[OT] mathematical query

[bramadan]

You guys should listen to 2WS-Steve he is probably the best game theorist in gaming this side of Johnatan Tweet. And also entirely correct in this issue.

I liked the Bayes' theorem explanation best and if noone minds I would like to put it in a bit more layman's terms.
Sigil is wrong on a very basic assumtion and that is a common assumption in proability which states that independent states are equally probable. If we had no information at all it would be the case and we would only have 25% chance for each of MM, MF, FM and FF combinations.
However, *with prior information* provided by the picture things change and we have to ask ourselves what is the likelyhood that a person who certanly has a son falls into each either of the categories and it would be:
MM - 50%
MF - 25%
FM - 25%
FF - 0%
the categories are not equivalent anymore because prior information has put bias into the way probabilities are allocated.

It is kind of like betting on the outcome on two dice if I do not know any of the dice then the best odds I can give at the result being 12 are 1:36. If I know that one of the dice has fallen on 6 the odds of a 12 change to 1:6 (odds of 11 change from 1:18 to 1:6 etc...) Simmilarily if you count M as a 1 and F as 0 chance of total being 2 if one "die" is known to be 1 is 1:2 as opposed to 1:4 if both dice are unknown.

Equal probabilty assumption is realy usefull and is used left and right in statistics but is only valid without prior information.

 

[ichabod]

You guys should listen to 2WS-Steve he is probably the best game theorist in gaming this side of Johnatan Tweet. And also entirely correct in this issue.

I liked the Bayes' theorem explanation best and if noone minds I would like to put it in a bit more layman's terms.
Sigil is wrong on a very basic assumtion and that is a common assumption in proability which states that independent states are equally probable. If we had no information at all it would be the case and we would only have 25% chance for each of MM, MF, FM and FF combinations.
However, *with prior information* provided by the picture things change and we have to ask ourselves what is the likelyhood that a person who certanly has a son falls into each either of the categories and it would be:
MM - 50%
MF - 25%
FM - 25%
FF - 0%
the categories are not equivalent anymore because prior information has put bias into the way probabilities are allocated.

No, that's the probability that if you take a picture of a random kid of his and it comes out male, that he has kids of two sexes. I saw nothing in the original question about the picture having been taken of a random child.

Since people are throwing around probability formulas, let's go with the definition of conditional probability, namely P(E\F)=P(EF)/P(F). In english, that is the probability that E given F is true is equal to the probability of (E and F) divided by the probability of F.

If the question is, given that we know he has at least one male child, what is the probability that he has children of different sexes, E = he has children of different sexes, and F = he has at least one male child.

P(F) = 0.75 (Out of MM, MF, FM, and FF; 3 of the four give him at least one male child)

P(EF) = 0.5 (Out of MM, MF, FM, and FF; 2 out of 4 give him at least one male child AND children of different sexes)

Thus P(E\F) = 0.5 / 0.75 = 2/3 = 0.67

I'm not sure why you are using Baye's theorem. You would use Bayes to find the probability of E by conditioning it on F and not F. You're getting the right answer for P(E), namely 0.5. However, the question was about P(E\F), not P(E).

 

[CRGreathouse]

I'm not sure why you are using Baye's theorem. You would use Bayes to find the probability of E by conditioning it on F and not F. You're getting the right answer for P(E), namely 0.5. However, the question was about P(E\F), not P(E).

You are right if you assume the problem is P(E|F), but that's not the problem.

The probability 2/3 works if the woman asked the man if he had a male child and he showed that picture. 1/2 works if he showed a picture of a random child.

There have been many correct explanation of this in this thread, but here's the simplest one (posted above, but I can't seem to find it):

Picture is of boy: 2/3 chance for different genders (2/3 chance other is girl)
Picture is of girl: 2/3 chance for different genders (2/3 chance other is boy)

Without a picture, the chance for the other being a boy is 1/2; with a picture - any picture - of the first child, the chances change.

Showing an unrelated picture changes probabilities?

 

[ichabod]

You are right if you assume the problem is P(E|F), but that's not the problem.

The probability 2/3 works if the woman asked the man if he had a male child and he showed that picture. 1/2 works if he showed a picture of a random child.

Okay, I see what you're saying. My bad.

 

[Darkness]

Nit-pick: Gender distributions are not exactly 50:50 (although they're damn close).

 

[2WS-Steve]

I think the paradoxical bit about this puzzle is that both versions seem the same but give different answers; for example:

Case 1
You ask someone to show you a picture of one of their kids (or they volunteer to do so). They show you a picture of a son.

>>> 50% chance the other kid is also a son.

Case 2
You ask someone if they have a son. They show you a picture of a son.

>>> 67% chance that the other kid is a daughter.

In both cases they respond in the exact same way and you appear to gain the exact same additional information. How could there possibly be a difference in the probability change?

However, you don’t gain the same evidence. In case 2 you placed a restriction on the way they could answer; they had to show you a son if they had one and if they didn’t have son you’d know precisely what gender both their children were. For instance, if they showed a picture of a daughter in case 2 you’d know the other child was also a daughter. But if they did the same in case 1 you wouldn’t know if the other child was a daughter or a son.

That might indicate that there’s a general rule that placing restrictions on the way one can answer a question increases the value of the evidence you gain. Perhaps when we think about it that should be obvious but it certainly doesn’t seem obvious. It seems like all that should matter is the actual physical evidence we receive.

 

[2WS-Steve]

Sorry, last post didn't get it quite right. This is the real reason case 1 is different from case 2, I promise!

Let's say the guy has pictures of each of his two kids in his wallet; he shows the top picture to you.

In case 1 these pictures are randomly arranged so we get these possibilities (top-bottom):

M-M
M-F
F-M
F-F

In case 2 the pictures also start out randomly arranged just like in case 1. However, since we specifically ask him to show us his son he rearranges the pictures if they're F-M. This gives us the following distribution:

M-M
M-F
M-F
F-F

The reason we can guess right more often in case two is because we're forcing the distribution to be less random.

As you can see, the "down card" has a flat 75% chance of being female after forcing the rearrangement. Essentially case 2 is just a magic trick.

 

[apsuman]

A man and a woman meet for the first time.

...

Ok, I started this thread... it was interresting, and I think I will reread it all a little later. I quoted by first post because I had to go back to it to make sure of what I posted before I replied.

Thanks everyone for the discussion, it was fun (in an odd sort of way).

But to the answer. In my problem statement, all I revealed (or rather tried to reveal) is that the man has *A* son. With that information and no other, I believe that the correct answer is that the other child is a girl 2/3 of the time.

(I am wrong often)

The island example I think is the best, with BoB, BiG, GaB, and GuG.

g!