# Combinatorics problem related to Birthday Problem from Introduction to Probability

Problem

A group of 50 people are comparing their birthdays (as usual, assume their birthdays are independent, are not February 29, etc.). Find the expected number of days in the year on which at least two of these people were born.

Solution

By linearity of expectation, the answer is 365 times the probability that at least two people were born on a given day. For a given day, there are 50 choose 2 or 1225 ways to choose two people who are born on that day and since the remaining people could be born on any day there are 365^48 choices for their birthdays. Dividing by 365^50, the number of possibilities with no restrictions, and multiplying by 365 yields, the expected number of days on which at least 2 people were born is 365(1225*365^48/365^50) = 1225/365, which is incorrect.

• Say A, B, and C share the same birthday. When you choose the 2 people, you can choose A and B, A and C, or B and C. Jul 30, 2020 at 1:24
• Your way of getting the probability that two people share a given birthday results in an over-count. I suggest going another route: Compute the probability that zero or one of the people has their birthday on a given day. Then subtract that from $1$ to get the probability that two or more share that birthday. Jul 30, 2020 at 1:34

I think the linearity of expectation approach is fine, but like I said in my comment, you didn't compute the probability of $$2$$ or more people sharing a given birthday correctly.

Take January $$1$$ as the birth date. This is a binomial distribution question. We can get the probability of $$0$$ people having that birthday by doing:

$$\binom{50}{0}\cdot(\frac{1}{365})^{0}\cdot(\frac{364}{365})^{50}\approx0.8718$$

And the probability of one person being born on that date as

$$\binom{50}{1}\cdot(\frac{1}{365})^{1}\cdot(\frac{364}{365})^{49}\approx0.1198$$

Then we get the probability of $$2$$ or more January $$1$$ birthdays by subtracting:

$$1-(0.8718+0.1198)\approx0.0084$$

Finally, we take your suggestion of linearity of expectation to get the answer:

$$365\cdot(0.0084)\approx3.0757$$ which is different from your value of $$\frac{1225}{365}\approx3.356$$

There are indeed $$1225$$ ways to choose two people out of the $$50$$ to be born on January $$1.$$

One of those ways is to choose Alice and Bob. Another way is to choose Bob and Carol. Yet another way is to choose Alice and Carol.

Now for each of those three choices (and each of the $$1222$$ others), you say there are $$365^{48}$$ ways to distribute the other people's birthdays. And to get all the possible ways to have at least two people born on January $$1,$$ you add all $$1225$$ sets of $$365^{48}$$ ways together.

This is fine if each way to choose two people and then distribute the other $$48$$ birthdays is a distinct outcome from every other. You can add the probabilities of disjoint events.

But these are not disjoint events. The case where Alice, Bob, and Carol all are born on January $$1$$ is counted when you choose Alice and Bob, then again when you choose Bob and Carol, and again when you choose Alice and Carol.