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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

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I calculated the odds last time I was in Las Vegas - Keno is by far the worst game you can choose to play in a casino (assuming your aim is to minimize the house's edge). The best game for most people to play, incidentally, is blackjack (but not video blackjack!) –  Chris Taylor May 1 '12 at 17:02
    
@ChrisTaylor: Blackjack odds depend on how you play, but if done correctly you can be close to even without counting. Craps is only -0.5% or so on the pass line, probably better than most people actually achieve at blackjack. –  Ross Millikan May 1 '12 at 17:40
    
@RossMillikan You're right that it depends on how you play. I was assuming basic strategy, which most people should be able to learn in an afternoon - and in any case, the dealer often tells you what the optimal play is! Where are you getting your figure for craps from? The figure I've seen quoted is >1% house edge for pass/come, e.g. here. –  Chris Taylor May 1 '12 at 18:13
    
@ChrisTaylor: I was remembering a definition of difference from 50%. You are right that if you play 495 on the pass line you expect to have 488 left, a loss of about 1.4% –  Ross Millikan May 1 '12 at 18:23

1 Answer 1

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.

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Thanks. Would you know how to get the probabilities of getting 0-4 matches, 5, 6, 7, 8 , 9 and 10 ? I'm stuck here now –  EulerChild May 1 '12 at 17:27
    
To get 5 right is ${20 \choose 5} \cdot {60 \choose 15}$ –  Ross Millikan May 1 '12 at 17:38
    
right, so to get the probability of getting 5 right we just divide by (80 C 20). is this correct? –  EulerChild May 1 '12 at 17:43
    
@EulerChild: that's right –  Ross Millikan May 1 '12 at 17:47
1  
@EulerChild: That isn't surprising. They are picking 1/4 of the numbers, so you would expect 1/4 of the numbers you pick to match. That is 2.5/10, so 2 or 3 should be most probable. –  Ross Millikan May 1 '12 at 18:18

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