What is the probability of 12 friends each having birthday on different months of each other. What is the probability of 12 friends each having birthday on different months of each other. Twelve people having birthday on every month of the year.
We are a group of 12 friends and each one of us, have a birthday on a different month, so we cover the whole year. What is the probability of that?
Thanks
Leo
 A: The probability is low.
Suppose we write the months of the birthdays of each person in a list, starting with person 1, then person 2 and so on (assuming the people are numbered in some way). We get $12^{12}$ possible lists, we assume they are all equally likely, which is not the case in real life since some months are shorter, and even ignoring this, births seem to be more popular in certain parts of the year.
In this case there are $12!$ acceptable lists, that is $12!$ lists in which each month appears.
Hence the probability is $\frac{12!}{12^{12}}\approx 0.000053723217$
A: If you would like an alternative answer to Gamamals that is equivalent but more intuitive, I think, consider the following. Number you and your friends from 1 to 12, where you are #1. The probability that #2's birth month is different from your own is 11/12. The probability that #3's birth month is different from your own and #2 is then 10/12. Keep iterating this, then the probability you all have distinct birth months can be solved by multiplying those probabilities (as all birth months must be distinct to span all 12 months):
$$ \frac{11}{12} \times \frac{10}{12} \times \frac{9}{12} \times \frac{8}{12} \times \frac{7}{12} \times \frac{6}{12} \times \frac{5}{12} \times \frac{4}{12} \times \frac{3}{12} \times \frac{2}{12} \times \frac{1}{12} = \frac{11!}{12^{11}} = \frac{12!}{12^{12}} $$
where the last equality comes from multiplying both the numerator and the denominator by 12. I hope this serves as an alternative (and perhaps more intuitive) way to come by Gamamal's answer.
