Let's start by seeing if we can agree with my simple premise above?

Again substitue the K in *all positions* to which it can legally be applied without violating the rules.

Sorry didn't see that last one

But we agree that the possibilities I listed are the only ones, right?

I know what you are saying. I agree with that much. I disagree with removing an extra occurence of BB or HH or whatever becaue constant K can occupy either position so it becomes KB/KH *and* BK/HK.

There is a 1 in 3 chance that a combination of kids containing at least one boy has two boys. But once the establishment of a kid being a boy is set, as it is by the rules, we are left with the other kid which makes it positional and means there is a 50/50 chance the other kid, be it in position one or position two, is a boy.

Ok, I get that. I'm going to try to convince you otherwise, like I convinced myself earlier today.

If we agree that there are only those three possibilities, then does marking a "K" on one of the coins change the number of possibilities? Of they are both heads, I get to choose which one I mark, without telling you, but it doesn't mean that it was a different outcome, you can't change the number of outcomes that are possible just by writing the letter "K" on a penny. Whether I marked the left one or the right one, it's still the same outcome.

That's what cleared it up for me after I had confused myself

@soc: any critique of my explanation of the formula?

There are actually only three possibilities if we go positionally and to do that you must substitute K for every position it could hold. If K is non positional then there are only two possibilities B+G and B+B because non-positionally G+B is the same as B+G as there is no position for G or B.

@semck, if I could +1 this thread again, I would. Who would have thought that we would get another day of intellectually stimulating conversation out of this one. I'll definitely be asking this question to my smartest friends to see the discussion that ensues.

hey phil sorry for the absence. B is that child A is a boy and A is child B is a boy. so we have 0.25/0.5. I think Draug's point on position vs non positional is very very important undoubtedly. I think the fact that we know that one child is a boy means we can firm one child as being a boy and then the other is 50/50. or rather because they are independent we can ignore the first one. and I think that if you count GB and BG you must count BB twice, but really i think more that gb and bg are just equal. moreover as i said before they aren't both real possibilites as we pick one of the values to definitely be a boy. sorry if i'm not articulating this very clearly anymore, i'm rather tired, made my points better earlier, and now draug is back arguing everything i would just more eloquently.

I must say I much prefer these threads and conversations to the political ones, even with me having radical views I like to debate. Not just because they are more civil but they really provoke some real thought.

zultar got mad at me yesterday for not being more clear, so I will try today.

The first thing you all need to get straight is whether you are talking about the same question. I think semck maybe originally put it in a confusing way (maybe even on purpose). Does everyone agree that the question is equivalent to the following?

Mr. Jones has two children. If we know that he has at least one son, what is the probability that he has two sons?

philcore, i guess how i would argue it right now, given that i'm too tired to think, is that an axiom for this thought experiment is that the coin flips/sex of child is independent. that being the case if we know the outcome of one the other becomes irrelevant.

i think a difference is coming from the simple idea of if we drew a diagram illustrating all the possibilities. you are trying to eliminate only the gg/tt option, whilst really i think to be accurate you have to eliminate the side of the diagram that says the first one is not a boy/heads, as this ensures we have the 1 occurence of a head/boy and then asks about the other one. If we only deleted the gg/tt at the end we would be ignoring the fact that our diagram is really starting with boy and then boy/girl for the other child (on the basis that we know we have a boy). I know you are going to try and argue that by deleting the gg option we ensure that there is always going to be a boy, but we are then not looking at the true nature of the case - one that the situation ensures there is at least one boy (before we tinker with it at the end to ensure so). hopefully that was somewhat intelligible, either way i will post more in the morning in this interesting conversation

@soc, then do you agree that your assignment of variables is incorrect? Because A should be that both children are boys (1/4) and B should be that at least one is a boy (3/4). The essence of the problem is "what is the probability of both children being boys given that at least one of them is a boy?"

It isn't "What is the probability that both children are boys given that the oldest (or youngest) is a boy?" so I agree that position determines the difference between the two answers whole heartedly. But the question is ignoring position explicitly.

Aha, I posted while you were writing. It seems that we do indeed disagree on the essence of the question.

uclabb, i would phrase the question as Mr Jones has two children. One is a son. What is the probability that the other child is a son? But for now I suppose your way of wording the question can suffice. really though before I hear the old so there is BB BG GB line again, i really think this is the long way of looking at it, and that if you guys insist on disregarding that we can have B in a set position and then look, I would insist that we can have the first being boy and the second being either, or the second being boy and the first being either, without the apparent double counting applying only to the two boys. but as i said i will return in the morning

So we all agree that if position is determined, then it is 1/2. Do we also all agree if position is not determined then it is 1/3?

Damn Europeans and their offset clocks, it's only 5pm here and I just started drinking! That's when I do me best posting!

Ok then it can wait until tomorrow, I guess.

I hope soc dreams about coins and babies all night ;-)

"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."

Draug, your fallacy is saying that the boy is a known entity. You say that once we know "there is at least one boy," then we can take and label that child. But there is no such child to label. If there are two boys, doing so would entail making a choice and treating the children as asymmetric illegitimately. Any choice you make would assume the presence of information you were not given, which is what is causing you to come up with the wrong answer.

In effect, you are double-counting BB because the boy you fix in that case could be either (but in the other cases, it could only be the one). You are not allowed to do so.

Let's do this correctly. I tell you, "Draug, Mr. Jones has two children. At least one of them is a boy." You say, "OK. Fix that boy, call him A." That's where you're making a mistake. You should say, "OK. If there is only one boy, fix the boy and call him A. If there are two boys, fix the older boy and call him A.

"In the first case, A may be younger or older -- GA or AG. There is one of the second case. So there is a 1/3 chance we are in BB (i.e. AB)."

The reason you still have to count GA and AG is because your labeling does not reflect any known persons or information. It is just a choice that was broken down into cases, and you can't just assert the probability of the different cases (A and a girl, or A and a boy). You still have to compute them.

@SD: what you are saying just goes to show that probabilistic independence is very counter-intuitive. The result of 1/3 in this case is BECAUSE the children are independent. It's just that humans are very bad at making these computations intuitively.

In any case, Draug, I'm back to asking you the following very explicit question. Please answer it.

If I tell you I tossed two fair coins and that at least one of them came up heads, what is the probability that they both came up heads?

According to your reasoning, we can label the heads we know about to be the first one, K. So we either have KH or KT. So it's 50/50.

What do you say the probability is in this exact (coin) situation?

My nightmares about coins and babies has forced me to return for a bit.

Semck, if it is independent and we can discount the one then it is just up to the other one, and I agree humans are bad at these things intuitively (reading a paper on it at the moment), but that affects us with the monty hall problem more than this where it seems you guys are too busy trying to outsmart the situation!

How about this: the crux of your argument can be said to apply to a situation where there are two coins, where we know at least one is heads, and in fact you point it out to me, and then ask what the other is, but add the stipulation that you don't know whether to call this coin the second coin or the first coin, from this you deduce that there is a 1/3 chance the other coin is a head because of the '3' possibilities we now have because you have chosen to say that you don't know whether to call this the first coin or the second coin. Surely you agree that this is fallacious reasoning and that just by saying that the probability doesn't change from 1/2 to a 1/3?

Socrates,

If I point out the coin to you, then yes, it's 1/2.

But if I toss two coins, and tell you (without telling you anything more) that at least one was heads, then the probability that they are both heads is 1/3.

One might try the following. "OK, there is at least one heads, say K. Now if K is first, then we can have KH or KT. If K was tails, then we can have HK or TK. So that's KH, KT, HK, or TK, and half of them are two heads."

The problem is that we double-counted HH: KH and HK are both two heads.

but semck, you point out the coin to me, but we don't know the order, so if it goes first then we have a H and then eitheer a T or H (so we have HH and HT) or we have H second and either a t or h first (HH or TH) again, but we've supposedly double counted the HH and therefore only have HT TH HH so 1/3 it is heads. or am i missing something here?

and KT and TK are both one of each, we've double counted by assuming the order is important!

Either the coin on the left is head, so HT and HH or the coin on the right is so TH and HH. It isn't double counting it because HH occurs as two distinct possibilities even though we don't know which coin was is the referred to head (or more to the point *because* we don't know which coin is). This si not an abstract but a concrete real world situation so we can know for certain the Left coin is heads or the Right coin is Heads and there fore we can figure all four possibilities involving those two positions.

Draug, I asked you a very simple, concrete question. Please answer it.

Draug, you are smarter than this. This isn't a political discussion; it is math. There is a right answer and semck is explaining it. It is possible that you are misunderstanding the question. If you are convinced that is not the case, I encourage you to start thinking in the frame "How am I confused?" rather than "Let's convince semck he is wrong"

"and KT and TK are both one of each, we've double counted by assuming the order is important!"

The order is always important. That's part of what independence means. If the order weren't important, then 1/3 of 2-child families would have 2 boys.

"but semck, you point out the coin to me, but we don't know the order, so if it goes first then we have a H and then eitheer a T or H (so we have HH and HT) or we have H second and either a t or h first (HH or TH) again, but we've supposedly double counted the HH and therefore only have HT TH HH so 1/3 it is heads. or am i missing something here?"

If I'm pointing out the coin to you, then we need more specification so we don't end up in a Monty Hall type problem (which we're now getting very close to).

If the agreement beforehand was that I would toss two coins and, if there is at least one heads, point a heads out to you and ask you the probability, then 1/3 remains the answer.

But if I just tossed two coins, and told you, "Hey, look, this one is heads" (keeping the other one covered), the answer is 1/2.