Why would you assume a silent frog is equally likely to be male as female? If a male sometimes croaks and the female never croaks, and male frogs and female frogs are evenly distributed, then a non-croaking frog is more likely to be female.
Possibilities of a random 2 two frogs — assuming a male is just as likely to croak as not:
1. Male (L), Female ® --25%
Which is either:
Silent Male (L), Female® --12.5%
or
Croaking Male (L), Female® – 12.5%
**2. Female (L), Male ® --25% **
Which is either:
Female (L), Silent Male ® --12.5%
or
Female (L), Croaking Male ® --12.5%
**3. Female (L), Female ® --25% **
**4. Male (L), Male ® --25% **
Which is one of these four:
Silent Male (L), Silent Male ® --6.25%
Croaking Male (L), Croaking Male ® --6.25%
Silent Male (L), Croaking Male ® --6.25%
Croaking Male (L), Silent Male ® – 6.25%
Now you hear a single croaking frog and turn to see two frogs. The chance is two out of three that one of these frogs is female.
However, it should be noted on this basis that the single non- croaking frog on the rock is therefore 75% likely to be female. So you should really lick that one instead!
Er… scratch that. The single frog is also 2 out of 3.
So now it seems to me it doesn’t matter which you do. The odds are 67% in your favor either way you lick.
After thinking more about this, I have a new answer depending on the following conditions.
1.) A male frog will croak at most once in the time allowed.
2.) All male frogs paired with another frog have an equal chance of croaking, and that chance is not affected by the sex of the other frog.
3.) You cannot tell if the croaking you hear from the pair on the rock is from one or both frogs.
4.) I am assuming a frog by itself will not croak cuz it hurts my brain to think about it otherwise.
If the odds of a male frog croaking in the allowed time interval are 1/n*, *where n is greater than or equal to 1, then if you look at all possible combinations, you will survive 2n times, die 2n-1 times, so the odds of surviving by licking the pair are
2n/(4n-1).
So, in the case where a paired male always croaks, then n=1, and the odds of survival are 2/3 :
FemaleFemale (Discarded because you never hear anything)
FemaleMale - Survive
MaleFemale - Survive
MaleMale - Die
In the case where the odds of a paired male croaking are 1/2, then the odds of surviving are 4/7:
M=croaking male
m=silent male
F=female that would have croaked had it been male
f=female that would have been silent had it been male
**ff - discarded
fF - discarded
Ff - discarded
FF - discarded
fm - discarded
fM - survive
Fm - discarded
FM - survive
mf - discarded
mF - discarded
Mf - survive
MF - survive
mm - discarded
mM - die
Mm - die
MM - die**
Survive 4 times, die 3 times, odds of surviving are 4/7.
In the case of the odds of a paired male croaking are 1/10, in your head you can work out that, in 400 possible combinations
you will survive 20 times
you will die 19 times
you will discard the remaining 361 cases where you hear nothing
Odds of surviving drop to 20/39; you are starting to approach 50:50.
If the odds are extremely high against a paired frog croaking, then
as n approaches infinity, limit{2n/(4n-1)} = 1/2
So, worst case, it is 50:50 if you lick the pair, but could be 2 outta 3.
Because those are the basic premises of the puzzle.
The video puzzle is obviously trying, as wolfman said, to simply restate/redo the classic boy/girl puzzle. However, the problem is not so simple as given because we don’t know how often male frogs croak.
If you know, because I tell you, that at least one of two frogs is male, than yes the odds are two of three that the other is female. But here you know because of something a male frog might do. That changes things.
I’m already lost in all the logic trees, but I’d just like to point out that this is irrelevant for the purposes of the puzzle, because a non-croaking frog exists in both directions.
Right, but a male frog *could have croaked *. Some of the male frogs you meet will croak. Some will not. None of the female frogs will croak. You will over time meet the same number of male frogs as female. Therefore, any particular non-croaking frog you meet has a higher chance of being female.
Clearly, without any knowledge of the gender of the frogs, choosing the two frogs wins 75% of the time.
Leaving aside the issue of croaking for the time being, if we somehow magically know that at least one of the pair of frogs is male (like in the Boy/Girl Paradox) then this chance falls to ~67%, as in the official answer.
The frequency of croaking is a problem. For someone to have noted that only male frogs croak, they must croak sometimes. But the OP says there is no pattern to the croaking. I think the only sensible way to interpret this is to assume that non-croaking does not imply anything about the gender of a frog. Not strictly true, as Biotop says, but if male frogs croaked often, that would constitute a pattern. So perhaps we should say that they croak so infrequently, once every ten years perhaps, that it does not materially affect the probabilities here. That is, the croak is merely this puzzle’s way of letting you know that the pair of frogs are not both female. Under that interpretation, the official answer stands.
This is where I am still struggling. Say a male frog only croaks once every ten years. You hear a croak and see two frogs. I would think that the likelihood that they are both male increases because* for you to have ever heard a croak at all is more likely with two male frogs around then one.* Yet only one frog actually croaked.
OK, I’m going to restate the original boy/girl puzzle, to clarify the issue. You meet a random person, and ask that person whether e has exactly two children. The person says yes. You then ask that person whether it is the case that e has at least one boy. The person says yes. Given the answers to those two questions, what is the probability that the person has two boys? The answer is 1 in 3.
Please note, here, the way that I phrased this. I did not ask “what is the probability that the other child is a boy”, because the phrase “the other child” is meaningless here. You can’t say “the one other than the one the parent was thinking of”, because if the parent had two boys, e was thinking of both of them when answering the question. And this distinction is crucial, because if the puzzle is ever rephrased in such a way that “the other child” becomes meaningful, then the odds change back to the intuitive 1 in 2. For instance, if I ask the parent “Think of one of your children. Is that child a boy?”, then I can refer to “the other child”, and the other child is equally likely to be a boy or a girl.
And the way the frog puzzle is phrased, there is indeed meaning to “the other frog”: It’s the one that didn’t croak. You can’t tell which one that is by looking at it, but by the same token, you can’t tell which child the parent was thinking of. It doesn’t matter: “the other frog” is well-defined, and so the odds are 50%.
This is true. We don’t know how much more likely, since we don’t know how often male frogs croak, but a non-croaking frog has some chance greater than 50% of being female. But this doesn’t change the puzzle: Each direction has exactly one non-croaking frog, and so whatever the chance is of a non-croaking frog being female, it’s the same in both directions.
Yes, playing around with Bayes’ theorem (and hoping that I haven’t messed it up), it seems that if you assume any particular probability of a random male frog croaking in a particular period, the period in this case being the time our protagonist has to decide, then it doesn’t in fact make any difference whether you choose the lone frog or the pair of frogs. The chance of getting a female is the same.
If we select two frogs randomly and observe them over some period, then
p(both frogs are male, given that exactly one of the frogs croaks during the period)
= p(exactly one of two randomly chosen frogs croaks during the period, given that both frogs are male) x p(both frogs are male)
divided by
p(exactly one of two randomly chosen frogs croaks during the period)
And
p(a frog is female, given that it didn’t croak during the period)
= p(randomly chosen frog didn’t croak during the period, given that it is female) x p(a randomly chosen frog is female)
divided by
p(randomly chosen frog didn’t croak during the period)
= 1 x 1/2 / p(randomly chosen frog didn’t croak during the period)
= 1 / (2 x p(randomly chosen frog didn’t croak during the period))
If we assume that a male frog has a 0.5 probability of croaking during the period (so a random frog has a 0.25 probability, because it could be female), then
p(both male | exactly one of the frogs croaks during the period) =
(2 x 0.5 x (1 - 0.5)) x 0.25 / (2 x 0.25 x (1 - 0.25)) = 0.333… . So the chance of getting a female frog in the pair is ~67%.
But the chance of the lone, non-croaking frog being female is
1 / 2(1 - 0.25) ~= 67%. It’s the same. That is, there’s no reason to prefer the pair of frogs to the lone frog.
If male frogs croak a lot, say p = 0.9, then naturally we suspect that a non-croaker is female, and the calculation bears that out:
p = (2 x 0.9 x (1 - 0.9)) x 0.25 / (2 x 0.45 x (1 - 0.45)) = 0.0909…
That is, they’re much less likely to both be male. So about a 91% chance of getting a female frog. But the chance of the lone frog being female is
1 / 2(1 - 0.45) = ~91%
Again, it’s the same. The answers are always the same, whatever probability to you choose for a male frog croaking. So there is no reason to prefer the pair of frogs to the lone frog.
Now, if I’d seen Chronos’s post in time, I wouldn’t have had to do all that calculation 
I think I get it now. If the riddler had said that it is known that two female frogs never travel together in pairs-- but they otherwise comingle, then the odds would be better to lick the two.
But if we actually have a frog proclaiming, “I am male,” then the non-speaking frog’s femininity odds are 50%.
Wait a minute…
Grr…
Suppose the person first looked to the left and saw two frogs which could be either sex, but then looked to the right and saw one frog. While he was looking to the right, he heard one of the two frogs on the left croak.
Yes you have exactly one non-speaking frog in each direction. But you do not know which is the non-speaking frog on the left. All you can eliminate is the f/f possibility.
Ok. I think I finally see now. While we don’t know which frog croaked, we do know that a specific frog did croak. We therefore we can also eliminate the chances where a frog that was not the specific frog croaked.
This puzzle makes me want to croak.
Sigh. Can someone smarter than me tell me where I am mistaken on post #22? Because I think I understand and then lose it again.
DING! Light bulb goes off!
Post#22 is correct (ignoring the 75% stupidity at the end), but just saying the same thing… that it doesn’t matter which way you go.:smack: