Welcome to the Party!

People arrive at a party one at a time. Those who are still waiting for all the other guests to arrive amuse themselves by seeing if they share their birthday with anyone else. Up until the Xth guest arrives, no matches have been made. Let X be the guest who shares a birthday with one of the guests who arrived before. Find P(X = 3 or X = 4).

Solution: First we'd like to find the PMF and then sum the series of terms with X = 3 and X = 4. It's easier to start by considering the first k - 1 guests and use the fact that no one shares a birthday in that group. We could write out an expression that demonstrates this fact easily enough. Let the kth guest be the one who does share a birthday with one of the k - 1 guests.

$P(X\leq k-1) = 365*\left ( \frac{1}{365} \right )*364*\left ( \frac{1}{365} \right )...\left ( 365-k+2 \right )*\left ( \frac{1}{365} \right )$

We have k - 1 guests, so we have k-1 factors in the numerator and k-1 factors in the denominator of the following simplification of the above equation:

$P(X\leq k-1) = \frac{365*364...(365-k+2)}{365^{k-1}}$

When the kth guest shows up, we expect to get a match. There are k-1 ways to get a match with this element, and the birthday probability is given, as usual, as 1/365. So, now we have:

$P(X\geq k) = \frac{365*364...(365-k+2)}{365^{k-1}}*\left ( \frac{k-1}{365} \right )$

P(X = 3 or X = 4) =

$\frac{365*364...(365-3+2)}{365^{3-1}}*\left ( \frac{3-1}{365} \right ) + \frac{365*364...(365-4+2)}{365^{4-1}}*\left ( \frac{4-1}{365} \right ) \approx 0.0136$

It's worth trying this exercise with an application of Bayes' Rule...

$P(X=k(match)|X< k(no-match)) = \frac{P(X<k (no-match)|X=k (match))P(X=k (match))}{P(X< k (no-match))}$