[bonanova]

In a previous puzzle we are told that in a two-child family one of the children is a girl and then asked the probability that both children are girls.

The original [older younger] gender distribution has four cases, each with probability of 1/4:

1. GG
   
2. GB
3. BG
4. BB

Knowing only that "one of the children is a girl" removes the fourth case, while leaving unaffected the relative probabilities of the first three. So the desired probability is 1/3. But clearly there were other true statements our reporter might have made. He may have been able to say one child is a boy. He might have described the older child, or the one nearest to him. What if we knew more about how the statement was chosen? Does the "filter" the reporter uses affect the answer?

Let's check that out by restating the problem differently:

A reporter with access to the family is allowed to select any true statement of the form "One of the children is a X" where X is boy or girl. We then desire the probability p[same] of both children being X.

Clearly before the reporter speaks, p[same] = 1/2. Now he says "One of the children is a girl."

If that's all we know, p[same] = 1/3, as in the original puzzle.

But now we ask the reporter: "How did you select your statement?" He replies:

Before I met the family, I decided that I would ...

1. Let the children play Go Fish and then describe the winner.
2. Describe the older child.
3. Say "one of the children is a girl" if I could; otherwise say "one of the children is a boy."

For each of these algorithms, what value of p[same] follows from the reporter's statement?

[Guest]

-_-I think

1) 1/3

2) 1/2

3) 1

Edited April 11, 2010 by donjar

[Guest]

1) 1/2

2) 1/2

3) 1/3

Trying to apply Bayes' Theorem:

P(A) = Probability that both children are girls (always 1/4)

P(B) = Probability that the reporter says one of them is a girl (varies)

P(B|A) = Probability that the reporter says one of them is a girl given that they are both girls (always 1)

P(A|B) = Probability that both of them are girls given that the reporter says one of them is a girl = P(B|A)*P(A)/P(B)

1) Assuming that each child has an equal chance of winning in a game of Go Fish:

1/4 of the time a girl will always win (GG)

1/2 of the time a girl has a 50% chance of winning (GB,BG)

1/4 of the time a girl never wins (BB)

so P(B) = 1*1/4 + 0.5*1/2 + 0*1/4 = 0.5

P(A|B) = 1 * 0.25 / 0.5 = 1/2

2) The older child has a 50% chance of being a girl so P(B) = 1*1/2 + 0*1/2 = 0.5

P(A|B) = 1 * 0.25 / 0.5 = 1/2

3) There are only no girls 1/4 of the time so P(B) = 1*3/4 + 0*1/4 = 0.75

P(A|B) = 1 * 0.25 / 0.75 = 1/3

[Glycereine]

1. Let the children play Go Fish and then describe the winner.

Knowing nothing else, the probability that either of the children win the game is the same as the probability that the reporter will randomly choose which child to describe. Therefore no change, 1/3.

2. Describe the older child.

Same situation as above. There is no higher probability for the older child to be boy or girl, we know only that there is at least 1 girl so we're back to the original situation, 1/3.

3. Say "one of the children is a girl" if I could; otherwise say "one of the children is a boy."

Also the same as above. We know only from this statement that both children are not boys. Meaning the other three cases all have equal probability with 2 girls being one of 3 remaning options, 1/3.

3a. However if the reporter were to say "one of the children is a boy" if he could; otherwise saying "one of the children is a girl" and the reporter said "one of the children is a girl," then we know with 100% probability that both children are girls, 1.

So with the 3 examples given the probability remains 1/3, however there are statements that could be made that affect the probability as explained in 3a.

Edited April 12, 2010 by Glycereine

[Glycereine]

2. is 1/2. By describing the older child we remove BG but not GB as an option so there are only 2 possibilities left with equal probability.

so:

1. 1/3

2. 1/2

3. 1/3

Edited April 12, 2010 by Glycereine

[Glycereine]

1) 1/2

2) 1/2

3) 1/3

Trying to apply Bayes' Theorem:

P(A) = Probability that both children are girls (always 1/4)

P(B) = Probability that the reporter says one of them is a girl (varies)

P(B|A) = Probability that the reporter says one of them is a girl given that they are both girls (always 1)

P(A|B) = Probability that both of them are girls given that the reporter says one of them is a girl = P(B|A)*P(A)/P(B)

1) Assuming that each child has an equal chance of winning in a game of Go Fish:

1/4 of the time a girl will always win (GG)

1/2 of the time a girl has a 50% chance of winning (GB,BG)

1/4 of the time a girl never wins (BB)

so P(B) = 1*1/4 + 0.5*1/2 + 0*1/4 = 0.5

P(A|B) = 1 * 0.25 / 0.5 = 1/2

2) The older child has a 50% chance of being a girl so P(B) = 1*1/2 + 0*1/2 = 0.5

P(A|B) = 1 * 0.25 / 0.5 = 1/2

3) There are only no girls 1/4 of the time so P(B) = 1*3/4 + 0*1/4 = 0.75

P(A|B) = 1 * 0.25 / 0.75 = 1/3

I believe your probability for case 1 is the probability that a girl will win prior to the revelation that one of the children is a girl, not the probability that both children are girls after that revelation.

[Guest]

I believe your probability for case 1 is the probability that a girl will win prior to the revelation that one of the children is a girl, not the probability that both children are girls after that revelation.

The probabilities need to be determined not only by what the reporter said, but by the fact that it was the reporter who said it. For example, lets look at 1). When the reporter says that at least one of them is a girl, he's also telling us a girl won the game of Go Fish. Now Since at least one is a girl we are reduced to the same 3 cases of Gg, Bg and Gb. But they are not equally probable because it's twice as likely for a girl to win if both children are girls than if one of them is a boy.

[bonanova]

I believe your probability for case 1 is the probability that a girl will win prior to the revelation that one of the children is a girl, not the probability that both children are girls after that revelation.

In case 1 the reporter tells us the gender [female] of the child that wins a game.

In case 2 the reporter tells us the gender [female] of the child born first.

In both cases we are asked the probability that the other child's gender is the same [female].

Can you make a case that these probabilities differ?

[Glycereine]

In case 1 the reporter tells us the gender [female] of the child that wins a game.

In case 2 the reporter tells us the gender [female] of the child born first.

In both cases we are asked the probability that the other child's gender is the same [female].

Can you make a case that these probabilities differ?

Yes and No. I believe Tuckleton's answer is correct, however I still have a hard time with it.

Edit: I follow now

In case 1 order does not matter, it could be either child that wins and there are still 2 situations where there is a boy even if the girl won (albeit not the same probability which is where I get confused and why I am probably wrong)

In case 2 order does matter, the child listed is always the first child and thus eliminates one of the two possibilities remaining where there is a boy sibling.

P(A) = probability of both sisters being girls. = .25

P(B) = probability that one of the sisters is reported to be a girl.

P(B|A) = probability that if both sisters are girls one sister is reported to be a girl. = 1

P(A|B) is the probability that both sisters are girls after the statement that 1 is a girl.

If I use Bayes Theorem (which it looks like I should have originally) I get the same answer so I'm not disputing it, but my logic can't follow it in this situation to a congruent solution.

Edit: in case 1 where order doesn't matter the chance for a girl to win in each of the situations where there is a boy and girl (BG and GB) is exactly half. This essentially removes one of the possibilities, or gives them half weight compared to the likelyhood that a girl will win if there are two girls (GG). So although it's twice as likely that there are a boy and a girl (BG or GB) than two girls (GG), a girl winning means that there is a 50% chance of GG and 25% chance each for GB and BG. That's what gets to the difference in my brain so I'm ok with it now.

Edited April 13, 2010 by Glycereine

[bonanova]

Yes and No. I believe Tuckleton's answer is correct, however I still have a hard time with it.

Edit: I follow now

In case 1 order does not matter, it could be either child that wins and there are still 2 situations where there is a boy even if the girl won (albeit not the same probability which is where I get confused and why I am probably wrong)

In case 2 order does matter, the child listed is always the first child and thus eliminates one of the two possibilities remaining where there is a boy sibling.

P(A) = probability of both sisters being girls. = .25

P(B) = probability that one of the sisters is reported to be a girl.

P(B|A) = probability that if both sisters are girls one sister is reported to be a girl. = 1

P(A|B) is the probability that both sisters are girls after the statement that 1 is a girl.

If I use Bayes Theorem (which it looks like I should have originally) I get the same answer so I'm not disputing it, but my logic can't follow it in this situation to a congruent solution.

Edit: in case 1 where order doesn't matter the chance for a girl to win in each of the situations where there is a boy and girl (BG and GB) is exactly half. This essentially removes one of the possibilities, or gives them half weight compared to the likelyhood that a girl will win if there are two girls (GG). So although it's twice as likely that there are a boy and a girl (BG or GB) than two girls (GG), a girl winning means that there is a 50% chance of GG and 25% chance each for GB and BG. That's what gets to the difference in my brain so I'm ok with it now.

Bayes aside ...

Keep in mind p[GG] given that a particular child is G, write p[GG|G] = p[bB|B] = p[same] = 1 - p[mixed].

Just as [older younger] = GG GB BG BB are equally likely, so that p[same] is 1/2 by inspection,

the four cases of [winner loser] = GG GB BG BB are also equally likely.

That is, winning, as it pertains to Go Fish, is not related to gender. In other matters it becomes problematic.

So p[same], when applied to winner and loser, is also 1/2 by inspection.

[Guest]

just do all the cases like with the original problem. Taking 1 again we have 8 equally probable cases going into it:

Gg G wins

Gg g wins

Gb G wins

Gb b wins

Bg B wins

Bg g wins

Bb B wins

Bb b wins

When the reporter says it's a girl we are left with only 4 cases which remain equally probable:

Gg G wins

Gg g wins

Gb G wins

Gb b wins

Bg B wins

Bg g wins

Bb B wins

Bb b wins