I will try to get control of the most standard interpretation of our question by using (at first) very informal language. Let us call someone *unhappy* if one or more people share his/her "birthday." We want to find the "expected number" of unhappy people. Define the random variable $X$ by saying that $X$ is the number of unhappy people. We want to find $\text{E}(X)$. Let $p_i$ be the probability that $X=i$. Then $$\text{E}(X)=\sum_{i=0}^{n} i\,p_i$$ That is roughly the approach that you took. That approach is correct, and a very reasonable thing to try. Indeed have been *trained* to use this approach, since that's exactly how you solved the exercises that followed the definition of expectation. Unfortunately, in this problem, finding the $p_i$ is very difficult. One could, as you did, decide that for a good approximation, only the first few $p_i$ really matter. That is sometimes true, but depends quite a bit on the values $N$ of "days in the year" and the number $n$ of people. Fortunately, in this problem, and many others like it, there is an alternative *very* effective approach. It involves a bit of theory, but the payoff is considerable. Line the people up in a row. Define the random variables $U_1,U_2,U_3,\dots,U_n$ by saying that $U_k=1$ if the $k$-th person is unhappy, and $U_k=0$ if the $k$-th person is not unhappy. The crucial observation is that $$X=U_1+U_2+U_3+\cdots + U_n$$ One way to interpret this is that you, the observer, go down the line of people, making a tick mark on your tally sheet if the person is unhappy, and making no mark if the person is not unhappy. The number of tick marks is $X$, the number of unhappy people. It is also the sum of the $U_k$. We next use the following very important theorem: **The expectation of a sum is the sum of the expectations**. This theorem holds "always." The random variables you are summing *need not be independent*. In our situation, the $U_k$ are not independent, but, for expectation of a sum, that does not matter. So we have $$\text{E}(X)=\text{E}(U_1) + \text{E}(U_2)+ \text{E}(U_3)+\cdots +\text{E}(U_n)$$ Finally, note that the probability that $U_k=1$ is, as carefully explained by @Henry, equal to $p$, where $$p=1-(1-1/N)^{n-1}$$ It follows that $\text{E}(U_k)=p$ for any $k$, and therefore $\text{E}(X)=np$.