the only TedEd video I’ve noticed flaws in.
(And there are more than a few of them)
Ive been searching for someone who noticed this. Thank you.
Exactly, I'm really sad that not 100% of comments are just saying it. On the left, knowing at least one of the two is male, only means one thing : the other can be male or female so 50/50 on the left, and 50/50 on the right.
Actually that is not the correct calculation. Since you do not know which combination is incorrect they are still possible. There is still an flaw however since they have not taken into account the two distinct cases of the male pair. Either one could croak and each state is equally likely. However for the Male-female pair there is only one state for each. The male is the one to croak either way. So the probability is still 50% but not because they counted one state to many but rather one state to few.
You:don't know what mushroom you just eat
*Also you:Know the type of frog that can cure the poison from the mushroom that you don't know*
Perhaps he didn’t know what specific mushroom species he ate until symptoms specific to that mushroom appeared.
@@lilacdragon44 he seen it before
@@lilacdragon44 if he knows that such a mushroom exists, why would he eat random mushrooms?
C'mon TED-Ed, this is incorrect. The odds are 50% either way.
There is an easy experiment, take 1000 tables and flip on each table 2 coins. you will habe round about 250 times 2 head/2 tails. and 500 times 1 head and 1 tail. Now you put all tables away with 2 heads. If you now chose a random table left what is the probability for get at least 1 head ?
You have 500 tables with 1 head and only 250 tables with no heads, so you chance is 67%.
@@goldenehimbeere You're over thinking this. I don't disagree with the math in your example, I disagree that your example is equitable to the riddle in the video. Each option in the video offers you the opportunity to lick one frog with an unidentified gender, thus each option offers 50% survival.
There are many other explanations further down in the comments.
@@jamesscott6527 they are all still wrong. i can simpify it for you:
just take 2 tables with 2 persons. the first person just flips 1 coin the other person flip two coins till he gets at least get 1 head. now you can chose with of them has fliped 1 tail. The person who fliped 2 coins has a 67% chance of a tail. ^^
@@goldenehimbeere "the other person flip two coins till he gets at least get 1 head" doesn't quite fit. You should discretely mark one of the coins to be the "croaking frog" coin.. and the other person should flip two coins until THAT coin is heads.
If that coin is tails and the other coin is heads, it would represent the silent frog being a male and the croaking frog being female, which should be disqualified from your sample set.
If there is a male croaking towards a female it could be a mating call. Since there is one male there the other cant be female as what's the point to calling to another one. So they would both be male trying to get the female.
The correct answer for this riddle is quite crazy. You don't even have a full chance of surviving it.
***** The problem is that this riddle has no definite answer. It only has a probability. It says which should you pick, not which has the probability of killing you. But there are still many other things that can contribute to the probability. So that's why this riddle's answer is incomplete. But it should have many different answers such as the one in the video.
It's a riddle. Ted is wrong due to this fact, as are you.
It's 50/50 chance. This is fact.
If, however, it was a question?
Great, then you could be right (though Ted would still be wrong).
why is it 50/50 chance if you go to the clearing with two frogs?
+xPhysicism
I'm going to hope that you're genuinely curious, rather than the usual whom is incapable of understanding and thus asks rhetorically:
It is 50/50 chance if you go to the clearing with two frogs. The clearing with two frogs has two frogs which can both be male or female. Now, use a probability tree (search online for an explanation of what one is, if you do not know what one is) and fill in the first line with "Frog A is the croaking Male" and "Frog B is the croaking Male", each with 50% chance - after all, it is guaranteed that one of them is the croaking male, we simply don't know which. Then, stemming from each of those add on "Frog B(Frog A stem) is Female/Male" and "Frog A(Frog B stem) is Female/Male", all 4 of the second lot should have 25% chance each due to each frog having an independent event, thus 50% chance of either gender. You now have shown that there is a 50%/50% chance in a far more time-consuming way than acknowledging that one is Male, and therefore excludable from the results to begin with.
;)
There shouldnt be a difference between MF and FM.
Thus leaving with the 2 possible outcomes MM and MF/FM.
In both situations is 50%
Süleyman Uluköylü if FM /MF is counted than MM/MM should too
No. There are two frogs that are independent entities. Name them Casey and Riley. Casey could be a boy and Riley a girl, or Casey could be a girl and Riley a boy. These aren't the same. On the other hand: Casey is a boy and Riley is a boy is the same as Riley is a boy and Casey is a boy. Similarly, Casey is a boy and Riley is a girl is the same as Riley is a girl and Casey is a boy. The order of facts isn't important, but the fact that the two frogs are different is. (Naming them just helps to preserve this fact when you start rearranging things.)
Bryan Stevens let’s separate the problem into two possible scenarios
Frog 1 is definitely male
So there are two possibilities
MF or MM
50%
Frog 2 is definitely male
Two possibilities
FM or MM
Either way it’s 50 %
@@olixx1213 I definitely agree! 🙂
@Everstruggling It's about probability. In one, frog 1 is definitely male while frog 2 happens to be male. In the other, frog 1 happens to be male while frog 2 is definitely male. That makes MM twice as likely to occur
i still dont get how it's a 67% chance of living when going to the clearing it should be a 50/50 shot going to either side right. when he was doing the math he included a duplicate pattern [male, female] [female,male] which made his answer of 67% incorrect or am I wrong and should both be included?
Since it is shown in the video that you can lick both frogs at the same time. Positioning of the Frog whether it's a Male and Female or Female and Male doesn't change the fact that you still need to find a Female regardless of it's position. So I believe that it should still be a 50% chance. Hence if you really look at it closely the one of those 2 frogs only has 2 possibilities either being a male or a female so it's a 50/50. Position should not affect your chances of it being a male or female frog.
@@taiyou2331 and yet there's somehow still people that argue that MF and FM are two different scenarios that are independent to each other. I find it astounding that people graduated high school without ever having learned the difference between a combination and a permutation at all, thus leading to so many pointless arguments in defense of the 67% chance
@@Bapringles but they are still the same combination
@@flyingonionring And that's what they don't realize and/or defend. People just can't comprehend this simple idea and believe MF and FM are different scenarios that should be counted as such, despite both leading to the same result
@@Bapringles Search the Monty Hall game or look at Bayes Theorem; you'll see the answer both theoretically and empirically is 2/3!
wait what? but wouldn't both have the equal amount of chances? because with the single frog, we know that there is a 1/2 chance. and with the double frogs, we know that at least one is male with basically leaves us with only one frog. and that frog has the same amount of chances as being male as the single one. the pictures of the two female and male frogs at 3:14 is the exact same combination but swapped around? that makes no difference whatsoever
It doesnt make sense logically its all mathematically while your logic would work if there where infinite numbers there is not. And as for the male female female male that goes for what is on the left and what is on the right
+Raivthx That's what I think too.
+Raivthx Remember, the frog on the treestump is probably a forever alone nice frog that's been friendzoned so it's probably male. Meanwhile the frogs on the clearing are likely a pair having sex which is why chadfrog was croaking so therefore there's a female there.
Seriously though, the guy on the video incorrectly used permutation instead of combination to calculate odds which is incorrect because frog positions do not matter. Licking both frogs on the clearing means you're taking a chance with only one frog, just like the tree stump frog. This is literally stuff you learn in day 1 of any college probability class.
+Raivthx There are 2 frogs so the probability of at least one of them being female is higher. When you toss a coin its 50% chance. You toss the coin twice, the chance of at LEAST of them is 75%.
+Raivthx Yeah I thought the same thing as well. I disagree with their solution, I think it's a 50% chance either way
"You ate a poisonous mushroom."
*jokes on you! I dont even like mushrooms!*
But this mushroom looked like a tide pod
@@surelock3221 😳😳😳😳😳😳😳😳😳😳😳 tidepod!1!1!1!!1!!1!1!!1!
Yea boiii
@@surelock3221 omg 😂
LOL :D
I know this is more of a PSA on conditional probability than an actual riddle, but you have a 50% chance of survival whichever way you go since the male frog is meaningless
a lot of people are trying to argue that the solution posed in the video is incorrect. I just wanted to point out that this riddle is just another version of a well known and researched paradox called the "monty hall' paradox. It's structured differenty (you must choose between one DOOR vs Two Doors instead of frogs, but the actual paradox is the same). I'm not sure why these guys decided to make up their own monty hall paradox instead of using the original, but if you have doubts, please do some research on the original paradox and you will see that it has been proven that the probabilities in this video are correct. Mythbusters even did an episode on it.
It's not 50%.
Think of this differently; what is actually being said is "if you grab two frogs at random, it's more likely you will grab a male and a female than two males".
If you phrase it that way it's easier to understand.
@@maxastro since 1 of the frogs are male then that means that only the other frog decides whether or not you live and thus only 1 frog matters, still 50/50
@@spoonythegamer21 That's not how flipping coins works. You can try this yourself very easily: Flip pairs of coins thirty or so times and record the results.
You will see that one heads and one tails, in any combination, happens about twice as often as two heads.
@@maxastro he just described the frog riddle and you came back with “that’s not how flipping coins works.” Right. It’s not the same problem. No one is arguing with you about your coin problem, we all get it. You can stop bringing it up. It doesn’t fit the video.
Help me please because i'm stuck with the idea that: you see 2 frogs, you KNOW that one of them is male, which means only one of the two can be female, which is a 50% chance. So.... 50% chance for me both ways :S
Here it shows "Female-Male" and "Male-Female" being 2 different things, but does it matter? The only dilemna is "Male-Male" or "Male-Female" in no order.
Why would knowing what the gender of one male is affect the gender of the other one? Perhaps is it like the Monty Hall paradox... but even the monty hall paradox I get to understand.
+Jack Scully I'm pretty sure you're wrong.
As you said - you are licking both frogs. So which one croaks has absolutely no relevance to this puzzle. So why should (m/m) be there twice? Think about it like 2 coin tosses - there's 4 possible outcomes each with the *exact same probability*:
(m/m)
(f/f)
(m/f)
(f/m)
Now because the "order" seems to confuse you we can also represent it as
and
Both still with the *same probability*
If we remove (f/f) from the "even" set we are now *twice as likely* to get an odd pair than an even pair. Which means our probability of getting an odd pair (and surviving) are 2/3 or 66.6..%
Oh, and one more thing that kinda bothers me. Maybe you should try "i think they're wrong" instead.
+Jack Scully I totally agree with you and think that the sample space was the issue. Can you explain what you are labeling as "a" and "b" in the equation though? Tried to work it out but I'm not sure which probabilities you're assigning to the letters. Thanks
Ralph Fischer
After having debating this, it appeared that TED is actually correct here. You have to consider the problem more like this: the 2 frogs are taken from a pool of female and male, and if the 2 are female, then the peer is rejected and you try again. And actually, FemaleMale and Male-Female, are actually different (although it took me a lot of time to agree to it), because you have to consider that the presence of this Male is not arbitrary (the boy could very well have found 2 females, but the croak says otherwise), and as such, if the first frog is a Male, there's only a 33% chance that the other one is too.
*****
I'm not a native english speaker and I have trouble explaining such a subtle thing, but the video is correct. You have to think that the presence of the croaking male is not arbitrary, it already took its part of the chances of having 0 female.
*****
I'm gonna repeat it: think again and you'll begin to consider that the presence of this croaking male diminished the chances of having another male.
If you know that a certain frog is the cure to a certain mushroom then why eat the mushroom ?
+saoirse casey so you have an excuse to lick frogs and get hige?
+matico24 yes i suppose 😄😄😄😄😅😅😂😂😂😂
isn't male and female and female and male the same outcome, shouldn't they be classified under the same label making it a 1/2 probability either way? I'm confused.
i think exactly the same,looks like ted ed made a mistake
That's what I thought. Going left or right doesn't matter, it's still a coin flip.
The ordering is irrelevant because you know already one of them is male. M+F = F+M. So only thing that is unknown is the 2nd frog (position doesn't matter), and it just has plain 50-50%. The ordering matters if you have to roll 11 with 2 dice for example. You can roll it with 5+6 or with 6+5, giving you more probability (since first die can be either one, second needs to be exact the other).
I was actually looking for a comment like this because I was wondering the same thing.
It... should be 50% chance both ways...?
Correct. They fucked up, Female/Male and Male/Female is the same combination and therefore would only count as one probability. Eliminating the Female/Female pair would leave you with a 50/50 chance either way.
+Isabella Taylor exactly what I was thinking
No, TED is correct.
For the initial sample space (ignoring the additional information from the croak), the probability of both frogs being Male is 25% (50% * 50%). The probability of both being Female is 25% (50% * 50%). The probability of a Male-Female combination is 2*50%*50% = 50% (as the chance for Female/Male is 25% and the chance for Male/Female is 25%, you just add them together to get 50%). This is a simple example of a binomial distribution (distribution in the form nCx*p^n*q^(n-x) ).
An alternative wording, it is clear as m=0.5 and f=0.5 for (m+f)^2 = m^2 + 2mf + f^2, where m is the likelihood of a male frog, f is the likelihood of a female frog. The term m^2 is the likelihood of two male frogs (0.5^2 = 0.25 or 25%), the term f^2 is the likelihood of two female frogs (0.5^2 = 0.25 or 25%), and the term 2mf is the likelihood of a male and female frog (2*0.5*0.5 = 0.5 or 50%).
When the additional information comes along, you eliminate the Female/Female pair leaving a 2/3rds chance of having a Female at the clearing.
+lenno 15697 But there is no actual difference between the m/f pair and the f/m pair. Unless there was something to distinguish one from the other, these would fundamentally be the same thing. If one of the frogs moved, you wouldn't know if it was the m or f frog, if both frogs WERE different genders. One female and one male is the same combination as one male and one female. The communatative property shouldn't come into play here. It is one possible outcome.
Huh. How does commutativity even relate?
The essential flaw in your argument is that you consider m/f and f/m the same pair AND assume that the probability for any pair is equal (thus an initial 33.3% chance each before eliminating the two female pairs making 50% chance each). This is simply not true. *
Think of it like this. What is the probability of getting a male? 50%. Getting it twice? 50%^2 = 25%. Same thing for two Females (25%).
The only remaining option is a male and a female. And since the probabilities must sum to 100%, the probability of a male and a female is 50% (or 100% - 25% - 25%).
Alternatively, the probability of getting a Male and then a Female is 50% * 50% = 25%. The probability of getting a Female and then a Male is 50% * 50% = 25%. If you don't care about the permutations (whether Male is before Female), then you add the two probabilities together (Male then Female + Female then Male) to get 25% + 25% = 50%.
Case in point: Flip two coins and record the pairings. You'll find that the pair HT/TH (heads and tails pair, don't care about order) occurs 50% of the time. You can try this by yourself if you don't believe me (or write a program which will be able to do this many times).
* Edit: Actually, you may be thinking along the lines of there is a guaranteed male and the next gender has a 50-50 chance of occurring. That would be true if you knew the position/order of the male (the first or the second in the pair), even though order doesn't have any effect on whether you die or live if you choose to go to those frogs.
A similar thing with conditional probability occurs in the Monty Hall Problem.
www.khanacademy.org/math/precalculus/prob-comb/dependent-events-precalc/v/monty-hall-problem
Here's a video that better explains conditional probability.
www.khanacademy.org/math/precalculus/prob-comb/dependent-events-precalc/v/bayes-theorem-visualized
wait a second... if he licks both frogs, we know 1 of them is male, so its a 50/50 if the other frog is the female you need.
that's what I thought, they are wrong
@Sophie Toma Male-Female and Female-Male r the same. Female-Female is impossible since 1 is male. its male-male or male-female
@Sophie Toma we know charlie is male. the video tells us that. (0:44) So...
Charlie is Male and Alex is Male
Charlie is Male and Alex is Female
@Sophie Toma ITS SAYS MALE FROG. AT LEAST ONE OF THE FROGS IN THE CLEARING IS MALE.
@Sophie Toma i see the light now
The problem I have with this is that, on the sample space, there was m/f m/m f/f and f/m. But m/f and f/m are the same, reducing it into a 1 in 2 chance. The same as the tree stump.
Those are different outcomes actually. Imagine there was frog 1 and frog 2. The first letter would be frog 1 and the second, frog 2. You don't know which frog croaked so M/F and F/M are different.
yeah thats wrong assumptions
Cindy Lei I get that, it just doesn't seem to fit right in my head :P
No they aren't different you can lick both
Both M\F and F\M will save you. However, that doesn't mean it's the same. Imagine the male frog was toxic and the first letter is the frog you licked. Would you still say they are the same?
But it's wrong though..
You specifically hear a male frog. So you're sample space would be
Male 1 (croak) - male 2 (no croak)
Male 1 (no croak) - male 2 (croak)
Male - female
Female - male
Its a 50% chance.
You’re wrongly assuming that Male 1 (croak) - male 2 (no croak) has the same chance of happening as male - female. The croak combination is a subset of the male - male possibility,
So male (croak) - male (no croak) and male (no croak) - male (croak) both have half the possibility of occurring from male - male
Leading to the answer in the video
@@chininckenwang6004 You're also wrongly assuming the probabilities of permutations, and missing the fact that Male (croak) - female (no croak) is also a subset of male-female. I would have written the sample space like this
(croak) male - (no croak) male
(croak) male - (no croak) female
(croak) female - (no croak) male
(croak) female - (no croak) female
by eliminating the combinations with a croaking female we're left with
(croak) male - (no croak) male
(croak) male - (no croak) female
Because we can't assume that (croak) male - (no croak) male has the same chance as (croak) male - (no croak) female, you still can't say your survival rate is 50%... it depends on the odds a silent frog will be female. But we know the survival percentage is the same as the single silent frog in the other direction.
(no croak) male
(no croak) female
There's a better way to understand this (:
It's known as the Monte Hall Paradox. There was an old game show where contestants had 3 doors and 2 of the doors had goats behind them while 1 door had a car.
You would pick a door, the game show host would SHOW YOU a door which HAD a goat behind it, and he would then ask you would you like to switch to the final door remaining.
Do you say yes?
Most people would say there's no difference and the chance of you getting the car is 50/50, but you should ALWAYS switch because you have a 67% chance of the other door having a car. There's an easier way to see this.
Let's play the game with a MILLION doors. You pick a door, the game host opens ALL of them EXCEPT door number 777,777, then he asks if you want to swap... Are you going to stay with door number 1, or are you swapping to door 777,777? Essentially, what are the chances that you picked the CORRECT door on the first try, or what is the probability you picked the correct decision to start? The information given helps you deduce which door can't have the car behind it.
@@fj7509 The only similarity between these two problems is that there are 3 of something and 1 is revealed. The most important part of the monty hall problem is that the host is not revealing goats at random... If you chose door 1 and they accidently revealed that there was a goat behind door 2 (like if you hear a goat behind door 2), there would be no reason to switch doors.
@@kfbr3923 Me when I forget that the *_male_* has a distinctive croak (0:32):
(croak) female - (no croak) male
(croak) female - (no croak) female
I love these riddles.
RainbowPie me too
RainbowPie ME TOO
oh my gosh xD
V
me 2
just because there are more positive outcomes doesnt mean theres a higher chance of a positive outcome because different outcomes can have different chances
the chance is 50/50
Exactly
Man my brain hurts, I thought male/female and female/male were the same thing.
TheDudeReviews You're right. _He_ was wrong.
I could be wrong, I just don't understand it enough yet.
+TheDudeReviews Nope. Think like they were your children. First kid could be m/f. second could be m/f. In all you are more likely to have m/f in any order than m/m (your second child was not born first). Congrats for not blindly listening to Jake.
ka da Jake? not the narrator, not me, not the creator of the riddle
Who's Jake?
You are. You are Jake now. Get the f*** used to it. Lol jk jk. You're wrong though :)
This riddle doesn't make sense, the probability of it being male/female or female/male is the same outcome, which would mean it could only be female/male or male/male making it a 50% chance for either direction. Usually Ted Ed has good riddles, but this one just isn't isn't one.
"Can you solve the frog riddle?" Apparently TED-Ed cannot.
haha
oof
haha
Prove it
No, apparently Derek Abbott cannot. The description says Derek Abbott show you how
But the male frog in the clearing wouldn't be croaking if he found the female.
Yup the croak is actually the clue to the answer.
...True, considering the croak attracts females. What if the female hopped down from a tree after you take a few seconds of the video to realize this? It'd make sense then.
But...then they should be fucking right?
We may be overthinking this.
I can't be the only person that used biology to solve this and not probability?
nope
I did that too. And this changes everything. You usually wouldn't see 2 females together. But it might be 2 males calling the female.
+Symphonia doll But that's unlikely since they would fight rather than hoping the female likes one of them more. Frogs only croack if they mark there area or call for pairing but they wouldn't if they had their business with another male
+fabske 1234 I won't say much since I'm not an expert. But I think they do go to a specific spot together and call so females coukd hear them better. Then she'll choose. I've seen it on a documentary.
+fabske 1234 I've just got one question. Is what you said true for sure or just what you think?
The frogs are separate from each other. You have the same probability either way. The chances of one frog being female is 50%, and there is one of these chances in the clearing and on the stump. The male frog does not matter to you, so there is essentially one frog on either side.
Nah. Think like this. A friend of yours tossed two coins, and ask you if he tossed any tails. But he also tells you that he did not toss two tails. What are the probability of having at least one tails?
Possibilities are these, each with the same chance of occuring.:
HH
TH
HT
TT - But this one is out of the question.
Chance sums neatly up to 2/3.
@@RegiRanka And what is the difference between TH and HT? Both scenarios have only one Tails
@@nomoiman the point is that there are two ways to flip one tails and one heads, but there is only one way to flip two heads and one way to flip two tails.
If you are playing craps, there are a lot of ways to roll a seven, but there is only one way of rolling a 12 and one way of rolling two. For the purposes of counting, you don't care if you roll a one and a six or a six and a one, but for the purposes of probability these are two distinct events.
@@ianhruday9584 No see, it doesn't matter in which order you get TH, your still left with one of each
@@nomoiman obviously, but that's not the claim. The claim is that heads tails and Tails heads are two distinct ways to get to the same outcome, but there is only one way to get to the outcome heads heads and there is only one way to get to the outcome Tails tails.
Monty hall problem???? ANYONE??????
probably
I know my first thought, also what I used to estimate my chances.
+Ludvig SC Games Original Monty Hall problem is more interesting imo, and funner to think about.
+Ludvig SC Games In Monty Hall you pick 1 door. You're not picking just 1 here.
Christopher Stone It is quite similiar with you "picking" the side with the 1 frog.
Monty opens the door and you get to see 1 of the frogs over there is male now it is your choice:
Do you stick or do you switch?
The list of possible combinations would actually be 4 times as large, and the remaining possible options would be:
Definitely Male Maybe Female
Definitely Male Maybe Male
Maybe Male Definitely Male
Maybe Female Definitely Male
So still a 50% chance of survival with the two frogs.
Thank you. I thought the same, but your explanation was much more clear than the one I came up with.
The problem is you counted the croak twice, but you only need to count it once. Your list is simply the wrong list, because you counted MM twice, when it is only counted once. Counting MM twice is like saying that the chances of flipping 2 coins and getting 2 heads (HH) are the same as getting one tails and one heads (HT, TH). The only way that could be vaild is if HH and HH are different, which is absurd. But HT and TH are of course different. If you don’t belive me, just try it. Flip 2 coins 100 times and disregard TT, and write everything down and see how many give you at least one T and put that number over 100-# of double tails. You will probably get a number between .55-.87.
@@annadoesroblox6205 This is true without taking into account that you KNOW in this case that one of the frogs is male. This means that in your example, you have to know that one of the coins is heads. meaning the results could be any of these:
Right coin: definitely heads left coin: maybe heads
Right coin definitely heads left coin: maybe tails
Right coin: maybe heads left coin: definitely heads
Right coin maybe tails left coin: definitely heads
Honestly it doesn't even need to be this complicated though. All you have to understand is that one of the coins being heads doesn't have an effect on the other's result. If you're looking for at least one tails, and you know one of the coins is heads, you can simply throw that one out and only worry about the result of the remaining one. You don't have to know which is heads to do that either, you just have to know that one of them is, and the other is still unknown. Only the results of that unknown coin matters.
Babrukus - except that actual computations of probability and actual experimental results both show that it isn't 50%, but rather 2/3.
The problem that you and _many_ other is having is that you think the following two scenarios are the same:
1. You know the left-most frog is male. What is the probability that the right-most frog is female?
2. You know at least one frog is male. What is the probability that one of the frogs is female?
These are different scenarios that result in different probabilities. The frog riddle deals with scenario 2, but you are saying that this is the same as scenario 1 so the correct answer is the answer to scenario 1.
This is an example of Martin Gardner's "Boy or Girl Paradox". These two scenarios are different. The computations show this. And experimental results also show this.
@@MuffinsAPlenty the problem with the boy or girl paradox is that it has 2 answers based on assumptions. If you take every pair of siblings with at least one boy, you'll find only 33% are both boys. However, if you ask every boy with one sibling if they have a brother, 50% will say yes. Similarly, if you select a random pair of siblings, then randomly identify the gender of one sibling, there is a 50% chance that the other sibling is the same gender. This is the scenario that applies to the frog riddle, as one of them is randomly identified as male, rather than searching for a pair with at least one male.
This is actually totally wrong. The tree stump is all right, but in the clearing, there is a 100% chance that one frog is male, and 1/2 x 2/2 is 50%. You have an equal probability going either way.
You don't know which one is male
Actually i think the stump is better because both frogs could be male in the clearing...
+cibrlx01 yeah and the one on the stump could be a male.
+Alex Wong you may not know which is the male, but you do now that one is a male so you have to account for the male.
Wouldn’t it be 50/50 either way because hearing the one frog croak eliminates it from the problem. Also female-male is the exact same as male-female
you don't konw which one of them it's the trick
@@sajeucettefoistunevaspasme no it`s not
@@lindwurm5976 ok then just make math
@@sajeucettefoistunevaspasme
I wrote a long comment with the correct math.
I can't solve any of these, so I just watch, cause it's fun...
EDIT :I actually got this one!!
Cookie? Same! The only one!
Rotten Apple Gaming worst part is, if you got the answer in the video, you're wrong....
Cookie? Me too but I didnt solve this.
me too
OneGuyTheGoat nice job!
"67% chance of getting a female"... wish that number also applied to human beings
But in the "sample space", wouldn't the MF and FM be the same? So wouldnt it be a 1 of 2 chance because it could either be MM, or MF/FM?
Charles Li yeh i think its wrong look at it this way we know one is male so actually we are only realy talking about the second one and that has a 1 in 2 of bein female so this sample space here in the vid is WRONG
Yes! My friend and I thought the same thing and were so upset that Ted dropped the ball. I normally love these riddles, and now my holidays are ruined. 😭
lets say the frogs come in different colours then, but it gives no indication of sex. The two frogs are yellow and red, the solo frog is blue... so the two could be... yellow male, red female... it could be yellow female, red male, or it could be red male, yellow male... all those are still possible after the croak right? So how can you combine 2?
nor do i
By making them yellow and red I was trying to set it up so that you didn't conflate them in your mind.
It's disappointing that, 7 years after a widely circulated refutation of this video was made (ua-cam.com/video/go3xtDdsNQM/v-deo.html), the makers have still not posted a correction or taken down the video.
For anyone who doesn't want to follow the link, the makers of this video chose a poor example, leading to wrong answer. The general principles regarding conditional probability were sound, but they were incorrectly applied here.
The fact that any number of refutation video exist doesnt mean this video is incorrect.
@@theeraphatsunthornwit6266 you're right, it the math in those videos, and the documentation on the way that this paradox can fail that means this video is incorrect.
Anyone else wondering why the guy was dumb to eat a poisonous mushroom yet smart enough to figure all that out
Perfectionist vs Pin
Lol exactly
Perfectionist vs Pin ya
solved the riddle...
that's enough work for today,
i'm going to bed.