So, in my statistics class we discussed the recent huge jackpot, but we decided to focus on something more "trackable"—scratch tickets.
A part of the assignment is to figure out how many tickets are remaining given the following data: chances of winning, number of prize brackets, number of prizes in each bracket, and number of prizes remaining in each bracket.
I've made the following example in order to figure out an approach to solve this: 1 in 3 chance of winning, 3 prize brackets (e.g. $50, $25, $5), 1, 3, 7, prizes in each bracket (respectively) and 1, 2, 5, prizes remaining (respectively).
Obviously, we know for a fact that at least 3 tickets were sold (3-2)+(7-5) = 3, meaning that there are 30 tickets left. However, this isn't the best conclusion for how many tickets are remaining given that the chances of winning are 1 in 3.
I'm having trouble figuring out how the 1 in 3, would factor in to show how possibly more than 3 tickets were sold given the number of prizes already claimed. Any suggestions?
The distribution of prizes between buckets does not matter when estimating the number of tickets sold so far, so the useful information is, as you say:
One approach might be to say that $3$ prizes have been won so far and a third of tickets overall win prizes, so a reasonable estimate here is that $3 / \frac13 =9$ have been sold and that $33-9=24$ remain. This approach will give an unbiased estimator.
Another approach, which happens to give the same answer in this particular question, is to consider the likelihood of seeing this after $n$ tickets have been sold, which is (based on the hypergeometric distribution) is $$\dfrac{\displaystyle {11 \choose 3}{22 \choose n-3}}{\displaystyle{33 \choose n}}$$ and this is positive for $3 \le n \le 25$ and is maximised when $n=9$. So $9$ is the maximum likelihood estimate of the number of tickets which have been sold given that $3$ out of $33$ prizes have already been won, corresponding to $24$ tickets remaining.