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Keno: an odd game

For the past 20 years, I’ve been going to Las Vegas for a weekend here and there and, just like everyone else, I couldn’t even stop gambling to eat; and why should I? The casino owners so graciously offer to take even more of my money as I munch on the buffet of the day with the lottery alternative, Keno.


In case you’re unfamiliar with the game, its basic version goes like this. On a grid with 80 spaces labeled with numbers 1 through 80, the player may choose between 1 and 20 spaces and mark them. The computer then randomly chooses 20 spaces (regardless of how many the player chose). A win is determined by how many spaces marked by the user were chosen by the computer (these are called matches).

Matches Payout
1-2 $0
3 $2
4 $5
5 $150
6 $3000
A possible payout table for picking 6 numbers in Keno.

People from all over the casino turn in $2 and a card containing their marks for the next game and the computer’s same 20 chosen spaces are played against all of them. Note that the more spaces the player marks, the more spaces must match in order for the player to win. In this example, the player chose 6 numbers, then hoped for 3 or more of those number to match. Please note the marked increase in the payout as the number of matches increases, this is what first got me interested in the game. I also wondered how the computer could control the total payout (in some cases a player can win over $100,000) when so many people are playing, and the computer must pick the same 20 numbers. I knew there was no way a casino would run a game where the odds weren’t in their favor, so I set out to determine the actual probabilites involved.



Problem: Given a Keno game as previously described, what is the probability that, of the X number of marks a player makes, Y of them are matches?


A helpful way to look at this problem was to reverse the order of play. The computer marks 20 numbers randomly (and secretly), and the player then chooses his X numbers, trying to pick those that the computer has already marked. We then break the problem into two parts: finding the probability of getting Y matches, and then of getting (X-Y) non-matches.


Notation: Before I get into the formulae, let’s go over some notation briefly:

  • N! = factorial, or N * (N-1) * (N-2) * … * 2 * 1
  • M(j=a,b)(F(j)) = Multiple, or F(a) * F(a+1) * F(a+2) * … * F(b-1) * F(b)
  • C(a,b) = Combination, or (a! / (b! * (a-b)!))

These are fairly standard math constructs, further discussion can be found in any Linear Algebra text.


Probability for Y matches:
The probability of the first match is fairly easy to determine. Basically, the player has a 1 in 4 (20/80) chance of picking one of the 20 computer marks. As we continue however, it is imperative to keep previous choices in mind. The probability of the next match is 19/79, then 18/78, etc. Combined, the chances of Y matches are: MatchProb(Y) = M(j=0,Y-1)((20-j)/(80-j)) which, through a little algebra, can be converted to: MatchProb(Y) = C(20,Y) / C(80,Y)


Next to be determined are the chances of the non-matches. This is similar, but the fact that we’ve already made Y choices cannot be forgotten. The probability of the first non-match is 60/(80-Y). The numerator is the number of possible choices that are non-matches, and the denominator takes into account the fact that there are only 80-Y choices yet available after all the matches have been made. The probabilities for the next non-matches are 59/79-Y, 58/78-Y, etc. Combined, the probability of (X-Y) non-matches is: NonMatchProb(X-Y) = M(j=0,X-Y-1)((60-j)/(80-Y-j)) which, again with the algebra, can be converted to: NonMatchProb(X-Y) = C(60,X-Y) / C(80-Y,X-Y)


By multiplying these two functions together, we get the probability that the player will choose Y matches, and then (X-Y) non-matches, when making his X choices. Unfortunately, we don’t care what order the matches come in, so we have a little more work to do. The formula for finding the number of ordered sequences that can be made from multiple pools of equal items is:

where

  • k1 = number of items in first pool
  • k2 = number of items in 2nd pool, etc.
  • kt = k1 + k2 + k3 … + kN

the total number of sequences = kt!/(k1! * k2! * k3! … * kN!)

In this case there are only 2 pools (matches and non), so the formula is: X!/(Y! * (X-Y)!). This just happens to be C(X,Y). Given this final factor, we can now offer a solution to the problem above: KenoProb(X,Y) = C(X,Y) * (C(20,Y) / C(80,Y)) * (C(60,X-Y) / C(80-Y,X-Y)). Finally, even more algebra can convert this to something a bit easier to read:


KenoProb(X,Y) = (C(X,Y) * C(80-X,20-Y)) / C(80,20)


Matches Probability Payout Product (est.)
1-2 N/A $0 $0
3 0.12982 $2 ~$0.26
4 0.028538 $5 ~$0.0856
5 0.0030956 $150 ~$0.46434
6 0.000129 $3000 ~$0.387
Total ~$1.19694

We can now take the payout table from before and find the payout percentage by multiplying the payout by the probability of the match. As shown here, the average return on a $2 bet would be about $1.20, roughly a 60% payout. I ran this on all the payouts tables on a Keno sheet from a casino (which shall remain nameless) and came to the staggering conclusion: Keno is not a good bet. The best payout I found was picking only one number, with a probability of match at 25% and a payout of $6 for a $2 bet, for a 75% return. Good luck!


The Count


Links:

  1. Keno on Wikipedia

3 comments to Keno: an odd game

  • Interesting that there were different payout rates.

  • Well, they seem to pick values for the payouts that had punch, but didn’t in any way leave them open to actually losing money. For larger numbers of selections, you’ll often see payouts of $50,000 or even $100,000 but, since the probability of winning is so small, there’s little danger to the casino.

  • i just started playing keno. its an interesting game. i am learning how to play. Your blog provides some useful info about keno. thanks for this info.

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