Can you solve the frog riddle? - Derek Abbott
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- Опубліковано 28 лют 2016
- View full lesson: ed.ted.com/lessons/can-you-sol...
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.
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
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.
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*
“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
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
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 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...
@@kylee6051 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.
"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...
*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
Step 1:confirm you have green eyes
Step 2: ask the poison to leave
TRUIEEEEEEEEEEEEEEEEEEEEEEEEEUEUEUEUEUEUEEJLUFSAGDFIYLSGDI:
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.
But he said every frog is individual at the start which makes you question are they really a mating partners? Then this also fall under 50% chances
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???
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....
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...
Enlighten me: He licks both frogs on the left and 1 is male, meaning the last is 50/50 of being male/female = 1 frog is uncertain and 1 is 50/50 so left is 50% since he licks both
Then the right side is also a 50/50
He argues that the combination of female + male is different from male + female, but it doesn't matter when both are licked anyways, so it is the same thing
Therefore both sides have a 50% chance of survival or am I missing something?
No you're fine. The vid itself is unintentionally wrong, since they argued that MF and FM are different situations when they lead to the same result
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 differently (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 this is not even remotely akin to a monty hall problem. It's a variation of the boy-girl paradox.
@@MrCruztm do you have anything to do with your life than ctrl c ctrl v the same wrong answer?
@@MrCruztm Flip a coin and do not look at it. 50% chance of tails, right? Take another coin, and put it heads up next to the first one. But make sure, that you do not know, which coin is on which side. Following the logic of this riddle, you have a 66% chance of getting tails up at least once now. Because the possibilities are HT, TH, HH...
You see, it has nothing to do with the monty hall paradox, because there is no moderator choosing a door to open for you
I just guessed that a female would be hanging around the male 😂😂
Same.
Same
@@heberthr.6978 ?
@@heberthr.6978 I think they assumed this because male frogs don’t have much reason to hang around each other. (Or maybe they do, and I just don’t know it.)
@@heberthr.6978 When male frogs croak it is either a mating call or a threat usually, so, the female could be answering the call. Meaning this assumption is right when it comes to frogs.
"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
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?
1:06
Imagine he does that and realizes he violated two frogs for nothing.
BRUH LOL 😂😂😂😂😂😂😂
Remember kids: no matter how high are the odds, they are still against you
@@InkyWinkDink also applies the other way
Yeah that's true.
wish my girl did that for me
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 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
I've played enough fire emblem to know a 67% chance ain't goin to save anyone's ass
XD true
SOOOO RELATABLE
R.I.P. Cherche
except with criticals
better than 50/ chances
how i felt trying to get Nah... nah...
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.
Ted Ed puzzles in a nutshell :
All disasters will stop themselves to give you time to think
Yup
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
lol I we just going to say the male frogs were trying to croak at the female frog on the stump
That's exactly what I thought!!
same XD
Yea I thought it was some type of mating call, and the two were more or less fighting for the female on the stump
lol..same..we're genius though..hahaha
Hanna Inoue
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.
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
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
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
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.
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.
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.
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.
This is actually completely wrong. It's still 50% because the probability is the other frog is female given one is male. They're independent events so P(B|A) = P(B).
Yes, this is correct. Another example: looking at all couples with two children, 75 % of them have at least one boy. But if you already have a girl and are now expecting again, you don’t have a 75 % chamce of getting a boy, but 50 %. The probability of having a boy resets for every pregnancy so it’s always 50 %. In the same way, the chance of a new-found frog being female is always 50 %.
You would be right if you supposed that he would lick one frog . But they supposed that he would lick the two frogs he gets on the ground.
@@user-mo8oh7kp6h I disagree with this answer. I think the probability of the left has 1F is approaching 50% while the population of the frog increasing. Here is my thought, we assume that there are 6 frogs in total and 3M 3F. We all know that 1M shown at the left and 1 unknown at the right. In this case, the probability of the left has 1F is 3 out of 4 (when the unknown is M), or 2 out of 4 (when the unknown is F). Therefore, the total probability of the left has 1F is (3/4+2/4)/2 which is 5/8. However, if we assume that there are 100 frogs, the total probability of the left has 1F is (50/98+49/98)/2 which is 99/196. As you can see, the probability of the left has 1F is approaching 50% while the population of the frog increasing. As the evidence has shown, we pick the left side since the probabilities of the left has 1F are always higher than right.
@@yuhaocupid3746 I think the riddle presupposes that there are enough frogs on the island that one frog being male won't throw the odds excessively, but even if that's true, you're incorrect. If the odds that the second frog on the left are 5/8, then the same is true for the frog on the right. You just run the numbers in the opposite direction.
@@Sir0chicken I believe that in the condition male and female have equal number doesn’t mean that the odd of a frog being female is 1/1. And using examples is the best way to fits the conditions.
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.
(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!
Plot twist: all three frogs jumps away and you die no matter what
Omg lol!
jump*
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
@@shadowbanningcommentssucks3551 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
@@shadowbanningcommentssucks3551 Search the Monty Hall game or look at Bayes Theorem; you'll see the answer both theoretically and empirically is 2/3!
Wait if you didn’t know if the mushroom was poisonous then how would you know your poisoned in the first place?
Or maybe don't be an idiot to eat a wild mushroom.
That doesnt make any sense
it does
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 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.
This guy doesn't have anough time to walk to both frogs but he has anought time to think this hole mathematical equation through... that's straight out of anime.
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
Or you can just not eat a random mushroom?
haha true
Ya simple
+Nicole l-/
|-/
Lel
exactly
I think the issue in this is that we dont care about the possible permutations, we just want a combination of frogs that contains a female. So MF and FM aren’t materially different possibilities which cuts down the sample space to 2 possibilities.
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.
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?
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
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.
Me: *doesn’t know that a mushroom is poisonous*
Also me: *knows the antidote*
HOW
copied
No time to question! Quick, lick some frogs! :D
I a
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.
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
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.
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
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!
“You feel your consciousness beginning to fade”
Yet you still solve this riddle?
Dzjt
It's really fast to solve in the moment... Obviously it'll be slower in an educational video where the narrator has to explain the math to a potentially mathematically inept audience.
Yeah i agree with a lot of the comments. The croaking does not need to be accounted for in the probability grid. You have one mystery frog in the clearing and one on the field, so 50/50 either choice.
false. You can map each possible case as shown by the video. It's objectively a higher chance. Some argued that it isn't a higher chance because two of the cases are functionally identical, but they missed the whole point which was to show that it had two occurrences in 3 possible cases showing that 2/3 times by choosing the two frogs you will get a female
The korean subtitle has a wrong info in 0:24, it says that the male frog produces a matter that make things worse
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
Yeah, but who the hell wants to lick a frog.
Lol
+Ernesto Silva are you saying you want to lick a frog?
Yes, if my life really depended on it.
So you would rather die than lick something disgusting and wash the slime off later? It's not permanent, death is permanent. It's your choice.
Errrr...... If that happens after I'm cured, then I'll rather just die lol.
if he didn't thinked about all the mathematic stuff and licked the two frogs, and the one on the tree he would be probably still alive
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.
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.
Wait... So you know the frog is the antidote to a mushroom but you don't know that the mushroom was poisonous?!?
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.
You can only go 1 direction..
*proceeds to think about the logical idea for 3 minutes*
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...
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
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
So if there are 4 frogs, and you hear 3 croaks (provided that each male can croak only once), that means the 4th frog sitting next to them has an 80% chance of being a female (4 out of 5 outcomes: *FMMM, MFMM, MMFM, MMMF,* MMMM)? But then if you switch the silent frog with the one sitting on the tree stump, then it's 50% again? How is the silent frog sitting in the clearing next to the croaking frogs different from the one sitting on the stump? I don't think the logic in this video is right.
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
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.
The math only works if each possibility in the sample space is weighted equally, which is why the detail about the equal amounts of male and female was included.
I don’t see the compulsory sacred meme of these riddle videos so I’ll go ahead and add it
Step 1: confirm the frogs have green eyes
Step 2: ask the frogs to leave
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
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.
@@kylee6051 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 will admit that the order of operations in this one makes it a little less sound logically. He heard a croak then looked, but since you already know there's a male frog, the 2nd frog gets added in separately with it's own separate probability in my opinion. You don't start with a set of 2 the way they're discovered, you start with a single male frog, then discover there's another with it.
sorry i know this comment is old but the trick is that you don't know which frog is male. if you knew which one was male and then another frog joined it, sure that new frog would be 50%. but because you don't know which of the two is male, there are three equal chances: frog 1 is male, 2 is female; frog 1 is female, 2 is male; or both are male, hence the 2/3 chance
@@pheebthedweeb5652 but since you get to lick *both* frogs M-F and F-M aren't distinct combinations. The question is "does the pair contain a female?" and there's only two unique combinations, M-F and M-M. Since the odds of the non-croaking frog (even if you don't know which one it is) are independent of the croaking frog, those two combinations are still 50/50.
If I flip two coins behind a screen and tell you "At least one of them is not heads, what are the odds one of them is heads?" you'd just use the 50/50 odds the other coin had.
@@ItsDan123 This puzzle is flawed for a different reason. Which is that the fact that one of the frogs on the left croaked, makes it significant that the frog on the right didn't. In your coin scenario, it would depend on whether you would tell me that there was at least one tails, provided it was true, regardless of what you had flipped. If that were so, then the odds of at least one heads would be 2/3.
I like how you teach something nice with riddles
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
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
Him: not having enough time to go one direction
Also him: having enough time to calculate all of this
There are actually 2 important hidden hypothesis behind this solution:
1/ that the probability of hearing one male frog within a group of 2 is the same whether there is one or two male frogs in this group
2/ that you couldn't be hearing 2 male frogs at the same time if they are 2 males (as they say you've heard just one, either that they cannot make noise at the same time, or that you wouldn't be able to tell).
If I told you: "'there is a probability p for each male frog to make a noise in the time interval you had", and that you only heard one frog of the 2, then it would mean that for a group male/female, the probability to happen was p, and for a group male/male, to have had only one it is 2p(1-p), so that overall the chance of the other one being a female is: 2p / (2p + 2p(1-p)) which simplifying by 2p is 1/(2-p). If you can't tell whether is was ONE frog who made a noise or the 2, then the probability if it's a male/male of hearing at least one frog is 1-(1-p)*(1-p) = 2p-p2 and the chance of the other one being a female is thus 2p / (2p + 2p - p2) which simplifies into 2/(4-p). So we can see that 2/3 is a specific case where actually you will always hear ONE AND JUST ONE male frog, if there is at least one within the group of 2... If you change the scenario in terms of whether you could hear 2 at the same time, or that a male would not have always made a noise, the probability is no longer 2/3 (but can be as low as 50%, and as high as 100%).
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.
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.
Plot twist: this a made up story of what happened to the neighbor’s kid so you don’t have to eat your mom’s mushroom soup.
Can't believe they didn't issue a correction
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'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).
i feel like you could also look at this problem as, 'you know that there is one male and one unidentified frog in one direction, and one unidentified frog in the other. eliminating the known male frog from the equation because it does nothing, you now once more have a 50/50 chance in either direction.
like you do have to jump through some hoops to get to a 67% probability..
Think of this as flipping 2 coins, and knowing one of them is heads. 67% chance of a tails because HT and TH are different but HH and HH are the same.
@@Owen_loves_Butters So, by the logic of this riddle, he also could simply catch one frog which he knows to be male. Then he could mix this frog with the frog on the stem (which has a 50% chance to be female), until he dows not know which frog is which any more... and BAM- his chance of having at least one female frog increases from 1/2 to 2/3!
... No, thats... flawed
@@trissebude2184 No, having 1 heads coin and mixing it with an unknown coin isn’t the same as knowing at least one coin of 2 is heads. It’s like the difference between the probability of blindly throwing a dart out of a plane and having it hit a target vs. throwing the dart then painting the target around it. Same outcome, different probabilities.
@@Owen_loves_Butters The probabilities are exactly the same, because the variables are all exactly the same. Same input, same results. Reality does not care, how the two frogs came together. What counts is, there is atl east one male frog and you do not know about the other one. As easy as that
@@trissebude2184 Different inputs.
Thanks for a funny and great video 🙏🏻
"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 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.
I have a question:
If you know 1 is male for certain, why does it matter which one is male?
I think in this context you have 2 frogs, one of them is male for sure. Isn’t it a 50% chance for the other one to be female? I understand the math behind 2/3 chance overall, but in this case isn’t female/male vs male/female the same outcome?
Think permutation vs combinations.
Any thoughts?
Yeah you're right, Ted Ed just messed up their logic in this riddle
He doesn't know which one is for certain
MF and FM are the same outcome, so combined they are twice as likely as MM
it doesn't matter which one is male, they are trying to find the probability of any one of the two being FEMALE
It is the likelyhood that we are interested in.
If you toss 2 coin 100 times. About how many times you get one head and one tail. ...??
From sample space you get 25hh 25 tt 25 ht and 25 th
Did you cancel ht and th out in this case? Then you would get 25hh 25tt and 25 th????
Try tossing two coin 100 times by yourself.
Gosh ... the correct subcomment has zero like before i liked.... the wrong one has 5
2:14 ted-ed doing math while i'm just like, "well, they're probably mates since there's two of them
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?
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.
This question resembles the Monty Hall problem
I went to the clearing since there was only one croak