WTF? The probability is 1/2. We already know that you have one boy, thus that does not matter in terms of calculating probability. There are only two possibilities then, that the other child is a boy, or the other child is a girl. Assuming having a girl is equally likely as having a boy, the chance you have two boys is one in two.

His methods are wrong, because he is adding a hidden variable (action sequence). This is a well known error.

No. This is not at all similar.

His use of "about" shows that he is not doing the same thing.

In the original case, we **KNOW** that one of the children is a boy. Whether the boy is older or younger is utterly irrelevant, because we know two facts:

1) You *already have* two children (so no variation of probability applies to the number of children)
2) We *already know* that one of the children is a boy (thus no variation of probability applies to whether *at least* one of the children is a boy.

Thus, the ONLY variable is the sex of the other child. And, barring very odd genetic variations, there are only two options -- boy and girl -- and these are generally considered to be equally likely (50/50).

Therefore, there is a 1/2 probability that you have two children.

In the example above, both black and green balls are randomized. To be equivalent to the ORIGINAL scenario, you must proceed as follows;

You first give one ball at random to all your 1,000 guests. Then everyone who has a black ball is asked to raise their hands.

At that point, anyone who has a green ball is asked to leave. This leaves you with ("about") 500 people in the room *known* to have one black ball. You then pass out to these five hundred one more ball each, at random. The number who will then raise their hands to say they have now got TWO black balls will be ("about") 250, or, as I said, 1/2.

THIS is the equivalent of the original case.

Put it like this - you have two coins. One is hidden. The one you can see is heads-up. What is the probability both are heads-up? Not 13/27, not 1/3. It's one in two, because it depends entirely on a 50/50 chance.

But if I have two coins which you can't see, and tell you that one of them is heads-up, the chance that the other is heads-up is 1/3 - because you cannot know which coin I am talking about. See the black/green snooker ball analogy.

I am slightly too thick to join in any mathematical discussion, but thought I'd pop in and say I just loved the Daleks in your home. As Eccleston would say, fan tastic.
quarsan

It all depends, of course, on how this particular problem came to be stated in the first place. It's if it was in response to a query like, "tell me how many children you have, and then pick one child at random and tell me their sex and the day of the week they were born" that we get the 13/27 answer.

It's partly a difference in conventions about stating problems. Mathematicians have a set of informal conventions when they state problems: objects are point masses, ropes are weightless and inextensible, people make only true statements (or negations of true statements), coin tosses and die rolls are fair, you can't use the principle of relevance when analysing a statement (only its literal content), and so on. Without these conventions, mathematical problems would largely consist of tedious boilerplate.

The trouble comes, of course, when you take a problem for mathematicians and present it to non-mathematicians, people who aren't familiar with the conventions. 90% of any ensuing conversation involves getting the non-mathematician up to the point where they understand the problem (in its intended form). My exchange above with seawasp show how difficult this can be, even for problems much simpler than the "Tuesday" problem.

The maths is so basic I'm pretty sure I even did it at O-level, but what confuses people is that it's such an artificial and meaningless scenario that one automatically thinks of the more natural, mathematically distinct question "I already have a son born on a Tuesday, what's the chance the next child will be a boy?".

Me too. The original question is not clearly stated. I assume this was deliberate as I understand it was posed at a conference to stimulate debate.

Oh, dear God. That link has somebody getting the Monte Hall problem wrong, over and over again. Isn't there a circle of hell like that?

I would beg anyone who is disagreeing with the numbers presented by Peter Birks, to please go away and *do the experiment* described by Gareth Rees in the thread here, to answer this question:

I flip two coins randomly and hide them from you. You ask me, "did at least one of the coins land heads?" I answer "yes". What is the probability that both coins landed heads?

The answer to this is 1/3. It really, really is. If you are convinced that it is not 1/3, then you are incorrect, *but* it will not be possible for us to convince you of this. The only way to convince yourself is to do an experiment. It will take about the same amount of time as replying to this message.

You can do this yourself, with paper, pen, and two actual coins:

1. repeat sixty times:
- flip both coins.
- write down HH, HT, TH, or TT on a fresh line.

2. now go down the column, crossing out any TT lines.

3. now go down the column, ticking HH lines and crossing HT and TH lines.

4. how many ticks and crosses do you have?

Many years ago I didn't understand probability very well. Despite a good maths degree, I used to get into arguments about Monte Hall on Usenet. Finally I did a Monte Hall experiment, and then I understood.