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GyroNinja
Nov 7, 2012
I just realized that Gaining Weight's argument would be correct if Bill were revealing the gender of one of his children at random, boy or girl, and not just telling you that one of his children is a girl. (This same distinction is actually also important for the Monty Hall problem).

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GyroNinja
Nov 7, 2012

GAINING WEIGHT... posted:

Yes. The second case seems more likely because you have 2 options per die, but you have to subtract out the cases where you get 2 sixes or 2 ones, which is not the outcome you're looking for. Both are 1/36.

No, the probability of getting a one and a six (in any order) is 2/36, not 1/36. Look, here's the list of (non-repeating) dice combinations;

1 1
1 2
1 3
1 4
1 5
1 6
2 2
2 3
2 4
2 5
2 6
3 3
3 4
3 5
3 6
4 4
4 5
4 6
5 5
6 6

Notice how there's only 21 possibilites? If all of them had a probability of 1/36, then what do you think happens the other 15/36ths of the time you roll the dice? Instead, the six doubles are only 1/36, while the fifteen non-doubles are 2/36, for a total probability of 36/36, which is what you'd expect.

Let me ask a (slightly) different question. Do you think that rolling a 5 and a 6 is as likely as rolling two 6s? And if so, do you realize that you're saying that the odds of getting a total roll of 11 are the same as a total roll of 12, which is not how observable reality works?

GyroNinja
Nov 7, 2012

GAINING WEIGHT... posted:

Of course they mean or?


I am fairly sure you're the one being inconsistent. We can talk about two possibilities ("contracted" view as above), which are: either both are girls, or there is one of each; or we can talk about four possibilities ("expanded" view), where order matters and GG is two of the possibilities, but we can't talk about 3 outcomes, because then you're expanding/contracting one and not the other.

This is easier to see if neither of the two options for the unknown child match the first. So let's add in a third gender and phrase the problem like this: there are two kids, one is a girl, the other is either a boy or a flibbertyjib. Then it's easy to see that there are four possibilities (taking birth order into account) :

GB
GF
BG
FG

Half of which are B, giving us the 50%

I think I see your problem. You're visualizing it as "Well, I know that one of the siblings is a girl, which means that either I know that the first child is a girl, or that the second child is a girl, and therefore there's only one child who's gender I don't know, so the probability of the unknown child being a boy is 50%", but in reality, you don't know the gender of EITHER child right now. That is to say, the first child could still be a girl or a boy, and the second child could also be a boy or a girl, so it's wrong to act as if one of the child's genders is already "known". They're both unknown.

Let me try a different way of looking at it. Let's not actually talk about whether his "other child" is a son, because I think that's confusing language that makes it sound like we know more than we do. Instead, what we're asking is, "Given that Bill has at least one daughter, what is the possibility he also has one son?" Instead of talking about BG or GB, let's just look at the number of daughters Bill has. With two children, he either has no daughters (1/4th of the time), one daughter (1/2), or two daughters (1/4th again). Notice that with no prior information, there is a 75% chance that Bill has a daughter, a 75% chance that Bill has a son, and that the probability of him having a daughter and a son is twice as high as him having two daughters (or two sons, for that matter).

So what happens if we ask Bill if he has a daughter, and he answers "Yes"? We can now ignore the 25% possibility that he has two sons. But this gives us no information to make us think that he has either one or two daughters; he would answer the same either way. So it would make no sense for his Yes answer to change the fact that he is still twice as likely to have one daughter as opposed to two, which is why the new probabilities become 2/3rds for one daughter, and 1/3rd for two. So learning that he has one daughter reduces the chances of him having a boy from 75% to 66%.

If we have some way of differentiating the two children, and we know that one of them specifically is a girl, then that would only leave one unknown child, so the answer would be 50%. Likewise, if Bill were to pick one of his two children at random, and reveal their gender to us, boy or girl, that would also make the answer 50%, since Bill would be twice as likely to show us a daughter if he had two of them instead of one, so him revealing a daughter would be new information that would counterbalance our original probability of him being twice as likely to have one daughter instead of two. But when we're only asking Bill if either one of his children is girl, the 2/3rds answer is correct.

GyroNinja fucked around with this message at 00:09 on Sep 5, 2016

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