In reply to Rob Pratt.

To illustrate what Rob said, consider the case of N=2. Either no one pulls her own name or both do. Thus the expected number of people who pull their own name is 1, even though the probability that someone (one or more people) pulls her own name is 0.5.

The probability that exactly N-1 people pull their own name is zero, right? If N-1 pull their own name, so does the Nth person.