13/27.

There are 14 different combinations of {gender,day}. So for a family of two as you describe, there are 196 possible combinations.

However, only 27 of them include a boy who was born on a Tuesday. In 14, the older child fits the description and the younger is any of the others. In 14, the younger fits and the older is any of the others. Since one combination got counted twice, there are 14+14-1=27.

Similarly, 7+7-1=13 have two boys.

Since a parent who answers “yes” to your question is equally likely to have any of these combinations, the answer is 13/27.

But your version isn’t how this puzzle originated. At a math-puzzle convention, a man named Gary Foshee announced “I have two children, and one of them is a boy who was born on a Tuesday.” He then asked for the same probability that he did. He said the answer was 13/27, but it really is 1/2.

Why is it not the same question? Because is all but the one case where he has two Tuesday boys, he could have formulated his question in two possible ways. We can only assume he chose randomly, so we can only count half of the cases where that was possible. The answer [(13+13)/2+1]/[(6+6)/2+1]=14/7=1/2.

This is the same logic that (correctly) solves the Monty Hall Problem, and the Bertrand Box Problem.

And the similar problem “You know that a certain woman has two children and at least one boy. What are the chances she has two boys?” Answer: 1/2, not 1/3.