Think you understand Monty Hall? Try the Tuesday boy problem.

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Unfortunately, these questions you ask are ambiguous, and it is the failure to recognize how they are ambiguous that causes the results to seem unexpected. Consider two versions of what led up to the first statement:

Case #1: A father is chosen at random. He is given a slip of paper as he is led onto a stage. The paper says "Pick one of your children. Tell the audience the number of children you have, the chosen child's gender, and the day of the week on which it was born."

Case #2: A father is chosen at random from all fathers who have two children, including one boy born on a Tuesday. He is also ushered onto a stage and given a slip of paper that instructs him to tell the audience the criteria used to select him.

Now shift scenes. You are in the audience when a man is ushered onto the stage. He looks at a slip of paper, thinks a moment, and says "I have two children and one of them is a boy born on Tuesday." What is the probability that he has two boys?

The answer to the question depends on which case applies to the man you listened to. In Case #1, it is 1/2. In case #2, it is 13/27. Your simulation only covered the second case. To get the first, after you have two children, flip a coin to see which one the father will tell about. If it is not a Tuesday Boy, don’t keep that trial even if the other child is a Tuesday Boy. You will find that the 27 cases where you have a Tuesday Boy reduce to 14 (just over half, since one father didn’t need to flip the coin), the 13 where you also have two boys reduces to 7, and the answer is exactly 1/2.

If you simulate the simpler problem, where you don’t worry about the day of the week, the answers are 1/2 and 1/3 for the two cases, respectively. The reason 13/27 seems unintuitive, is because the fact that a Tuesday Boy was REQUIRED in the second case is not intuitively obvious from the statement "one of them is a boy born on Tuesday." In fact, as you point out, the puzzle could equally well be named after either of your two children, which is probably two different names. You choose one, just like the father in case #1, so the better answer to your question is 1/2, not 13/27. It is still ambiguous, but there is no valid reason to assume that case #2 applies.

However, I still fail to comprehend how the "at least one is a boy" quirk maths out to a 1 in 3 chance that his second child is also a boy.

Taken literally, it does. I understand that, in a set of data, GB is different from BG. But for the sake of our comparison, the order the children were born in doesn't matter. We're seeking if the other child is a boy, or not.

In my mind, the bit about it being the younger or older sibling is irrelevant information. We're comparing gender, not age. Regardless of if she is the younger sister, or the older sister, she's still his sister, and therefore not a boy.

I think the introduction of age is convoluting the issue, unnecessarily.

I now standby, ready to be proven wrong. I'd really like to wrap my head around this one, but I must insist that the age information is irrelevant.

We have no knowledge of child B so the probability is even for it being a boy or a girl.

> I have two children, at least one of them is a boy who is born on a Tuesday. What is the probability that I have two boys?

We have incomplete knowledge of both children, we are constraining a probability space in 2 dimensions (i.e. the 2 children) rather than refining it down to 1 dimension as before. In this case the probability is 13/27 as these tables (http://news.ycombinator.com/item?id=3290349) nicely show.

When the father says "one is a son born on a tuesday" this can be read as him selecting a child and then asking you about the second or as giving you some information that can apply to either child and then asking you about both.

In my opinion the language used leads much more easily to the first interpretation, this is the reason surely for the confusion.