In the beginning, there was the 10-spot. The 10-spot ticket was the only ticket played in the traditional Chinese games, and for years this practice held in the Caucasian games as well.
When keno came to Nevada and assumed the guise of Horse Race Keno, a typical pay table might have looked like this:
Pick 10 Horses
If 5 of your horses are chosen, pays 2 for 1
If 6 of your horses are chosen, pays 18 for 1
If 7 of your horses are chosen, pays 142 for 1
If 8 of your horses are chosen, pays 800 for 1
If 9 of your horses are chosen, pays 1,600 for 1
If 10 of your horses are chosen, pays 10,000 for 1
These pay rates seem meager by modern standards, but it must be remembered that a new car could be purchased for $500 to $600 in the 1930s and 1940s when these tables were current.
The early operators of the keno games were not mathematical sophisticates; they did not know the actual odds against each particular catch on the 10-spot, but they did know that the traditional odds produced a profit for the game.
They did not possess computers, nor had they access to university mathematics departments, nor did they even have the pocket electronic calculators that are so common today.
They were, however, clever and resourceful. Given a 10-spot pay table that was known to produce a profit, the idea arose that other tickets might be produced using the 10-spot rates as a base.
On a keno ticket
It is sometimes useful to know how many distinct catches there are on a way ticket, particularly if one wants to compile a "cheat sheet."
Obviously, if there are several thousand catches on the ticket, such a project would not be feasible unless you have a computer program to automate the process. It is interesting to note that a way ticket always has the same number of total catches regardless of the ways that are played on it. The number of distinct catches on a way ticket may be calculated as follows:
Number of distinct catches = Combinations of (Size of group + number of groups) taken (Size of group) at a time.
EXAMPLE: A 10 way six, marked with five groups of three. The size of the group is three. The number of groups is five. 5 + 3 = 8. Eight things taken three at a time is equal to 56. There are 56 distinct catches on this ticket.
If the total catches on a mixed group ticket need to be determined, then the same procedure is followed, except that the results for each size group must be multiplied.
EXAMPLE: A way ticket with 3 groups of 3, 3 groups of 2, and 3 groups of 1. (3-3-3-2-2-2-1-1-1) Considering the groups of 3, 2 and 1 separately as above, and multiplying the results, we have (6 taken 3 at a time)x(5 taken 3 at a time)x(4 taken 3 at a time)= 20 x 10 x 4 = 400 total catches on the ticket.
Well that’s it for this week! Good luck, I’ll see you in the lounge! e-mail: [email protected]