Spotting the right strategy in keno

February 03, 2015 3:08 AM
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Any time you have a game based on chance all kinds of math are involved. It can be as simple as basic arithmetic or as complex as binomial theorem. Keno has it all. Let’s go to the math of marking tickets and figuring ways.Any time you have a game based on chance all kinds of math are involved. It can be as simple as basic arithmetic or as complex as binomial theorem. Keno has it all.

First, let’s go to the math of marking tickets and figuring ways. You write a ticket grouped as 2-2-2-1-1, giving a total of 8 spots with 3 deuces (groups of 2) and 2 kings (groups of one). To find the total number of spots, add up the total number of spots in each group.

Finding the number of 7-spots is easy, too. Since you need seven numbers for a 7-spot and you have two kings, all numbers except one king is a 7-spot, and all numbers with the other king is the other, for a total of two 7-spots.

A simple axiom in Keno Kalkulations: If the ticket has 8 spots and you select all the 5-spots (without changing the group configurations), then the number of 3-spots has to be the same as the number of 5-spots. How? If you take 5 from 8, you have 3 remaining, and for each 5-spot possible, there is one corresponding 3-spot.

Let’s now go to the 6-spots from our ticket. Using the above we know the number of 6-spots must be equal to the number of 2-spots. Thus you have four groups of 2, the three deuces plus the two kings taken together. Since you have four groups of 2 you must have four groups of 6.

It gets more involved seeing how many 5-spots you have, but it’s not that difficult. Three is an odd number, so you either must use an odd number by itself or an odd number plus an even number to have an odd number. You also must examine all the combining possibilities.

You have three groups of 2 and three groups of 1 in this ticket. For each 3 you have a group of 2 and a group of 1. Since 3x2 = 6 there are six ways to have a 3-spot here. By the first axiom there are also six ways to have a 5-spot.

We have the 4-spot left to figure. Four is an even number and since the largest individual group is a group of 2, it will take two groups of 2, or a deuce and the two kings to make a 4-spot. You have two ways of doing the calculation – from left to right or right to left.

You have groups of 2-2-2-1-1. Go 2-2-0-1-1, then 2-0-2-0-0, then 2-0-0-1-1, then 0-2-2-0-0, then 0-2-0-1-1, and finally, 0-0-2-1-1 – six ways to make a 4-spot from this ticket.

However, if you have more groups and/or ways it gets very complicated. So what are we looking to do? We know on this particular ticket with only deuces and kings you either have to have two deuces and a deuce and two kings to make a 4-spot. To make things easier think of the two kings as a group of two as well.

In essence you have four groups of 2 from which you need to choose two groups to make a 4-spot. In math this is called combinations and written with subscripts such as 4C2. What this means is you are choosing two out of a group of 4. The mathematical formula is written 4!/(2!2!). The “!” is a factorial: 4! = 4x3x2x1 and 2! = 2x1, giving you (4x3x2x1)/(2x1x2x1) = 24/4 = 6. It confirms six ways to choose 3 groups out of 4.

But, there is a shortcut. It can also be written: (4x3)x(2x1))/((2x1)x(2x1)). Since you have a factor of (2x1) in both the numerator and denominator, you can “cancel” them out, leaving you with (4x3)/(2x1). This has sometimes been called the “Keno Formula.” Basically it means you can figure out keno combinations in this easy way.

For example, you want to go after the $2 progressive at the Orleans or Gold Coast. You write a ticket with five groups of 4, playing the 8s at $2 each.

How many groups do you have? We know two groups of 4 would make the necessary 8-spots, so we need to choose two groups out of 5. Thus we have 5C2, or 5!/(2!3!). It simplifies to (5x4x3!)/(2!3!). We “cancel” out the 3! in both the numerator and denominator, giving us (5x4)/(2x1) = 10.

Thus, if you have five groups of 4 and choose two of them to make your 8-spot you have ten 8-spots to play at $2 each, a $20 ticket. It does not matter how many in each group for this kind of calculation. As long as the groups are the same number of spots, choosing 2 of 5 or 4 of 7 or 16 of 20, the result will be the same number of tickets required, whether they are 2-spots or 5-spots.

Getting back to our original 2-2-2-1-1 ticket, we have one 8, two 7s, four 6s, six 5s, six 4s, six 3s, four 2s, and two 1s for a total of 31 possible ways.

You may also know the number of ways are 2 to the nth power minus 1 (n being number of groups), or 25 (2x2x2x2x2) = 32-1 =31, showing we have the correct totals for our groups.

Is learning all the above useful? Of course it is as it gives you increased flexibility to figure out good way tickets, especially for tournaments.

Pesach Kremen is a former UNLV Masters Gaming student, has won and placed in multiple local keno tournaments, and has written several academic papers on Keno. You can reach him at PesachKremen@GamingToday.com.

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