Probability that two randomly chosen people out of 10 have birthday in the same month

Question

There are a group of 10 friends. What's the probability that two randomly chosen members of this group have birthday in the same month? How would it affect the probability if the group were composed of 100 members instead?

My intuition is that this probability is always $1/12$, no matter how many members this group had (of course more than one), because - assuming uniform probability distribution - more people, apart from more shared birthday possibility

Answer 1

Since there are only 12 possible birth months, we have a 100% chance of some two birth months being the same for a group of 13 people or more by The Pigeonhole Principle.

For groups of less than 13, the analysis can go as follows. What are the chances that 2 people don't have the same birth month? 11/12 yielding a probability of 1/12 that they do.

There are 12 ways of picking a birthday for 1 person. There remain only 11 ways to pick a a second person so that you don't have matching birth months. There's a total of 12^2 =144 ways to pick two different birth months.

In general the formula is: $P(k)=\frac{12!}{(12-k)!}\frac{1}{12^k}$ where $k$

is the number of people in the group. Where $P(k)$ is the probability of not having a matching pair. So the probability of there being a matching pair is $1-P(k).$

So for 10 people the probability of two having the same birth month is 99.6132%.

This assumes all months have the same number of days and all months have the same chances of people being born in them. Both assumptions are slightly off.

The chances hit over 50% for any group over 4 members large.

If $k=2$, $P(2)=12*11/144$, and $1-P(2)=1/12$.

So prob of 2 people no matter what sized group they are chosen from is 1/12 by the same analysis.

Answer 2

I'm sure that your answer of 1/12 is correct. When you randomly choose 2 people, the probability of person 2 having a birthday in the same month as person 1 would just be 1/12, assuming that the 12 months have an equal chance at being someone's birth month.

Answer 3

The first person $A$(say) is free to have any month as his/her birthday month so $P(A)=\frac{12}{12}$ . But the second person $B$(say) is compelled to possess the same birthday month as that of $A$'s so $P(B)=\frac{1}{12}.$ So required probability is $P(A).P(B)=\frac{1}{12}$