Birthday Problem: Why isn't the probability 253/365 Consider a set of $23$ unrelated people. Because each 23  pair of people shares the same birthday with probability $1/365$, and there are $\binom{23}2 = 253$ pairs, why isn’t the probability that at least two people have the same birthday equal to $253/365$?

Let $A$ be the event that some two people have the same birthday. For $i < j$, let $A_{i,j}$ be the event that persons $i$ and $j$ have the same birthday. Then, $\text{Pr}(A_{i,j}) = \frac{1}{365}$, and your calculation is essentially that
$$
\sum_{1 \le i <j \le 23} \text{Pr}(A_{i,j}) = \sum_{i,j} \frac{1}{365}  = \frac{\binom{23}{2} }{365} = \frac{253}{365}.
$$
But unfortunately, $\text{Pr}(A) \ne \sum_{1 \le i <j \le 23} \text{Pr}(A_{i,j})$,
because even though $A = \bigcup_{i,j} A_{i,j}$, the events $A_{i,j}$ are NOT disjoint: it could be that multiple pairs of people have the same birthday.
On the other hand, the total number of pairs sharing a birthday is $1$ birthday for each $A_{i,j}$ that occurs; therefore the expected number of pairs sharing a birthday is exactly what you have calculated:
$\sum_{1 \le i <j \le 23} \text{Pr}(A_{i,j}) = \frac{253}{365}$.
