# Twelve-Spot ticket has dozens of pluses

Sep 24, 2002 6:23 AM

For some reason, twelve spots are good tickets to play. I like them almost as much as five spots. Twelve spots provide a chance to hit something big, but are safer to play than most any ticket, provided you have a hundred dollars or so in your bankroll.

If you can, try to find a twelve spot that pays for 0 or 5, they are usually better than average tickets. I don’t recommend adding the two spots to the twelve, if you have that kind of money to play, play it on more twelve spots.

Since the twelve spot can be divided into many different tickets it is very versatile. A simple grouping of 4-4-2-2 gives you two tens, three eights, four sixes, three fours, and two deuces in addition to the twelve. Or, 3-3-2-2-2 gives you one twelve, three tens, two nines, three eights, six sevens, two sixes, six fives, three fours, two threes and three twos. I kind of like the six way seven and six way five.

You can play deuce-way twelves also. Take 14 numbers, group them in sets of two, and you’ll have a seven way twelve. Or take eight groups of two-that’ll give you a 28 way 12.

Trey way twelves, constructed of groups of three: Five groups of three gives you a five way 12, while six groups of three gives you a 15 way 12.

Four groups of four make a four way twelve, and when played with the six way eight makes a very good ticket.

If you decide to give twelve spots a try, be patient. They don’t pay off too often but the pay outs can be good.

The cutting edge

Yes, I know that keno is often derided by gambling "experts" in the popular press, mainly because the "house percentage" is higher than blackjack, roulette or craps, and some slot machines. However, if we learned nothing else this week, it is the fact that even a small edge, (or even no edge at all) can and does work against the gambler’s bank roll over a long series of bets. Since the essence of the game of keno is, like a lottery, the attempt to win a large amount of money for a relatively small wager, this process actually works in keno’s favor.

Suppose that we have a modest (well very modest) bank roll of only \$2, and we are determined to win the sum of \$1,000 or go home broke. In reality, we are probably doomed to go home broke in this case anyway, it IS a long shot, but the utility of the round sum of a grand twinkles in our consciousness like Orion on a crisp desert night.

Suppose also that we want to try our luck at 21 in order to win the \$1,000. We sit down at a \$2 table, smile at the dealer and place our bet. We know that to win the thousand dollars, we must at some point have achieved a surplus of winning hands over losing hands of +500 units to achieve our goal. Now remembering from last week, IF the game of 21 is a 50-50 proposition, our chances of NOT achieving our goal (by going broke) are equal to 1- 2/1000, or 99.8 percent. In other words, we have about 1 chance in 500 of winning the grand.

But look at the situation in keno. Quite a few keno games offer a special rate five spot that pays \$1,000 for a solid five played for a dollar, and we know from previous calculations that given a keno ticket with a 25 percent house percentage, we can, on the average, play 8 games before we go broke. Since the odds against a solid five are 1551 for 1, by playing eight times we can cut the odds to 1551/8, or about 1 chance in 194 of winning the thousand dollars. Is this a magical trick? No, for two reasons: (1) You are avoiding the inevitable grinding effect of thousands of decisions upon your bank roll, and (2) keno also offers ADDITIONAL pay outs other than the \$1,000.

If you need more convincing, imagine that the house percentage at 21 is a mere 1 percent in the house’s favor. The odds, then, of getting to a grand without going broke (while playing only \$2 bets) are one chance in 11,895,701,628. (!!)

So on a scale of one to five, with five being the highest, if you want to win a large amount of money with a small bankroll, Keno Lil gives keno a five.

Here is the math:

If a player is currently holding h dollars, his probability of going broke before reaching g dollars is:

p{h} = r^h - r^g/1 - r^g

Well, that’s it for this week, good luck! I’ll see you in line!