Even the use of the term "other" could be interpretted two ways. That is why it is best to ask "what are the odds both are boys" because "other" carries the implication you are looking at one of the boys or somehow know them personally.

@Mintyboy4, I find question as worded above is ambiguous and depends on what you mean by "the second one". Does "second one" mean the child that is NOT the one you know is a boy, or does it simply mean the younger child, and that in fact all you know is that one of the children is a boy but you don't know which one. As in this scenario..

"We we are walking down the street and we meet Mr Jones and he introduces us to his son. Later we are told that Mr Jones has two children. What is the probability that Mr Jones's youngest son child is a boy?

^ the answer to *that* question is 2/3.

WAIT! I got it! A 5 pointed star uses ten intersections and has 4 intersections each!

lol wrong thread I think Draug

what the hell?

Surely it is 50%? Don't get me wrong I like your BB BG GB trick, and it looks neat, but surely it only works spuriously? Just because conditional probability is worked out as:

P(A|B)= P(A and B)/ P(B)

which is rearranged to P(A and B) = P(B) X P(A|B)

which is simplified to P(A) = P(A|B)

Which in out case is 50%?

Or to just put in the numbers it should be P(A|B) = 0.25/0.5=0.5?

Where am I going wrong? I still see the GB BB BG answer but I think it is a neat trick rather than the right answer....

The thing is we know that child A is a boy, and are now asking what child B is (we don't know that child A is the second or first child but that doesn't matter for our labelling). And if I wanted to take this ad absurdum I would give you the case of 3 kids where we know 2 are boys, giving us bbg bgb gbb bbb, or 3/4 girl, and then 4 kids (all the way to infinity), with the odds apparently increasing towards certainty for an independent event of 50%! This idea reminds me very much of the monty hall problem in the sense that I believe you are looking at the case from the wrong position, but I would be very intrigued to see where I am wrong

@socDis: I don't think that listing out all possible combinations then circling the favorable ones is a trick. It's the definition of probability, afterall. The math should agree, and if it doesn't, I question the setup of the problem before I question the brute force approach. Of course the brute force approach gets extremely difficult, as the possible outcomes and the number of events increases, but for tossing a couple dice, or flipping a few babies, its still easy enough to enumerate the possible outcomes.

I think where I screwed up in my tom example with 4 possibilities is this

TB = BB
TG = BG
BT = BB
GT = GB

So two of my possibilities were actually the same possible outcomes, therefore I have to eliminate one from both the possible and the favorable. Which gives 1/3 again.

@socDis: also you ad absurdum example isn't absurd at all. Except that you reversed the problem. There is a 1/4 chance of all boys if we know that 2 are boys, you said (correctly, but backwards for this problem) that its 3/4 prob, that there is a girl in the family. Just 1-p(no girl). So that still makes sense. And to carry it to inf, the prob of a girl knowing that all but one are boys, does approach 1, since the prob of all boys approaches 0 as n approaches inf.

But looking at the actual conditional probability formula, this 'logic' proves false so must be a trick? You guys are seemingly to me looking at this the wrong way; either the sex of the child is independent or it is not. If the sex of the child is independent then the fact that one child is a boy matters not one iota to the sex of the other child. I definitely think that you guys are misunderstanding conditional probability, but perhaps I'm wrong.

And maybe it isn't absurd to you guys, but the idea that an independent variable of P(0.5) becomes P(1) because of factors of no influence is one that befuddles me.

Let me ask it this way, do you agree that even though the prob of each coin toss is .5, the prob of getting 2 in a row is 1/4 and the prob of getting 3 in a row is 1/8?

And that you can generalize this to 1/(2^N) for gettin all heads on consecutive tosses? And that 1/2^N approaches 0 as N approaches infinity?

That's all your example is saying, except that it is boys instead of heads.

I don't get what you're saying here phil... we aren't talking about getting the probability of two boys, but really the probability of getting one boy; or one boy given that we already have one boy if you would prefer. With the coin tosses we aren't asking about a specific head, you are asking about all of them together, here we know we have a boy already so there isn't this carry on effect.

I've actually leaned back towards the 50/50 proposition because exactly what SocDis says is true. The fact is we have a known entity, a boy. We will label him K. As he is a known entity he comes first, period, leaving only two option KB and KG. There is no GK because K is a known entity and therefore comes first. It is like dice in that if I throw a 6, I know I have a 6 so my odds of the second dice being a 6 are 1 in 6. It doesn't matter that I know which die is the 6, just that a die is a 6. We take that die as being the equivalent of die #1 (first position). It doesn't matter if we know which kid is the boy, just that a kid is the boy, he becomes the known and must be placed in position 1. The knowledge of his existence and sex removes him from the equation in this case.

So yes, I have gone back to 50/5 and no, the experimental code would not prove anything because it removes the fact that the boy is a known entity and therefore occupies position 1.

But it is though, the only thing the clue which says at least one is a boy does, is eliminates one of the possibilities, the one where both are girls. So instead of 4 possible outcomes, you only have 3. And of those 3, only one includes 2 boys.

@draug: ill stick with coins because its easier and its analogous to babies. Let's say that someone just flipped two coins and can see the outcomes. So far that doesn't change the 4 possibilities, right? There is one in four chance that they are both heads. Now he decides to give you a hint, and tells you that at least one of them is heads, and asks what the prob is that both are heads. All he has done for you is eliminated the possibility that there were 2 tails. So you now can say that there is 1 in three chances that it is 2 heads.

Totally agree with Draug here (amazing though it is), the argument for a third is relying on the ambiguity of the position of the boy constant (K), but there is no need for that, the situation isn't that we know that one is a boy, but either A or B is a boy, the situation is that one of them is a boy, and the other we don't know, so we can safely label A or B a boy, and then we only have two possibilities. In other words if we pick A as a boy it is BB or BG, and if we pick B as a boy it is BB or GB, but we have to pick which of A or B is a boy first (or maybe double count the BB option), it is no good to ignore the fact that we have a constant here.

I think you are misapplying the logic of the Monty Hall problem, or failing to understand the value of the constant and the basic formula for working this out, that said I still think your position is interesting and sounds good (it just fails to hold up IMO)

@phil - No it eliminates two possibilities because you can assert that the first child will be the boy leaving only the option for the second child (order is irrelevant but consider it order of discovery of gender) which means BB and BG are the only options.

Ok, still sticking with coins ... There's a divider between us and I can see the two coins one on the left and one on the right. If I tell you that the one on the left is heads, then you're correct, it's 50/50 that the one on the right is heads. But if I just tell you that one is heads, then the possibility that the one on the left is heads, the one on the right is heads, or their both heads

it doesn't matter if you say the one on the left or not, the fact that we know one is means we can have a constant of one and then the other one by itself. just use the formula! like how draug said consider it order of discovery, which you establish yourself. there is only really the illusion of 3 options, we just need to remember that it is an independent conditional probability problem. I was going to try and suggest an analogy but i have a strange feeling that you wont change your mind

@soc: I think I've figured out where your conditional probability formula is flawed. When you say P(A|B) what are A and B?your saying that A is "the other child is a boy" and B is "at least one child is a boy" then your math works out to 1/2. But if you make A be "both children are boys" then look what we have


P(both are boys) = 1/4
P(at least one boy) = 3/4
P(both are boys AND at least one is boy) = 1/4

So P(A|B) = P(A and B)/P(B) = (1/4)/(3/4) = 1/3

Q E to the muthafukin D

But the question is originally stated was "the second child" which was then revised to "the other child" so you just *proved* that "the other child" is 50/50 by your own admission!

In short, the moment you establish one childs sex, even if we don't know which child it is, that child becomes the constant K and is in the first column eliminating GB and GG. Otherwise you have:
KB
KG
GK
BK

which still works out to 50/50. you can't double up the position of the constant against the girl additional child without doubling it up against the boy additional child.

By the way, as I've said before, if the formula doesn't give me what the brute force method gives me, I'm going to trust brute force, when it's feasible, because that goes to the definition of probability. If the formula gives me a different result, its always because I haven't properly defined A and B

It's not an admission draug, we aren't in court. It was conceded long ago that with the original wording, it is definitely 1/2. But the intent was to say that both are boys, not that the other one is a boy. Hence the revision in the 6th reply

@draug, the listing you provided was exactly the post I started the day with that confused me again. I called K Tom, but I agree that is a reasonable way to think about it, however, I realized where I went wrong by equating them with the unknown combinations, and discovered that I repeated BB so I had to remove it ... Both from the possibles, and from the favorables, giving 1/3 again.

Even that revision said "the other". I was the first to bring up "both are boys" much later. But it really does come down to the fact that the known entity of one boy could be in either position of a two boy family, so BB counts twice as BK and KB.

Want to run a test?

Run X number assuming a boy in position one and a 50/50 chance of a boy in position two. Do the same number with the boy in position two and a 50/50 of another boy in position 1. 2X runs will return approximately X of other child is a boy and X of other child is a girl. The only time probability of combinations can reduce options is when you eliminate absolutes with a statement like "both cannot be girls" however this does not work in binary combination of two things with two options. It works if your two items have more than two possible results for each item or if you have more than two items with two or more possible results each.

Back to the game of coins behind the divider; I flip 2 coins and tell you one is heads, there are now three possibilities either the one on the left is heads and the one on the right is tails, or the one on the right is heads and the one on the left is tails, or they are both heads. Can we agree at least that those are the possible outcomes?

But removing the double BB is where the problem resides. BK is one alternative and KB is another. Because we don't know the position of K, we have to have a possibility that includes K in any slot that K fits the rule. So BK KB, GK and KG.