When you play a lottery and win a pari-mutuel Jackpot, there's always the possibility that one or more other players will also have won, and you'll be sharing (splitting) the prize money. Can you calculate what the Expected Value is of the Jackpot prize, given you are a winner? Certainly!
We need to assume something first: every ticket bought is a "Quickpick" - so any given set of numbers (a single ticket) can be randomly replicated on any other ticket. Suppose there are "N" such tickets sold (1 of which is yours, matching the winning combination). Suppose the probability of a ticket matching the winning combination is "p". Then, there are a total of "N" possible outcomes of "k" winners *besides* you, all the way from k=0 to N-1. Obviously "k" being more than a few is going to be extremely unlikely, but these terms are still part of the equation.
To explain Expected Value briefly, it is the sum of the product of each outcome by its probability (in other words, a weighted average). In our case the first part of the product is our share of the Jackpot (e.g. if one other winner, we get $Jackpot/2, if two other winners, we get $Jackpot/3, and so on. Let's use the symbol "J" for the nominal (advertised) Jackpot - what we would win (before taxes of course) if nobody else has a winning ticket.
EV(Jackpot-$return | I am a winner) = Sigma_(k=0 to N-1){[J/(k+1)] * C(N-1,k) * [(1-p)^(N-k-1)] * p^k}
Here "*" means multiply and "^" means to the power of. Sigma means sum each term (in this case, with the index k from 0 to n-1).
Also, C(n,r) = n! / [r! * (n-r)!] , the number of combinations of r items taken from n (where r <= n).
Try the formula yourself on any lottery. Assume a Jackpot and N. The problem you will likely run into with C(n,r) is that for any large N, say 100000000 for example, a math overflow error will result. This is because the factorials are calculated first, and anything above ~70! will be too large of a number to store. The way to get around it is to create function that will calculate the logarithm of C(n,r). Thus, for example, LN(N!) becomes LN(N) + LN(N-1) + LN(N-2) + ... + LN(2) + LN(1). Do this for the numerator and denominator on the right-hand side, and then convert the overall result back to C(n,r) with the EXP() function. For a typical lottery, anything after the ~5th term in the EV summation will probably be small that it can be ignored.
