Question about probability riddle: doesn't make sense

[benstt]

Sometimes taking the example to an extreme can help humans wrap their brains around it. Suppose instead of two frogs, there were 100 frogs on one side, and one on the other.

You know at least one of the 100 is male, because you heard a croak. However, all you need is one female in that 100 to survive if you choose that side.

Which would you choose? The side with the 100 or the side with the one?

 

[rhuarc29]

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.

Correct either way. You know with 100% certainty that one frog in the clearing is male, so the probability the other is female is 50/50, the same odds as the frog on the stump.

When people try to solve this problem, they assume there are three possible outcomes for the clearing:

Male, Male
Male, Female
Female, Male

Thus giving a 67% of one being a female.

But there are actually four (as Fuji points out):

Male(1), Male(2)
Male(2), Male(1)
Male, Female
Female, Male

This gives a 50% chance. It's a logic fail to include only one combination of two males, but two combinations of one male / one female. The video is wrong.

 

[barnacler]

This is similar to the "Let's make a deal" problem

At the end of the show, there was one contestant, and three curtains. Behind one, there was the big prize, between the other two were much lesser prizes.

So the contestant is asked to choose one curtain.

AFTER the contestant chooses, the host then unveils one curtain, (because he knows where the big prize is), which is NOT the big prize. The contestant then gets to switch curtains if he wishes.

Should he switch?

 

[fuji]

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

It doesn't matter, the fact is that one did croak. That is reality even though you don't know which.

Consider this: probability isn't relative. Suppose you have a video camera filming which DID see which frog croaked. After your choice you will watch it and learn which frog croaked. Watching the video you will know for a fact THAT frog is male, but still face 50/50 uncertainty over the sex of the uncroaking frog. There is a reality to the identity of Croakie even though you don't know it at the time you make your choice.

Run this experiment over and over many times. It will not be the case that you experience a 50% rate while watching the video but a 67% rate at the time of the choice. You will experience the same events with the same frequency of the outcomes.

If I flip two coins and have them behind a screen, you can ask me to give you one that came to heads, but until I give it you don't know if there is one. If I do give you a head there really will be a 67% chance that the remaining coin is a tail. If you have a video tape showing you what's behind the screen that you can watch later, you will experience that 67% rate watching the video later as well, because it's reality.

Counting the cases, they are different:

Croakie-male
male-Croakie

Those are different cases in reality, whereas with coins there is no identity information to create the second case, it's just

Heads-Heads

Even with a video showing you all the information later there is no way to distinguish the heads in a way to create the second case, it's just two heads. The full list of cases even with left/right included is:

Left head, right head
Left head, right tail
Left tail, right head
Left tail, right tail

Left head isn't the same as right head, there's no identity to link them the way that Croakie has an identity and can be said to be either in the left or the right.

The extra case created by the identity of Croakie makes the probability 50/50.

 

[fuji]

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.

Except that isn't the right list.

It's

Croakie-M
M-Croakie
Croakie-F
F-Croakie

Because you have the information that ONE croaked. Unless it's possible that TWO croaked in unison you have additional information that there is a difference between the croaker and the non croaker and that additional information lowered the probability of a female.

In Bayes terms, there is an additional information. Not only do you know that there is a male, you ALSO know that male croaked and is a different frog from any other male that may be present.

 

[Yoga Face]

if it takes 5 machines 5 minutes to make 5 widgets

how long does it take 100 machines to make 100 widgets?

 

[benstt]

Except that isn't the right list.

It's

Croakie-M
M-Croakie
Croakie-F
F-Croakie

Because you have the information that ONE croaked. Unless it's possible that TWO croaked in unison you have additional information that there is a difference between the croaker and the non croaker and that additional information lowered the probability of a female.

In Bayes terms, there is an additional information. Not only do you know that there is a male, you ALSO know that male croaked and is a different frog from any other male that may be present.

I'm thinking this through in my spare time, but I don't think this is right. I think you're double accounting the croak info.

The sample space is constructed before you know the additional information. This can be tricky about what to go back and condition on. He is saying, before you know that one croaked, is that MF are equally likely per individual frog, and there are two frogs in the clearing. The starting sample space for the clearing is then MM, MF, FM, FF.

If they then tell you that one of them croaked, you can only eliminate the FF as a possibility. There's no additional information in my opinion.

If they had told you up front that one of the possible frogs was a male named Croakie, and that Croakie could be in the clearing with other males or females, you could construct a different sample space as you are doing. To complete the problem, they'd have to tell you the probability that Croakie would be chosen vs other males and females.

 

[hearthewind]

In probability,

let event A = two frogs are males
so the complement of A, Ac = at least one frog is female

Your observation is B = at least one frog is male (based on what you have heard) so P(B) = 3/4

So, the probability of A given B, the probability of at least one female, given the observed
P(A|B) = P(A, B)/ P(B) = P(two males) / P(B) = 1/3

As a result
P(Ac|B) = 2/3

Probability is a weird thing

On another weird topic, if you actually saw that the FIRST frog is male (he made the sound)

then you have a 50/50 chance

 

[Yoga Face]

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.

a deeper truth can be found in the two ways we think

1 Intuitive (easy and often wrong)

2 deep thinking (hard and more likely to achieve positive results)

 

[Timbit]

It's math, people. It's ok to not get it, but please don't insist you're right when you are wrong. Take a statistics course or, if you want less effort, try to understand the explanation in the video.

Timbit

 

[fuji]

If they then tell you that one of them croaked, you can only eliminate the FF as a possibility. There's no additional information in my opinion.
.

But there is more information.

"One of them croaked AND is male" is more information than "at least one male is present".

You identified a specific individual, and I know you did, because I can name him Croakie, and we both know who I'm talking about even if I'm unable to figure out which one is Croakie. One of them IS Croakie, and the other isn't, even if male.

In the coin toss example, however, there is no meaningful individual to point to. Right coin could be heads or tails. Left coin could be heads it tails. There's not enough information for me to name a head. I have less information.

This is important because it enables me to separate the two frogs on a factor other than male. Not all male frogs croaked. There may be a male frog that didn't croak.

With heads and tails I can't make that separation.

In Bayes terms, we have.

P(Male|Croaked)=1 but P(Croaked|Male) is not 1. This additional conditional probability is relevant.

 

[Yoga Face]

It's math, people. It's ok to not get it, but please don't insist you're right when you are wrong. Take a statistics course or, if you want less effort, try to understand the explanation in the video.

Timbit

Another one people will close their mind to is the Monty Hall paradox

You must accept the freaking answer then take it from there

 

[Yoga Face]

on the other hand......

 

[Wingman88]

if it takes 5 machines 5 minutes to make 5 widgets

how long does it take 100 machines to make 100 widgets?

5 minutes

 

[benstt]

But there is more information.

"One of them croaked AND is male" is more information than "at least one male is present".

P(Male|Croaked)=1 but P(Croaked|Male) is not 1. This additional conditional probability is relevant.

I think you're right about their being an issue with the problem definition. He intended it to be simple but made it complex by the random nature of the croak. He should have just said that you are informed that one of the two frogs is male.

I thought more about this, and extended to a larger sample space, and using 'p' for the probability that a male frog croaks (zero for females.)

I found these guys online. I'll point to them instead of trying to reproduce it. Look for the answer that shows 1/(2-p) as I did for the probability of survival for either choice. So, no good choice either way.

Ie if the prob that a male frog croaks is 100%, you have a 100% chance of survival both options. This assumes only one frog croaked, you see. If you hear a single croak on one side, you know you have a female frog on either side. You win.

If the probability that a male frog croaks is only 0%, both sides devolve to a 50% chance of survival, which makes sense. Ie if the probability of a male frog croaking was very slim, hearing one croak on one side doesn't tell you much.

http://math.stackexchange.com/questions/1683658/the-frog-puzzle

 

[GPIDEAL]

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.

Sorry Stinky but how can you say there are two frogs on Side A when in the previous sentence you said there's a frog on Side A, and two on Side B?

 

[Yoga Face]

5 minutes

correct

the answer is not 100 minutes

 

[rhuarc29]

Someone go run this experiment:

Get two friends to flip coins in another room. Friend A flips one coin and Friend B flips two. If neither of the coins Friend B flips are heads, both friends re-flip their coins. If Friend B gets at least one heads, he calls out to you and says so. Now, you win a dream vacation if you pick a friend who has flipped a tails. According to what the video in this thread has told you, it makes the most sense to pick Friend B every time. Now run this experiment 1.0x10^20 times. If the video is right, you should win ~6.7*10^19 dream vacations.

But don't be surprised when you actually won roughly 5.0x10^19 vacations. The reason is because there are actually two distinct and equally possible combinations of HH [call it H1H2 and H2H1] (you lose in both cases), and one each of HT and TH (you win both times).

 

[fuji]

You can try this yourself in a pub.

Get a friend to flip two coins behind a screen. You get to pick heads or tails. If he has not got the coin you guessed it's a bye, and you play again.

If he has the coin you guessed, he slides it over to you, keeping the remaining coin secret. Now you have to guess that coin and if you guess right he buys the next round. If you guess wrong, you buy the round.

If you bet the second coin is the opposite of the first one, you win the bet 2/3rds of the time and wind up paying for only a third of the drinks.

It's really going to work.

 

[stinkynuts]

Sorry Stinky but how can you say there are two frogs on Side A when in the previous sentence you said there's a frog on Side A, and two on Side B?

Sorry, that is a mistake. It should be reversed.