Is Keno a fair game?

Question

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.

Answer 1

This is what I think could be correct...

$P$ (win) = The probability that all $3$ of his numbers are among the $20$ chosen is as you said = $20c3/80c3 = 0.01388$

$P$ (break even)= (not sure if this is correct) The probability that 2 of his numbers are successful = ($20c2 \times 20c1 \times3c2)/80c3$ $=$ $0.1388$

The way I calculated this was because there are 20c2 ways of choosing 2 correct combinations, and then 20c1 ways of choosing a wrong number, but there are 3c2 ways of this whole event happening anyway (correct correct wrong, correct wrong correct, wrong correct correct)

$P$ (lose) = 1- ($0.01388+0.1388$) $= 0.8474$

So, if he does say $100$ bets, each betting $\$1$, on average $84.74$ of the bets will be lost, thus $\$84.74$ is lost.

On average $13.88$ bets will result in no net outcome.

On average, $1.388$ bets will result in a win of $41 dollars.

So, net outcome = $41 \times 1.388 - 84.74 = - \$27.83$ dollars.

So this game isn't fair.

In order to make this game fair, we solve this equation:

$1.388x-84.74=0$

$x=61.05 $

Thus for the game to be fair, one must win $\$61.05$ (be paid $\$62.05$) if all $3$ of his numbers come in the chosen $20$.

Answer 2

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)