This is Problem 2.2 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.
You go to a party with 500 guests. What is the probability that
exactly one other guest has the same birthday as you? Calculate this
exactly and also approximately by using the Poisson PMF. For
simplicity, exclude birthdays on February 29.
To solve it, I used the equation $${499 \choose 1} \left({364 \over 365}\right)^{498} \left({1 \over 365}\right) = .348$$
I made a tree-and-leaf diagram where the root is your particular birthday and then there are 499 branches. Each corresponds to being matched with a particular person and each of those branches has two outcomes: the person has the same birthday or the person has a different birthday.
This seems like a total probability problem. But, the binomial formula suggests to multiply each of these possible outcomes together, instead of adding them.
Why isn't the binomial formula $$\sum_{i=1}^{n \choose k} p^k (1-p)^{n-k}$$
or, specific, to this case: $$P(B) = \sum_{i=1}^{n=499} (1/499 * 1/365)$$
The problem, as diagrammed in the tree-and-leaf visual, appears to be the same as a total probability theorem problem.
Best Answer
In this case, let $B$ be the event that exactly one other person has your birthday, and let $A_i$ be the event that the $i$-th person (besides you) has your birthday. Then $P(A_i) = 1/365$, and $P(B | A_i)$ is the probability that no other person besides the $i$th has that birthday, which is $(364/365)^{498}$ by the multiplication rule, since the chance of any particular person not having your birthday is 364/365 and these events are independent. If $B$ is true than the $A_i$ are mutually exclusive (since only one other person can have your birthday). Thus
\begin{align*} P(B) &= \sum_{i=1}^{499} P(B \cap A_i)\\ &= \sum_{i=1}^{499} P(A_i) P(B | A_i) \\ &= \sum_{i=1}^{499} \left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{498} \\&= 499\left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{498} \\ &= {499 \choose 1}\left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{498}. \end{align*}