This question is from the book "Heard on the Street."
There are N = 25 people at a party. One person asks everybody to announce their birthdays and for anyone who has the same birthday as someone else to raise a hand. How many hands do you expect to be raised? For example, if John, Jon, Steph and Mark all have the same birthday, but no one else has a matching birthday, the number of hands is 4.
The solution process starts by determining the probability that a person does not have the same birthday as the rest of the group. The probability that another person doesn't have the same birthday is
$$ P(X_i \neq x_1) = \bigg(\frac{364}{365}\bigg) \ \ i = {2,3,...,N} \\ $$
The probability that N-1 people doesn't have the same birthday as person 1 is thus $$ \bigg(\frac{364}{365}\bigg)^{N-1} $$
The probability that person 1 has at least 1 matching birthday with N-1 people is thus $$ 1 - \bigg(\frac{364}{365}\bigg)^{N-1} \ \ i = {2,3,...,N} $$
This is true for each person, so the solution states that the expected number of hands is
$$ E[H[N]] = N [1 - \bigg(\frac{364}{365}\bigg)^{N-1}] $$
I'm a bit confused why this is the correct solution. Because I'm thinking that this solution overpredicts the number of hands. If we consider the situation where person 1,2,3 all have the same birthdays. When we do the expectation for person 1, we're counting the number of hands from persons 2...N. When we do the expectation for person 2, we're counting the number of hands from 3...N and also from person 1. For person 3, we're counting the number of hands from 4..N and 1,2. Wouldn't persons 1,2,3 each be counted twice?
EDIT 1
It look like I may have misunderstood the solution description. The solution states that $$ 1 - \bigg(\frac{364}{365}\bigg)^{N-1} \ \ i = {2,3,...,N} $$ is the probability that person 1 raises their hand, and not the probability that at least 1 person raises their hand from persons 2...N.
