Can you solve the frog riddle? – Derek Abbott



View full lesson: http://ed.ted.com/lessons/can-you-solve-the-frog-riddle-derek-abbott

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.

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Fahad Hameed

Fahad Hashmi is one of the known Software Engineer and blogger likes to blog about design resources. He is passionate about collecting the awe-inspiring design tools, to help designers.He blogs only for Designers & Photographers.

42 thoughts on “Can you solve the frog riddle? – Derek Abbott

  • September 17, 2017 at 1:31 am
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    Has higher probability but doesn't eliminate the fact that it could be both male.

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  • September 17, 2017 at 1:31 am
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    But doesn't the fact that one is male eliminate it from the situation?
    And both combinations are the same they are just in a different order.
    I think it would still just be a fifty percent chance.

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  • September 17, 2017 at 1:31 am
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    The problem is that a Female / Male combination and a Male / Female combination is the exact same thing in this case. It doesn't matter which order they're in, therefore it's a 50% chance for the frog to be female on EITHER the stump or the clearing.

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  • September 17, 2017 at 1:31 am
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    The riddle restriction that you do not know which frog has crocked is unnecessary and misleading.

    Even if you had seen clearly that it was the left frog that crocked. The chances of the right frog to be a female would be 67%!.
    How can it be? …. Because at a MM situation the chances that the left frog will crock are only ½ of the chance (100%) it will crock at a MF situation.
    So if the left frog has crock. It is twice as probable that its partner is a female.

    To show it clearly, lest make a additional arbitrary rule:
    When 2 males are sitting together, the older frog will crock and the young will stay silent.
    Now we have 6 possible scenarios.
    M-old-crock / F-young
    M-young-crock / F-old
    F-young / M-old-Crock
    F-old / M-young-crock
    M-old-Crock / M-young
    M-young / M-old-Crock

    If the left frog had crocked then it have equal probability to be:
    M-old-crock / F-young
    M-young-crock / F-old
    M-old-Crock / M-young

    And 2/3rd of the possibilities involve the 2nd frog to be a female.

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  • September 17, 2017 at 1:31 am
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    the three frogs there arentthe onlyones alive so the math here does notmatter at all

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  • September 17, 2017 at 1:31 am
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    I said the opening because frogs aren't pack animals so if they were together they would be mating but then I realised the actual answer because we are doing stuff like this in maths

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  • September 17, 2017 at 1:31 am
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    Jesus christ if u have time to think about where u go you have time to go in both directions xD

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  • September 17, 2017 at 1:31 am
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    For those Calling this wrong check the monty hall problem…..this problem is identical to it

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  • September 17, 2017 at 1:31 am
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    So if I have 2 frogs and one croaks the other one is a 50% chance of being female. But if a blind person has the same frogs, it's a 67% chance?

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  • September 17, 2017 at 1:31 am
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    if i don't have time to go in both directions what makes you think i have the time to sit there strategically figuring it out… lmao

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  • September 17, 2017 at 1:31 am
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    Yeah so it's basically like the Monty Hall problem, right?

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  • September 17, 2017 at 1:31 am
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    Bottom line: you'd die. There is no way you'd be able to catch the frog/s regardless of which direction you took.

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  • September 17, 2017 at 1:31 am
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    As mentionned by others, the odds are actually nearly 50%.
    Short explanation : A duo of male croak artists nearly doubles the odds that one of them actually croaks, so this possibility is nearly as likely as both croaking-male-with-female-groupie possibilities combined.
    (The "nearly" comes from the unlikely possibility of a simultaneous double croak.)

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  • September 17, 2017 at 1:31 am
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    even if i have a 67% chance, i would probably die because that's how bad my luck is.

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  • September 17, 2017 at 1:31 am
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    i picked the 2 frogs because i thought that a male and female would have more reason to be by each other than 2 males

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  • September 17, 2017 at 1:31 am
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    If he's croaking he's looking for a mate so the one next to him can't be a female

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  • September 17, 2017 at 1:31 am
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    in my Opinion the correct answer is 58% because,
    first you have to look at all the possibilitys of how the genders of the frogs are distributed, and because you already know that atleast 1 frog is male the possibilities are:(all 3 male ; 2 male one female ;and 1 male and 2 female) because ther is a 50/50 distrubution of male anf female the oddes are (25;50;25) and in the case of 3 male and the case of 2 female the outcome is the same independent of your dessision ,just the case of 2 male and one female is important. In that case the odds of the female sitting on the clearing is 2/3 you get the propibility of survival by add the two possibility and you get (1/4)+(1/2)*(2/3)=7/12 ~58%
    Sorry for my english

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  • September 17, 2017 at 1:31 am
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    I think I figured out why this video is wrong. The odds of having a female frog in a pair when choosing a random pair of frogs in which one is male is 2/3. Because MF FM MM. But when you have one specific pair of frogs in which you know one is male it's 50/50 because if the first frog is male it's MF or MM, and if it was the second frog that's male it's FM or MM.

    The first case is like flipping 1000 pairs of coins and then looking at all the combinations that include a heads. Two thirds of them will contain a tails, but it's because you're ignoring the 25% of pairs that had both tails. The case prevented in the video is flipping one coin and getting a heads and then flipping another. The second coin is a regular 50/50 because the randomness comes after one coin is flipped, not before as in the first case. You don't know which coin was flipped heads, but it's entirely arbitrary.

    Anyone care to tell me why I'm wrong? I did well with math in school and don't have trouble with the Monty Hall Problem, but I'm no mathematician.

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  • September 17, 2017 at 1:31 am
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    i was just thinking that the one on tree stump is female and the two in the clearing were just croaking for mating season lol

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  • September 17, 2017 at 1:31 am
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    There is a wrong…. If we assumed that A is the event: the first frog is male, and B is the event: the second one is male. We are sure that one of them is male so let's choose A to be the male for example….Then, P(A)=1(A is of course a male ) and P(B)=0.5 P(A ∩ B)=P(A).P(B)=1*0.5=0.5 (although it is a conditional probability but both of the events are independent and we do know that one of the frogs is surely male so it is absolutely wrong to give its probability the value (0.5) ….The right value is (1)…Thank you.

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  • September 17, 2017 at 1:31 am
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    I'm already be dead cuz I need to compute such things over the time

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  • September 17, 2017 at 1:31 am
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    Anyone else get it right? I thought it would be a 2/3 chance.

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  • September 17, 2017 at 1:31 am
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    I said go to the tree stump because I thought the frogs were croaking at the one on the tree stump

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  • September 17, 2017 at 1:31 am
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    Or you could go to the female that didn't croak….or not eat random mushrooms

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  • September 17, 2017 at 1:31 am
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    I thinked about this problem, and i'm pretty sure this answer is biased and wrong, and that both sides have an equal probability of success.
    The solution of the video states that since two of the 3 possible combinations leads to a success, but the sheer number of possibilities can't always tell the true odds, an immediate example are the monopoly rolls, using two dices, you can roll any number between 2 and 12, so there are 11 possible outcomes, but not every number has an equal chance to be rolled, so a 2 doesn't have the same odds as a 7, the same occours in this problem, we have 3 possible solution, but not all the solution have the same odds, this because even if we don't know what is the male frog, we know that one of them is 100% male and the sequential probability formula is P1+(1-P1)*P2 where Px is a number between 0 and 1, so let's assume the first (P1) frog is the male one we have:
    0 + (1 – 0)*0.5 = 0.5 so a 50% chance of success, if the second frog is male we have:
    0.5 +(1 – 0.5)*0 = 0.5 so again a 50% chance, so we can assume that:
    if the first frog is the frog who croaked, the second has a 50% chance to beign female
    if the second frog is the one who croacked, the first will have a 50% chance to beign female, so we have the following possibilities:

    Male Female
    Male Male
    Female Male
    Male Male

    so since male male has a 2/4 chance we have the following odds:
    Male Male 50% (2/4)
    Male Female 25% (1/4)
    Female Male 25% (1/4)

    So even if there are 3 possibilities, those possibilities have different odds, like the monopoly dice, so to prove this with the dices, let's assume that we have two dices, one of them has three 1 and three 2, while the other has six 1, what are the odds of rolling a 3?
    the possibilities are:
    1 + 1
    1 + 2
    2 + 1
    but the odds aren't 2/3 because the true possibilities are:
    1+1
    1+1
    2+1
    1+2
    this because 2+1 and 1+2 are basically the same result, since is just a permutation of the position of the dices, so basically 1+1 is the permutation of 1+1, so we still have a 50% chance of rolling a 3 since one of the dices will always roll 1, and the position of the dice doesn't matter.

    Reply

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