Expectations Galore: Keno

Keno is the next game we examine based on expectation. About six months ago a message was posted on rec.gambling.poker, by someone whose name I've forgotten (just as well), in which the calculation for the probability of getting six numbers out of six was wrong by a large factor of 10. Thus, we start by showing how to make this kind of calculation. In fact, let's use a ticket with six numbers selected as our example. First, the total number of possible draws is C(80,20) since we are drawing 20 numbers from a total of 80 numbers. In order to determine the probability of having all six numbers drawn, we simply count how many draws have our six numbers included. Since our six numbers are to be included in the draw, we need to count how many ways the remaining 14 numbers may appear. These 14 numbers are being chosen from seventy-four numbers because essentially we have set aside our six selected numbers to appear in the draw. This means there are C(74,14)draws which will include our six numbers. Therefore, the probability of all six of our selected numbers appearing among the 20 numbers drawn is C(74,14) divided by C(80,20). If one carries out the simple calculation, the probability obtained is .000129 which is about 1/7,753.

Let's determine the probability of getting exactly five of our six numbers drawn. There are C(6,5) = 6 ways of choosing which set of five numbers will be in the draw. The remaining 15 numbers must then be chosen from the 74 numbers not on our ticket because we must exclude the possibility of all six numbers appearing. Therefore, the number of draws with exactly five of our numbers is 6C(74,15). Dividing this by C(80,20) gives us a probability of .00310 which is about 1/323.

Similarly, to determine the probability of getting four of the six numbers, we observe there are C(6,4) = 15 ways of choosing four of the six numbers and there are C(74,16) ways to choose the remaining sixteen numbers. Hence, there are 15C(74,16) draws in which precisely four of our numbers appear. Dividing by C(80,20) gives us a probability of .02854 that precisely four numbers appear in the draw. This is about 1/35.

Finally, for precisely three numbers in the draw, there are C(6,3) = 20 ways to choose the three numbers and C(74,17) ways to choose the remaining 17 numbers. Multiplying these two numbers and dividing by C(80,20) yields a probability of .1298 that exactly three numbers appear in the draw. This is slightly better than 1/8.

The payouts for keno are not the same for all casinos. However, the differences are insufficient to make much difference in the expected values. I am going to use the payouts I found at a major Las Vegas casino in August 1999. For all six numbers you receive $2,000; for five numbers $80; for four numbers $4; and for three numbers $1. Be aware these amounts are not your winnings since you have already paid $1 for the ticket. You must subtract one dollar from each to get your winnings. Multiplying the winnings by the various probabilities gives an expectation of

The expectation just worked out is very bad for the player and very good for the casino. In spite of this many people play keno. This column deals with mathematics and poker (or more generally mathematics and gambling). I have no intention of discussing psychology and gambling but the results just obtained and the fact the game is popular certainly indicate the psychology of gambling is an important topic. Somehow the possibility of a large payout for a small wager makes people willing to play a game with a terrible expectation.

The table below gives the probabilities and expectations for various choices of numbers played for keno. The expectations shown are the expectations for playing that many spots and not just the expectation for the line on which it is written.