Is Keno a fair game? This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with probability, which perhaps yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

In the game of Keno, from the numbers $1-80$, a player chooses three, and makes a $\$1$ bet.
Then twenty numbers are drawn.
If all three of his numbers are among the twenty, he is paid $\$42$ (gain of $\$41$).
If two of his numbers are among the twenty, he is paid $\$1$ (break even).
If fewer than two of his numbers are among the twenty, he loses.
What are his chances ?
How can this be made a fair game ?

My thoughts:
A player's entry can be any of $C(80,3)$ combinations $= 82160$
From the the $20$ numbers there are $C(20,3)$ winning combinations $= 1140$
$\frac{1140}{82160} = .0138753 =$ Probability of having three winning numbers.
But now I am stuck.

P(win) = P(choose 3 correct #s) = ${20\choose 3}/{80\choose 3} = \frac{57}{4108}$
P(break even) = P(choose 2 correct and 1 wrong #) = ${20\choose 2}\cdot{60\choose 1}/{80\choose 3} = \frac{570}{4108} $
P(lose) = $1 - \frac{57}{4108} - \frac{570}{4108} = \frac{3481}{4108}$
Let x dollars be the net winnings if all 3 #s are guessed correctly.
For a fair game, net winnings = 0 = $\frac{57x}{4108} - \frac{3481}{4108}$, which yields
x = $\frac{3481}{57}, \approx 61.07 dollars $ , (paid $62.07)
