Expected return from casino "Keno" game Keno is a popular game in many gambling casinos. In one version of this game the casino selects 20 numbers at random from the set of numbers 1 through 80. A player selects 10 numbers. A win occurs if at least one of the player's chosen numbers match any of the 20 numbers selected by the casino. The payoffs are as follows: 
Keno Payoffs
Number of matches         Dollars won for each bet
-----------------         ------------------------
0-4                           0
5                             1
6                            17
7                           179
8                          1299
9                          2599
10                        24999

What is the expected net gain for the player? Is this a fair game?
Attempt of a solution: The probability that the player will select 1 winning number is .05 ÷ $80 \choose 20$ which .05 for the number he selects (1 number out of 20) divided by $80 \choose 20$ (numbers drawn by casino) which is very unlikely, about $1.41\cdot10^{-20}$. 
So take this probability a multiply by 1, 2, ..., 10 (for the ten chances he will get a number 1, 2, ..., 10 times correct. Now we just multiply each of these by the payoff in dollars and add to get his expected net gain. The game is NOT fair because the odds are stacked so much against him. 
Can this be correct? I'm getting answers which converge toward 0, so I think I'm off somewhere

You are right the number of possible draws is $80 \choose 20$.  The number of ways to get 1 number right is $20 \choose 1$(right numbers to choose)*$60 \choose 19$ (choosing the wrong numbers).  This is much higher than you got.  My memory from calculating this (maybe with different numbers) was the house edge was around 21%, which explains why they work so hard to get you to play Keno.
