Trying to help my dad stop gambling using combinatorics My dad plays keno but always (mostly) loses. I want to explain to him in a way the odds are clear.
20 numbers are picked per game from the draw and the player picks 10. There are 80 numbers total to choose from. I hope I explained that clearly.
So, I thought it would be 80 choose 10 but I’ve seen another answer say it’s (equivalently) 20 choose 10/ 80 choose 10. Much appreciation to get an answer for this.
Here is the payout table:
Keno payout table
 A: For your dad to win the biggest prize, he would have to pick ten numbers of the winning twenty, so there are $20\choose 10$ ways he can win. The total number of ways he can make his pick is $80\choose 10$, so his chance of winning a million dollars is ${20\choose 10}/{80\choose 10}\approx 1.12\cdot 10^{-7}$.
However, that's not the only way he can win. He might also have anywhere from four to nine of the winning numbers. For him to have $k$ of the winning numbers, he would have $20\choose k$ ways to pick those numbers, and then $60\choose 10-k$ ways to pick the rest. Note that when $k=10$, that ${60\choose 10-k}={60\choose 0}=1$, so this generalizes the previous formula. There are still $80\choose 10$ ways he can make his choice.
With this, we can compute the expected value of this game, which means for each winning outcome, computing the probability that he wins, multiplying the probability by the prize money, and adding these values up. Let $p_k$ denote the prize money for having $k$ winning numbers (assuming $p_{10}=10^6$). Then the expected value is
$$\frac{\sum_{k=4}^{10} {20\choose k}{60\choose 10-k}p_k}{80\choose 10}$$
By running a short python script, we can see that the expected value is around $\$0.65$. If your dad is paying more than $65$ cents when he buys ten numbers (which he probably is), then on average, he'll lose money per ticket.
