birthday problem - which solution for expected value of collisions is correct?

Both expressions are correct but they are measuring different things, and the issue is how you count cases where three birthdays or more coincide.

Take a simplified example of three people and two possible birthdays. There is a $\frac34$ probability that only two birthdays coincide and a $\frac14$ probability that all three birthdays coincide.

• The first expression says that the expected number of people who share a birthday with somebody else is $3\left(1-(1-1/2)^{2}\right) =\frac94=2.25$; this is also $2 \times \frac34+3\times \frac14$.

• The second expression says that the expected number of birthday pairs is $\frac{3 \times 2}{2\times 2} =\frac32 = 1.5$; this is also $1 \times \frac34+3 \times \frac14$.

So in this small example, you can see that both expressions are correct, but the first is less than double the second because of what happens when all three people share the same birthday.

The same is true with your example, with the possibility of three items colliding causing the difference (plus the very small probability of more than three items colliding on the same value).

If you wanted the number of days where two or more people had birthdays, the expression would be different again: it would be $N - N \left(1-\frac1N\right)^n - n \left(1-\frac1N\right)^{n-1}$.

• In the small example this would be $2 - 2 \left(1-\frac12\right)^3 - 3 \left(1-\frac12\right)^2 = 1$ which is also $1 \times \frac34 + 1 \times \frac14$. In your large example it would be about $116.4045$.