When trying to break down the question into possible “outcomes”, so many (vos Savant et al) included either too few or too many possible combinations. So Mr. Smith has 2 children and one is a boy. What is the probability that the other child is a boy. The possible combinations are:

B/B
B/G

And that is all. Order does not matter.

Now, if one is so inclined to analyze it further, the same answer holds true. Most have presented the following combinations:

B/B
B/G
G/B
G/G

Eliminating G/G, leaves 3 possibilities, with only one including a second boy. 1/3. However, if one is going to include both B/G and G/B, implying birth order, then the possibilities must be:

B/b
b/B
B/G
G/B
G/G

Where B = the “known” boy. And eliminating G/G leaves 4 possibilities, with 2 including a second boy. 1/2.

Mark: yes, I pride myself on my stubborn-ness! I don’t think that the scenario you pose does actually create a paradox. It’s just that the questioner has reasoned incorrectly. The key point is that asking questions limits the options for Mr Smith. If we don’t know what has been asked, then options have not been limited. Assuming that the translator has no difficulty with the word ‘boy’, but doesn’t know which day was used, you are back to the classical version of the simple Gardner problem and the probability of two boys is 1/3 (it’s 1/2 if there’s also confusion about the words ‘boy’ and ‘girl’). If the uncertainty was just whether the day used by the translator was ‘Tuesday’ or ‘Wednesday’ then we’ve restricted Mr Smith’s options to these two days. This is not the same as restricting to one or other of these days. In fact, it’s not too hard to determine that the probability of two boys is 6/13. When we are asking questions, if we just ask whether Mr Smith has one boy, the probability of two boys is 1/3. The more restrictive the rest of our question is (i.e. the closer it gets to identifying one of the children), the closer the boy-boy probability gets to 1/2. An extreme example is if we ask ‘Is your elder child a boy?’ and Mr Smith says ‘yes’. Since restricting to only Tuesday or Wednesday is less restrictive than restricting to Tuesday, the result we get, 6/13, is closer to 1/3 than 13/27 is (the probability for the classical Tuesday’s child puzzle).

So, I still see no paradoxes arising with a Tuesday’s child probability of 13/27 when we ask questions and 1/2 if Mr Smith volunteers the information.

Mark: in which case I don’t see how Tanya’s paradox arises in the case where you ask the question.

Mule: You aren’t called S. Mule for nothing!

To make a puzzle that, even under your interpretations, generates Tanya’s paradox, we proceed as follows: You pose the question “Do you have two children, one of whom is a son born on a Tuesday?” to Mr Smith through an interpreter. He responds and the interpreter translates his response as “yes”. Mr Smith leaves and at this point you think the probability of two sons is 13/27. But then the interpreter says ‘Hmm, there may have been a translation problem there. I definitely used the word for one day of the week, and Mr Smith understood me. But I’m not sure it was Tuesday. Maybe it was Wednesday.’ You consider this and realise that if Mr Smith understood Wednesday then everything is essentially the same. The probability is still 13/27. But then the interpreter says ‘Well actually, I really don’t know which day of the week I said. I always get them confused. We don’t have these stupid weekdays in my native language’. You ask the interpreter what his best guess is as to which day he said and he responds that ‘It could equally well have been any day of the week’.

Now no matter what day you think Mr Smith understood, you will always get 13/27. But all you really know is that he has two children and one is a boy, which gives a probability of 1/3. The paradox is alive and well, living happily in the sunny exile of another rewording.

I have spent my spare time trying to independently reason why in the simplified two child probability puzzle in the classical solution we eliminated the GG and not the GB and BG false combinations to arrive at 1/3 as the answer with classical reasoning? Instead of 1/4? I was wondering why it was considered to be good reasoning to separately eliminate just one known false combination from the probability calculation in the classical solution?

If we don’t eliminate any of the false combinations we once again arrive at 1/4 or if we eliminate all the extraneous variables even eliminating the known sibling himself from the start and we only try to calculate the probability for the unknown child we arrive at 1/2.

Not considering myself a good mathematician this has been somewhat of a challenge but very interesting as an exercise in reasoning. When I came back to ask this question I noticed that you have already gone a long way to answering it for me he he :)

You have great timing :)

Cheers