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I never was never able to figure Keno odds, but after placing a few $1 bets in Vegas a few years back I vowed to stay away. My intuition was the payoff was ridiculously low. I since have seen that was a wise move on my part.

In case you've never done the calculation, I'll offer you the option of four different games and ask you to determine the expected return. There are 80 numbers, from which you pick from 1 to 4, and from which the casino picks 20, at random. Your winnings depend on the number of "matches" between your numbers and the casino's.

In each case you bet $1, which the house keeps.

Your winnings are as follows:

  1. You pick 1 number.
    If it matches one of the 20 picked by the casino, you are paid $3.

  2. You pick 2 numbers.
    If they both are matches (with the casino's 20 numbers) you are paid $12.
    There is no payoff in only one of your numbers is a match.

  3. You pick 3 numbers.
    If all three are matches you are paid $42.
    If exactly two are matches you are paid $1, breaking even.

  4. You pick 4 numbers.
    If all four are matches you are paid $120.
    If exactly three are matches you are paid $3.
    If exactly two are matches you are paid $1, breaking even.

If you play ten $1 games of your choosing, what is your best expected return?

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We need to calculate the expected value for each case. The main part is calculating the probabilities.

For 1st case -> 79C19/80C20

For 2nd -> 78C18/80C20 when both nos. match

For 3rd -> 77C17/80C20 when all 3 nos. match, 77C18*3C2/80C20 when 2 nos. match

For 4th -> 76C18/80C20 when all 4 nos. match, 76C17*4C3/80C20 when 3 nos. match and 76C18*4C2/80C20 when only 2 nos. match

I think the 1st case is most beneficial (relatively) where expected value is -0.25 per game.

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