Just how bad is keno’s house edge?

Nov 17, 2009 5:03 PM
by Keno Lil |

There are some who insist that the house percentage is by far the most important criterion by which you should judge a keno ticket, and by extension the keno lounge/casino that books it. I hold the opinion that the house percentage is far more important to the casino than to the individual player, and I will try to explain my position.

The house percentage is really just an average of what the house should win over millions of dollars bet. So if you have a 25percent house percentage keno ticket, and a million dollars is wagered, the house should, on the average, win about \$250,000.

If you have a 30percent house percentage ticket, the house should win, on the average, about \$300,000 on a million dollars of action. For the house, this difference is a substantial \$50,000. For the player, with a bankroll of \$100, the difference is, on the average, about \$5. Right? Well, not really.

If you take 10 people and poll them on their incomes, and nine of them make \$10,000 per year and one of them makes \$1,000,000 per year, it is mathematically correct to say that their average income is \$109,000 per year.

Mathematically correct, but hardly coinciding with reality! Life just isn’t fair. In situations like this, that have a greatly unbalanced distribution, "skewed" in statistical parlance, the proper statistic to use is the median, which is the income that is exactly in the middle of the range of incomes. In this case, that figure would be \$10,000, which is much closer to reality than the mathematical average.

The distribution of keno winnings is of course highly skewed just like the income example above. Let’s work a little thought problem. A 6-spot "cycles" about every 7,753 games, so let’s say that we have 7,753 players, each playing a \$1 6-spot one time. We will, on the average have:

• one solid six, paying \$1,480

• 24 five out of sixes, each paying \$100

• 221 four out of sixes, each paying \$3

• 1,007 three out of sixes, each paying \$1

• 2,393 two out of sixes, each paying nothing

• 2,819 one out of sixes, each paying nothing

• 1,292 zero out of sixes, each paying nothing

If we add together the total wins, we see that the total is \$5,550, which is an average win of 71.5¢per ticket. The house percentage is thus \$1.00-.715, or about 28.5 percent.

The MEDIAN win, however is that catch which is 3876 catches from the top or the bottom, if all these payouts were arranged in order. That catch would be a one out of six, which pays nothing. Thus the MEDIAN payout on this ticket is zero, which is much closer to reality for the typical player than the average, or house percentage!

If you have a keno question that you would like answered, please write to me care of this paper, or contact me on the web via email at [email protected].com. Well, that’s it for now. Good luck! I’ll see you in line!

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