# The Tuesday Birthday Problem - why does the probability change when the father specifies the birthday of a son? [duplicate]

I've most recently read about the Tuesday Boy Problem via twitter and I, as probably most other people, was sure that the probability has to be 1/2. After having read through a lot of solutions which were not identical at all, I've come to the conclusion that $P = \frac{13}{27}$ sounds the most reasonable. The corresponding argument was as follows:

Say the older child is the boy born on Tuesday. Then, if the younger child is female, there are seven possibilities and analogous if the younger child is male. In case the younger child is the boy born on Tuesday, for an older daughter, again there are seven possibilities, but for an older son, there are only six, because the case that the older son was born on Tuesday has already been counted. Thus,

$$P = \frac{6+7}{6+7+7+7} = \frac{13}{27}.$$

My first question is: Is this the correct solution? I've found other websites giving different solutions, however I could never agree with any of those.

In case this is the correct solution: Why does the probability change when the father specifies the birthday of a son? (does it actually change? A lot of answers/posts stated that the statement does matter) What I mean is: It is clear that (in case he has a son) his son is born on some day of the week. I could replace Tuesday with any day of the week and the probability would be the same. Say the father would have stated:

I have two children. (At least) One of them is a son.

Then, without loss of generality, we could say that this son is born on a Tuesday and again we would have $P = \frac{13}{27}$. But looking at an equivalent (?) problem, we get a completely different probability: Let's toss two coins consecutively and say heads is equivalent to "son" and tails to "daughter". Then, if we know that we tossed at least one heads, if it was the first tossed coin, the probability for another heads is 1/2. If the second coin was heads, then we can only take the case in account where the first coin was tails. So the probability for two heads is $P = \frac{1}{3}$.

So my second (actually third) question is: Where did I go wrong?

Lastly I want to ask (as my knowledge in probability/statistics is limited to what I have been taught in high school) whether the argument which gives $P=\frac{13}{27}$ for the original Tuesday Birthday Problem already takes the possibility of twins in consideration. Can it be regarded as a special case of two boys both born on Tuesday (I ask because in every argument leading to this probability there was a distinction between older child = boy vs. younger child = boy)?

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## marked as duplicate by Jonas Meyer, Lord_Farin, hardmath, N. F. Taussig, Mark BennetMar 6 at 23:51

–  Zev Chonoles Dec 1 '11 at 19:32
The older/younger distinction only matters insofar as it is a distinction. You could use other distinction you want: child who is better at math, child who is worse at soccer, etc. just as long as it distinguishes between the two children as two separate events. –  David Zhang Oct 8 '13 at 23:30
The wording of the problem is very important. The article you linked to included several versions of this problem, which one are you talking about? –  Flimm Mar 6 at 17:42

As Jason Rosenhouse points out in the blog post to which you linked, the correct answer depends very much on the assumptions made about the sample space. Specifically, it depends on what the speaker would say if he had a different set of children. Look at the three scenarios provided by Tanya Khovanova: in one of them the correct answer is $1/2$, in another it’s $1/3$, and in the third it’s $13/27$. In particular, if you assume that the speaker was randomly chosen from the pool of all men who could honestly say ‘I have two children, and one is a son born on a Tuesday’, $13/27$ is the right answer.

If the man says simply ‘I have two children, at least one of whom is a son’, the probability that the other child is a boy again depends on the sample space $-$ on the assumptions made about how the speaker was chosen. If he was chosen at random from the pool of all men who could honestly say ‘I have two children, and one is a son born on a Tuesday’, but simply made the weaker statement ‘I have two children, at least one of whom is a son’, the correct answer is $13/27$, as before. If, however, he was chosen at random from the pool of all men who could honestly say ‘I have two children, at least one of whom is a son’, the correct answer is $1/3$. And if he was picked at random from the pool of all fathers of two children and told you the sex of one of his children picked at random, then the correct answer is $1/2$. (These are, in reverse order, Tanya Khovanova’s three scenarios, modified for the revised statement by the father.)

These are bad puzzles, in the sense that they can’t be answered without making assumptions that go beyond what’s actually stated in the problem. Thus, there really is no single right answer. Rather, there are several answers that are right for different background assumptions.

For your final question, note that even with twins you normally have an elder and a younger child, even if it’s only by a very small time interval, so the elder/younger argument still works (in the settings in which it’s the appropriate interpretation).

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It all depends on how we got the statement from the father. So let's say we always knew the father has two children, and he made the statement "One is a son born on a Tuesday".

Now assume we gave the man a list with the 14 statements "one is a son born on a monday", "one is a son born on a tuesday", ..., "one is a girl born on a saturday", "one is a girl born on a sunday". And we asked him to read the first statement on the list that is true. Or the last statement on the list that is true. Or pick random statements from the list until he finds a true one, and read that.

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