# A pair of 8-spot way tickets offer mixed bag of wins

Aug 23, 2005 12:44 AM

This week’s column is a tale of two tickets.

I have used two different 8-spot payout rates to demonstrate the importance of frequency of win to the player.

I adjusted the payouts on each ticket to provide the same house percentage ( 25.25%) on each ticket. First, a little refresher on the odds against each catch on an 8-spot ticket:

Catch: Odds for one:

4 / 8 12.27

5 / 8 54.64

6 / 8 422.53

7 / 8 6,232.27

8 / 8 230,114.61

Let’s assume that under all circumstances you plan to play keno until you go broke, until your bankroll is exhausted.

Let’s also agree that this analysis will not be rigorous, but will be a "rough and ready" approximation to the real world.

Since both tickets below have the same house percentage and the same price, it follows that given an equal bankroll, both tickets will deplete your bankroll in the same number of games, on the average.

In this case, given a bankroll of \$100, the average player can expect to play 396 games before depleting his or her bankroll, on either ticket.

Ticket A Ticket B

8 SPOT TICKET 8 SPOT TICKET

PRICE \$1.00 PRICE \$1.00

0 / 8 0.00 0 / 8 0.00

1 / 8 0.00 1 / 8 0.00

2 / 8 0.00 2 / 8 0.00

3 / 8 0.00 3 / 8 0.00

4 / 8 0.00 4 / 8 1.00

5 / 8 9.00 5 / 8 8.00

6 / 8 100.00 6 / 8 90.00

7 / 8 1,480.00 7 / 8 1,480.00

8 / 8 25,000.00 8 / 8 15,900.00

House Pct.= 25.25% House Pct.= 25.25%

The problem is that the mathematical average in this case does not represent reality very well. The average includes those lucky players who hit 6-, 7- and even 8-out-of-8. To get a better picture of the real world, it is better to use the median figure in this case.

Let’s assume that you are a lunch-time keno player, and that your bankroll is \$12. On either ticket, the mathematical average says that you can expect to play 47 games before tapping out. But playing ticket A, which pays only on 5-out-of-8 or more, you have only a 25% chance of winning anything within 12 games, and a 75% chance of tapping out.

Thus, the median player will play 12 games on this ticket and go broke. If you play ticket B, however, the median player will expect to win one push for 4-out-of-8, and will thus be able to play a 13th game, and consequently have an 8% better chance of hitting the big one!

We could also make the assumption that you are a payday player, and have a bankroll of about \$55, and a few hours to play. Here the mathematical average states that you should be able to play about 217 games before tapping out.

If you are playing ticket A, you might expect to hit one 5-out-of-8 in this period, giving you an additional nine games to play. If you are playing ticket B, you can expect about four 4-out-of-8 hits and one 5-out-of-8 hit, for a total of 12 more games to play.

Or maybe you’ve decided to take a Saturday and play all day, and you have a bankroll of \$422. By playing ticket A, you might expect to hit one 6-out-of-8 and about eight 5-out-of-8s, and this will give you an additional \$172 to play with.

By playing ticket B, you will expect to hit about 34 4-out-of-8s, eight 5-out-of-8s, and one 6-out-of-8. This will give you an additional \$188 to play with, or about 16 more chances than with ticket A.

This is important because each additional chance that you have is one more chance at hitting "the big one." In this case, the big payoff is somewhat less on ticket B, but the 7-out-of-8 is the same on both tickets, and most 8-spot players are playing for the 7-out-of-8. All other things being equal, choose the ticket with the higher frequency of wins, since you, like me, undoubtedly have a limited bankroll!

Well, that’s it for now. Good luck! I’ll see you in line!