Birthday problem where all days have different probabilities So, the site seems to be filled with birthday problem questions, but most presume uniform probability for all days as far as I can tell.
I am looking for a way to calculate the probability of no collisions in the set, decent approximations are good enough for this.
$p =(1-q)^{\frac {N^2}2}$ has previously been given to me as a way of approximately calculating this probability, where $q$ is the probability of a pair matching, and $N$ is the number of people, which seems simple enough where all days are equally likely.
My question therefore is, where the days all vary in their likelihoods, is $q$ simply an average of all the probabilities of a collision at each day, and does this seem a reasonable mathematical approach to take?   
 A: If each birthday is selected at random using the same probability distribution, this is a reasonable approach.  The $q$, however, should be the chance of a match for any given pair, so should be the sum of the squares of the probabilities of each date.  So if each date has probability $r_i, i=1\dots n, q=\sum_{i=1}^nr_i^2$.  In the uniform case this gives $q=\frac 1n$, but in the nonuniform case it will be higher.  This explains Peter's first observation.
A: Important quantities in this problem are probabilities of $n$ independent variables from the distribution of birthdays to be the same, i.e. 
$$
   \mathcal{p}_n = \Pr(X_1=X_2=\ldots=X_n)
$$
Large $n$ approximation of the probability of no duplicates was discussed in the paper of Shigeru Mase, "Apprimxations to the birthday problem with unequal occurrence probabilities and their application to the surname problem in Japan," Ann. Inst. Stat. Math., vol. 44, no. 3 (1992) pp. 479-499.
Here is the next order approximation for probability of finding a duplicate (see p. 492):
$$
    \Pr(\text{duplicate}) = \exp\left(-\frac{n(n-1)}{2} p_2\right) \cdot \exp\left(-n(n-1)(n-2) \left(\frac{p_2^2}{2} - \frac{p_3}{3}\right)\right)
$$
