Second pair of matching birthdays The "birthday problem" is well-known and well-studied.  There are many versions of it and many questions one might ask.  For example, "how many people do we need in a room to obtain at least a 50% chance that some pair shares a birthday?"  (Answer: 23)
Another is this: "Given $M$ bins, what is the expected number of balls I must toss uniformly at random into bins before some bin will contain 2 balls?"  (Answer: $\sqrt{M \pi/2} +2/3$)
Here is my question: what is the expected number of balls I must toss into $M$ bins to get two  collisions?  More precisely, how many expected balls must I toss to obtain the event "ball lands in occupied bin" twice?
I need an answer for very large $M$, so solutions including summations are not helpful.

Silly Observation:
The birthday problem predicts we need about 25 US Presidents for them to share a birthday.  It actually took 28 presidents to happen (Harding and Polk were both born on Nov 2).  We see from the answers below that after about 37 US Presidents we should have a 2nd collision.  However Obama is the 43rd and it still hasn't happened (nor would it have happened if McCain had won or Romney had won; nor will it happen if H. Clinton wins in 2016).
 A: Suppose there are $n$ people, and we want to allow $0$ or $1$ collisions only.
$0$ collisions is the birthday problem: $$\frac{M^{\underline{n}}}{M^n}$$
For 1 collision, we first choose which two people collide, ${n\choose 2}$, then the 2nd person must agree with the first $\frac{1}{M}$, then avoid collisions for the remaining people, getting $${n \choose 2}\frac{M^{\underline{n-1}}}{M^{n}}$$
Hence the desired answer is $$1-\frac{M^{\underline{n}}}{M^n}-{n \choose 2}\frac{M^{\underline{n-1}}}{M^{n}}$$
or
$$ 1-\frac{M^{\underline{n-1}}(M-n+1+{n\choose 2})}{M^n}$$
When $M=365$, the minimum $n$ to get at least a 50% chance of more than 1 collision is $n=36$.
A: We will change the problem slightly and make some approximations.  Instead of the expected value of $n$ to get two collisions, we will think about the value of $n$ to get a $50\%$ chance of two collisions.  They should be very close, as the probability of two collisions will rise quickly from near $0$ to near $1$.  In the single collision case, it is the difference between a factor of $\sqrt{2 \log 2}\approx 1.177$ and $\sqrt {\frac \pi 2}\approx 1.253$.  The approximation is that each pair of chosen elements has the same probability to match, $\frac 1M$.  This ignores the correlations between the pairs.
In this case, the distribution of number of collisions is Poisson, with $\lambda$ being the expected number of collisions.  The chance that we will not have two collisions is $(1+\lambda)\exp(-\lambda)$ with solution $\lambda \approx 1.67835$ so we want $\frac {n^2}{2M}=1.67835$ or $n=\sqrt{2M1.67835}$ or about $55\%$ more than the number to get the first collision.
