Draug, list the outcomes explicitly. There are four outcomes, we've gotten to this point repeatedly. HH, HT, TH, TT

Thats the bottom line, if we can't agree on that, then all is lost.

We don't agree because the outcomes are hH Hh HT TH tT Tt. if you make the H+T positional then you must also make the TT and HH positional.

@Spy - I don't alternate, I randomly pick which coin is tossed and which is our constant.

Sorry, I meant randomize. Why does it matter which coin is tossed first?

Quick review, because I agree that the odds of picking a BB out of the population once the GGs are removed is 1/3, then you say Smith is part of the sample so there is the same chance the Smith family will be, yet the moment we meet one of Smith's kids it becomes 50/50 the unmet kid will be a boy? That makes no sense. If the unmet kid is 50/50 then the "other kid" will also be 50/50 which was the wording of the revised question.

@Spy - it doesn't actually matter but I wanted to include the variable of where the static or constant toss result appears.

"yet the moment we meet one of Smith's kids it becomes 50/50 the unmet kid will be a boy? That makes no sense. If the unmet kid is 50/50 then the "other kid" will also be 50/50 which was the wording of the revised question. "

Yes, because then you know more about Smith than just that he's an element of S. You know that he's an element of S, and that when one of his children was randomly selected, it was a boy.

Then all is lost. If you can't agree that the only 4 outcomes of 2 coins being tosses are HH, HT, TH, TT then we can't go any further. Toss a few coins, have a few babies, write a truth table or whatever it takes to convince you of that fundamental aspect of the problem.

If both coins land on heads then in no way does it matter which is tossed first, so why on earth would you count it twice?

I am bit lost. Would someone please restate the revised question, in simple clear terms.

But seriously, Draug, just sleep on it. You're smart and, more importantly, a gambler. You'll realize your mistake once you've had time away from this current rut you're stuck in.

The current question, spy, has two equivalent formulations:

If everything we know about Mr. Jones and his family is that (a) he has two children, and (b) at least one of them is a boy, what is the probability that he has two boys?

Alternatively:

If you know only that (a) two fair coins were tossed, and that (b) at least one of them turned up heads, what is the probability that both of them turned up heads?

Spyman read post 6 that is the revised question. I'll boil it down to this:

What is the probability that a family with 2 children has both boys, knowing that they at least have one boy.

That's the original question that Draug has already agreed is 1/3. No wonder I am confused. That is also the same question that my vba routine models. Whereas I can't see how Draug's VB routine model's that question alt all.

I think Draug has another question in mind.
Draug would you please confirm?

No, spyman, Draug changed his mind on the original question. See page 7, 5:37 PM EDT.

If you know only that (a) two fair coins were tossed, and that (b) at least one of them turned up heads, what is the probability that both of them turned up heads?

33% because there are 3 possible outcomes (HH, HT, TH) , of which one is HH.

NOTE THAT the original question says two coins *were* tossed; it is not the case that we're at the point where we've seen heads on the first toss and are about to toss the coin again. So we cannot assume that the first coin toss turned up heads. (If we knew the first one was heads then the probability of HH would be 50%, but that is not the question.)

1. Mr Jones has two children. What is the probability they are both boys. Answer = 1/4
2. Mr Jones has two children. One is a boy. What is the probability they are both boys? Answer = 1/3
3. Mr Jones has two children. The older child is a boy. What is the probability that they are both boys? Answer = 1/2

As we go from statements 1 to 3, with each, we know just a little more about the family. Hence we the probability changes.

Can we all see that statement 3 has more information than statement 3? (And by all I mean Draug :-) )

typo... statement 3 has more information than statement *2

If a family has at least one boy then it is 1/3 they have two. I do agree with this. But please reread post reply 6. It clearly states the "other" child which implies knowledge specific to one of the children making the "other child" have a 50/50 shot. Doesn't matter which child is the one we have knowledge on. With the coins: 2 coins have a 1/3 chance of being heads assuming at least one is a head, but the moment one of the coins is revealed as heads the odds become 50/50. My basic argument is that someone saw the coins and told me.one was heads so the coin was revealed, not to me, but to someone. So at that point the odds of the other coin go back to 50/50. But I guess if it is viewed that the person.looked at both coins before reporting to me that at least one was heads then the odds are 1/3 that both are heads.

God Texas Hold'em poker odds are so much easier... Counts the outs and.multiple by 100 then divide by the number of unexposed cards at that point in time (50, 47, or 46 unless a card was accidentally exposed). That tells you the odds the next card will complete your draw or improve your hand somehow.

What is the probability that a family with 2,000,000 children has 2,000,000 boys, knowing they have at least 1,999,999 boys?

Quote directly from.semck in reply 6...

" Suppose you know ONLY the following two facts about Mr. Jones and his family:

a) He has two children.
b) At least one of them is a boy.

Assuming equal sex distribution in society, what is the probability that the other one is a boy?"

He used other boy here implying knowledge of one.of them.

I did not imply knowledge of the other one, Draug, and even if you suggest that I did, I have at least 7 times since then been very clear about what I am talking about. As I pointed out quite awhile ago, the time to claim that you thought I was implying knowledge of a specific boy passed forever a good long while ago.

I'm glad you have come around to admitting you were wrong, as here:

"If a family has at least one boy then it is 1/3 they have two. I do agree with this."

However, I strongly resent your attempt to suggest that it was unclarity on my part that got you there. Many times I posted the following coin formulation:

"You know only that (a) two coins were tossed and (b) at least one was heads"

You also agreed explicitly to this:

"If you know only that (a) two coins were tossed and (b) at least one was heads, there is a 50% chance that both coins were heads."

Both contradict what you are now saying. Don't try to save face by pinning it on some alleged ambiguity in post 6.

What is the probability that a family with 2,000,000 children has 2,000,000 boys, knowing they have at least 1,999,999 boys?

Massive number of outcomes for 2000000 children… but the only ones allowed are
2000000 Boys
Girl, then 1999999 Boys
1999999 Boys, then Girl

3 outcomes, one with all-boys. Still 1/3.

"but the moment one of the coins is revealed as heads the odds become 50/50."

That question is equivalent to the first coin always being heads, in which case the probability of the other coin being heads is 50%.

Oh wait I totally confused myself with my own example :) ! Someone solve my riddle I actually am thinking through it without cheating but combinatorics has receded from my mind so much...

Tendmote, why does the girl have to be either the oldest or the youngest?

The probability is actually 1 / 2000001.

@tendmote: 1/(2^200,000 - 1)

No, philcore, that would be if we didn't know anything at all. If we know there are at least 1999999 boys, then the odds of 2000000 boys is 1/2000001.