Splitting the birthday paradox into months

Question

One of the biggest studies in my universities' statistics class is the Birthday Paradox. It main goal is to determine that, given n students in a class, what's the probability that at least two of these students share a birthday.

Now, suppose there are k students in a class and every student is equally likely to be born in any of the twelve months of the year. One of the questions we were asked is:

What is the minimum value for k such that the probability there are two (or more) students born in the same month is at least 1/2?

I modified the original equation to try and account for the fact that this takes count of months and not the entire year.

P(A) = Event that two people have the same birthday in the month.

Thus, $$P(A') = \frac{k! \cdot365 {365 \choose 30}}{365^k}$$

Where the ${365 \choose 30}$ is taking into account a specific given month. (This is where I believe I'm making an error)

Then, the answer can be achieved by $$P(A) = 1-P(A')$$ Finally, I can use some bounding and solve for $$P\left(A \geq \frac{1}{2}\right) $$

However, I'm missing lots of calculations in between the lines which I am not sure how to achieve. How can I go about solving this problem appropriately?

Answer 1

First you should understand how to solve the original problem: Given $k$ students in a class, what is the probability that there are two or more students born on the same day? Let $A$ be the event that there are two or more students who have the same birthday. Then $$P(A^C)= \frac{365}{365}\cdot \frac{364}{365}\cdots \frac{365-k+1}{365}= \frac{365\cdot 364 \cdots (365-k+1)}{365^k}=\frac{365!}{(365-k)!} \frac{1}{365^k}$$ because the $1$st person can have any birthday, the $2$nd must have a different birthday, the $3$rd must have a birthday different from the first two, and so on.

We could also write this as $$P(A^C)= \frac{365!}{k!(365-k)!} \cdot k! \cdot \frac{1}{365^k}= {365 \choose k} k! \frac{1}{365^k}.$$ It is also important to note that this formula only works for $k$ up to $365$. After that, the probability is always $1$.

Now we want to switch from days to months. $365$ represented the number of "categories" each person could fall into. If we switch to months, there are only $12$ "categories," so the $365$ should change to $12$. But the rest of the derivation remains the same. Do you see how to continue from here?

Answer 2

If you understand the solution of the birthday problem, just replace the number 365 (days) by 12 (months). This with the approximation that months are equiprobable even though they have different number of days (as well as other caveats such as season, Valentine-day effect and so on!).

You want $$ p=\frac{12^n-\frac{12!}{(12-n)!}}{12^n}\geq .5 $$ which happens for $n=5$ people ($p=.618$).