Maximizing raffle EV

Question

Given a raffle where tickets cost A, B tickets are already in the barrel, and C winners are chosen (you can only win once) for a D prize, is there an equation for A, B, C, D that tells you how many tickets to buy to maximize your EV?

Answer 1

There's not enough information to answer the question. It depends on how many of the existing tickets belong to each person. In the extreme case that they all belong to the same person and more than $1$ winner is chosen, you should just buy a single ticket.

The calculation for the fully specified problem with arbitrary given numbers of tickets per owner is complicated and unenlightening, so I’ll solve the problem for the case (which you may even have intended to imply) of one ticket per person except ourselves.

Buying $n$ tickets costs $nA$ and yields an expected return of

$$ \left(1-\frac{\binom BC}{\binom{B+n}C}\right)D\;. $$

Taking the difference for $n+1$ and $n$ yields

$$ \binom BCD\left(\frac1{\binom{B+n}C}-\frac1{\binom{B+n+1}C}\right)=\frac{CDB!(B+n-C)!}{(B-C)!(B+n+1)!}\;. $$

The least $n$ for which this is less than $A$ is the optimal number of tickets to buy. For instance, if there’s a single winner, $C=1$, this is

$$ \frac{DB}{(B+n+1)(B+n)}\;, $$

and setting it equal to $A$ yields

$$ A(B+n^*+1)(B+n^*)=DB $$

and thus

$$ n^*=\frac{\sqrt{1+\frac{4BD}A}-2B-1}2\;. $$

If this is negative, we shouldn’t buy any tickets at all, and if it’s positive, we should buy $\left\lfloor n^*\right\rfloor$ tickets.