Computing probabilities is difficult.  The following two problems are widely discussed and the results often not believed, because they are counter-intuitive.

In general, the best thing to do is to write out the set of all the possible end states, and then see with what frequencies the desired ones occur in that set.  This may be tedious or very difficult, but it does lead to the right answer.

The Monty Hall problem.

You are on a televised game show, faced with three doors.  By selecting a door you will "win" whatever is behind it.

One of the doors hides a car, the other two hide goats.  You select one of the three doors (but are not allowed to open it).  Then the show host opens one of the other two doors, and it reveals a goat.  You are now allowed either to stick with your first choice, or to switch to the remaining closed door.  If you want to win the car, what should you do?

Let's first write up all the steps with precision:

• the car and goats were placed randomly behind the three doors,
• the show host knows what is behind the doors,
• the show host always opens one of doors after your first choice,
• the door he opens is always one of the two that are not your first choice,
• the door opened always reveals a goat (because that is always possible, there being two goats)
• the show host has a poker face, so you cannot deduce from his expression whether your first choice was a car or one of the goats.

The set of all end states is composed of two parts:  those states in which you decided to stick to your first choice, and those in which you switch to the remaining door.  First the part where you stick to your original choice:

If you make a random choice between the three doors, you have a 1 in 3 probability to choose the door with the car.  It does not matter what the show host does with the other doors.  He might as well not open them.

Now the part where you switch away from your first choice.  Note that you can only switch to one door:  the one that was not opened and also not your choice.

Note also that the show host is not entirely free to open any door:  he can only open one of the two doors that do not correspond to your first choice.  If you chose a goat, he has no other choice than to open the door that contains the second goat;  his hand is forced.  If on the other hand you chose the door hiding the car, he then can open either one of the two remaining doors.  In that case he does have a choice, but as he wears a poker face, none of this gives you any information.

However, by opening a door which reveals a goat (and it always will), one of the goat choices has been eliminated.  You have more information than before.

There are three possibilities now:

• you originally chose the car, the remaining door hides a goat, you switch and you lose,
• you originally chose one of the goats, the remaining door must hide the car, you switch and you win,
• you originally chose the other goat, the remaining door must again hide the car, you switch and you win.

Therefore you win 2 times out of 3 if you switch, and only 1 time out of 3 if you stick to your original choice.

It is the elimination of one of the goat-options that changes the probability, and because in two of the three cases the show host's hand is forced, the resulting probability is that you have 2 chances in 3 to win.

The Sibling Brothers Problem

You are again on a game show.  The show host tells you the audience is composed of people who have exactly two children.  One person from the audience comes up and tells you:  "I have two children, one of which is a boy." The host now asks you to bet on the probability that the other child is also a boy.  What are the odds?

To make this problem tractable, we will assume a few points about children.  These assumptions do not correspond to reality, in which there are twins, children whose gender is not unambiguously determined, and where there are about 104 boys born for every 100 girls, and so on.  Instead we will asume a simpler world:

• there are exactly as many boys as girls born,
• the gender of a child is unambiguously male or female,
• there are no twins, children are born one after the other,
• the gender of the second child is independent of the gender of the first child,
• the audience represents a statistically valid sample of the entire population.

Under these assumptions, the audience can be divided into four parts:  those who have two boys, those who have two girls, those who have a boy followed by a girl and those who have a girl followed by a boy.

That means there are 50% who have children of the same gender and 50% who have children of different gender.

But it also means that only 25% have two boys, and that there are twice as many with mixed gender offspring than with two boys.

Note carefully that the order in which the children are born is of no importance in the statement of the parent, nor in the answer to the question.

To make the result clearer:  suppose the statement of the parent is only one of the four following scenarios:

1. The parent gets up and says nothing.  Then you have no information whatever, and the probability that he has two boys is 25%.
2. The parent says one of his children is a boy, but you do not know if it is the first child or the second.  You can only exclude the parents with two girls, leaving you with three equal parts, among which only a third actually have two boys.  The probability is now 33.333…%.
3. The parent says his first child is a boy.  That is more information than in case 2, because you can now not only exclude the cases with two girls, but also the cases where a girl was the firstborn.  Effectively there is a 50% chance that the second child is a boy and therefore that both children are boys.
4. The parent says both his children are boys.  Then of course the probability is 100%.

So the correct answer to the original question is:  you have 33.33…% probability that both children are boys.

As you get more information, you can narrow down the cases and get a higher probability of guessing correctly.