Question about probability riddle: doesn't make sense

[stinkynuts]

I came across a video on YouTube: https://www.youtube.com/watch?v=cpwSGsb-rTs

The riddle is essentially this:

You are in the forest and about to die from a disease. The only thing that can save you is if you lick the female species of a frog. Unfortunately, the only way to tell males and females apart is that males croak.

You see a frog on side A, and on side B there are two frogs. You only have time to reach one side. Suddenly, one of the two frogs on side A croaks, so you know it is male, and not going to save you.

You are feeling very sick. Hurry! Which side would you go to??


My logic is very simple. It makes no difference. On side B, the chances are 50/50 that the sole frog is male/female. On side A, one is male for sure. The other frog is 50/50 male or female.

So it should make absolutely no difference. But they came up with a very weird explanation in the video.

I'm not sure the video is right, and would like to know if there are any math experts who can clarify.

Why aren't the odds the same?? It should make no difference that one of the frogs is male. The other still has a 50/50 chance of being male or female. Just like the sole frog on the other side.

 

[pussyluver]

I watched the video and the explanation makes sense to me. Don't think I could explain it any better than they did.

 

[stinkynuts]

I watched the video and the explanation makes sense to me. Don't think I could explain it any better than they did.

I know, the explanation seems to make sense, but it is very counter intuitive, and I just don't think it's right....

 

[fuji]

I know, the explanation seems to make sense, but it is very counter intuitive, and I just don't think it's right....

Human brains aren't wired to think about probability properly, so it is counter intuitive.

I took a lot of courses in statistics in my university days and the only reason I got the answer right is that after coming up with a wrong answer I thought "no one would make a YouTube video about the obvious answer" so I went back and enumerated the possibilities and counted them and realized the "trick".

But if this question came up in normal life and I didn't know the answer was meant to be surprising my instinct was the odds were 50/50. I'd get it wrong if I wasn't alerted to work it through methodically.

 

[benstt]

The video part where they enumerate the sample space is the best way to understand it. The frog croak tells you a bit more information about the side with two frogs, and correctly using that information can be tricky.

The small interesting detail was he said you could lick both frogs on the two-frog side. That might be a flaw in his problem structure, but he may have meant to say that as part of a larger problem.

What I mean is, if you had not heard the croak, which side would you pick? You are licking only one frog on the one side, and two on the other, and will survive if one of the frogs you lick is female. The more frogs the better. One side gives you a 50% survival chance, while the other is 75%.

Details matter in probability.

 

[Polaris]

If this is the problem, that there are 2 frogs on side A, and 2 frogs on side B, and you only have time to go to one side, and that one frog croaked on side A.

Therefore, there is only 1 frog on side A, and two frogs on side B. What are your chances?

The answer is obvious ... if you're a baseball fan.

Let's say it is a tie game, bottom of the 9th or extra innings, home team at bat, lead off hitter gets on base. Man on first nobody out.

Manager will sacrifice the running into scoring position, giving them two chances to bring him home.

If there was one out and runner on first, there usually will not be a sacrifice because going to sacrifice an out for only one shot at it. However, have seen this happen in National League games that went deep into extra innings where there were no one left on the bench to bat.

One time, extra inning a lot of them, no one left on the bench, SF manager Bochy behind 1 run and two outs no one base, sent pitcher Madison Bumgartner out to bat for their last chance, and they loss that game.

 

[TeeJay]

I'm not sure the video is right, and would like to know if there are any math experts who can clarify.

Why aren't the odds the same?? It should make no difference that one of the frogs is male. The other still has a 50/50 chance of being male or female. Just like the sole frog on the other side.

The video is wrong because they screwed up there own sample space
To calculate the correct odds, there are always 3 frogs available, not 2

If you only saw the 2 frogs in the clearing then the math is correct
Too bad they seem to just forget about the single frog on the tree stump because he does change the odds of finding the correct frog

(This also is ignoring the obvious fact that you hear the croak behind you while looking at the frog on the tree stump
If you do not ever view/confirm one of those frogs in the clearing was the actual one who croaked, his entire explanation of eliminating the 2x female is dead wrong)

 

[RemyMartin]

From the sample space, he should also eliminate one of the (male-female), since (male-Female) is same as (Female-male).

by doing that you only have two choices left , which is still 50%.

 

[RemyMartin]

The answer is obvious ... if you're a baseball fan.

I don't think you can use baseball as example because to sacrifice or not, depends on other factors as well, like who is the hitter and who is the runners on base.

 

[Occasionally]

I'm no mathematician, but I think the OP meant 2 frogs on either side, where 1 frog on side A is male (so exclude that).

To me, it makes no difference as any frog you lick is a 50/50 chance.

One side can have one mystery frog, while the other has a million frogs. Each disgusting lick you do has a 50/50 chance of success.

 

[fuji]

From the sample space, he should also eliminate one of the (male-female), since (male-Female) is same as (Female-male).

by doing that you only have two choices left , which is still 50%.

That's intuitive, but wrong. There are in fact twice as many ways to pick a M-F combination as a M-M combination.

 

[fuji]

I'm no mathematician, but I think the OP meant 2 frogs on either side, where 1 frog on side A is male (so exclude that).

To me, it makes no difference as any frog you lick is a 50/50 chance.

One side can have one mystery frog, while the other has a million frogs. Each disgusting lick you do has a 50/50 chance of success.

The bit your are missing is that on the two frog side you get to lick BOTH frogs.

Without the croak the two frog side had a 75% chance of containing at least one female. The croak lowered the chances to 67%.

 

[johnhenrygalt]

It's similar to the old grade school probability question: "A boy has one sibling. What is the chance he has a sister? It's 2/3. But if the question is: "A boy has one younger sibling. What is the chance he has a sister? Now the probability is 1/2.

 

[fuji]

There's a great Wikipedia article about how the exact wording of the question determines whether the answer is 2/3 or 1/2.

https://en.m.wikipedia.org/wiki/Boy_or_Girl_paradox#Bayesian_analysis

 

[stinkynuts]

That's intuitive, but wrong. There are in fact twice as many ways to pick a M-F combination as a M-M combination.

Hmm... I can't seem to wrap my head around it. MF and FM should be the same, since we know that if one frog is Male the other has to be female.

 

[fuji]

On reflection, the video screwed up by saying one of the frogs croaked.

I don't know which, but I'm going to name him Croakie. Croakie could be either the left frog, or the right frog. The possibilities are:

Croakie, male
Croakie, female
male, Croakie
female, Croakie

There are four cases here, and there is a female here in half of those cases.

The supposed paradox collapsed in this example because I was able to distinguish the frogs: one croaked! So they were not interchangeable.

But here's where it does work:

You flip two coins behind a screen. You either have HT, HH, TH, or TT. I ask you to give me a head, if there is one, and you do give me a coin.

I bet you remaining coin is tail and I win that bet 2/3rds of the time, because:

HH: you gave me either head, there is no tail, I Iose
HT: you gave me the left head, keeping the right tail
TH: you gave me the right head, keeping the left tail

Odds are actually 2/3 that your remaining coin is a tail.

The paradox works in this case because the heads are indistinguishable from one another, there are only three cases here, not four.

When you can name one of the frogs or sons by any means, there are four orders, not three. The extra information, that there is a difference between them, alters the probability.

 

[stinkynuts]

I believe this is correct, Fuji.


But in simpler terms, and more intuitively, what I said in the beginning:

There are two frogs. Croakie the male croaked. Forget about him. The other frog is either male or female, so 50/50.

On reflection, the video screwed up by saying one of the frogs croaked.

I don't know which, but I'm going to name him Croakie. Croakie could be either the left frog, or the right frog. The possibilities are:

Croakie, male
Croakie, female
male, Croakie
female, Croakie

There are four cases here, and there is a female here in half of those cases.

The supposed paradox collapsed in this example because I was able to distinguish the frogs: one croaked! So they were not interchangeable.

But here's where it does work:

You flip two coins behind a screen. You either have HT, HH, TH, or TT. I ask you to give me a head, if there is one, and you do give me a coin.

I bet you remaining coin is tail and I win that bet 2/3rds of the time, because:

HH: you gave me either head, there is no tail, I Iose
HT: you gave me the left head, keeping the right tail
TH: you gave me the right head, keeping the left tail

Odds are actually 2/3 that your remaining coin is a tail.

The paradox works in this case because the heads are indistinguishable from one another, there are only three cases here, not four.

When you can name one of the frogs or sons by any means, there are four orders, not three. The extra information, that there is a difference between them, alters the probability.

 

[benstt]

On reflection, the video screwed up by saying one of the frogs croaked.

When you can name one of the frogs or sons by any means, there are four orders, not three. The extra information, that there is a difference between them, alters the probability.

I think they were clear that exactly which frog in the clearing croaked was not known. Their sample space was good in that scenario.

 

[benstt]

Hmm... I can't seem to wrap my head around it. MF and FM should be the same, since we know that if one frog is Male the other has to be female.

The sample space refers to all possible combinations, before you know anything more. You simply know the two are either male or female up front, so their are four possible permutations. MM, MF, FM, FF. Then you start eliminating.

 

[benstt]

The video is wrong because they screwed up there own sample space
To calculate the correct odds, there are always 3 frogs available, not 2

If you only saw the 2 frogs in the clearing then the math is correct
Too bad they seem to just forget about the single frog on the tree stump because he does change the odds of finding the correct frog

(This also is ignoring the obvious fact that you hear the croak behind you while looking at the frog on the tree stump
If you do not ever view/confirm one of those frogs in the clearing was the actual one who croaked, his entire explanation of eliminating the 2x female is dead wrong)

They are asking to analyze which decision is best, going one-frog side or two-frog side. So they are examining the conditional probabilities given those choices, not the overall odds of surviving. Limiting the sample spaces and comparing the 67% to the 50% is correct in analyzing the decision.