# Formula defines keno bankroll

Jan 6, 2009 5:04 PM
Keno Lil |

There is a simple formula which I discovered a few years ago. It tells you how many betting propositions you can make, on average, given a particular bankroll, house percentage, and size of wager, before you go broke.

This formula applies to events in the long run and does not necessarily reflect reality in the short term. It does give us some clues as to how we should behave when playing a Keno ticket.

The astute gambler will understand immediately why this formula must apply. The mathematically challenged will have to take my word for it!

Here it is: Bankroll divided by house percentage times size of wager. Example: Given a \$100 bankroll, a house percentage of 30, and a wager of \$1, the formula looks like this: \$100 / .30 X \$1 = 333.33. This means you can wager on the average 333.33 times before you go broke.

The house percentage must be stated in its decimal equivalent. Given the eight spot ticket that pays \$20,000 for 8/8, \$1,480 for 7/8, \$100 for 6/8, and \$9.00 for 5/8, we have an eight spot ticket with a house percentage of 27.42. Applying the above formula for a \$1 ticket, we can expect to play 36,469.73 games before going broke, given a \$10,000 bankroll.

I programmed my computer to play \$10,000 for each of 5,000 players, with an initial bankroll combined of \$50 million. You see, my Monte Carlo simulation took about a half hour, and resulted in an average of games played of 35,988 per player, very close to the theoretical, and a total played of \$179,940,000.

The player with the fewest games played had 14,648 games, a ratio of 1.46 games for every dollar of his original bankroll. One lucky player was able to play 925,522 games before going broke, a ratio of 92.5 games for every initial dollar. He obviously hit several solid eights.

The median number of games played was 26,604.5, almost 10,000 less than the arithmetic average. Bottom line: The median figure more closely describes the "real world" in this case, a "real world" alas in which most won’t hit that solid eight.

Given the median figure, what are the realistic chances of hitting a solid 8 with a \$10,000 bankroll on a \$1 ticket? About 1 in 9.16, compared to 1 in 6.82 using the theoretical figure.

If you have a Keno question write to me c/o GamingToday or e-mail me at [email protected].