Understanding the "Birthday Problem"

If you accept that having more than $2$ people with the same birthday is (for small numbers of people) much less likely than a single collision and can be ignored to within the accuracy used for this explanation/calculation, then the probability of a collision and the expected number of collisions are the same. The latter is (number of unordered pairs of people)/365 = $\frac{n(n-1)}{2} \cdot \frac{1}{365}$ for $n$ people. This will exceed $1/2$ when $n$ is about $20$.

The point is that the answer is approximately the square root of the number of days in the year, which much less than the intuitive estimate of half the number of days.