Fremont the route to keno 'Good Vibes'
July 25, 2017 3:00 AM
by Pesach Kremen
There are times you need to figure the number of ways on a ticket to see how much it will cost to satisfy the number of ways required for a minimum bet per way and to write the ticket correctly.
I shall use the Fremont in downtown for this example. They require a minimum of 10 ways if you wish to play at a quarter a way (that is two bits or 25 pennies). They are holding their annual Trick or Treat tournament Oct. 9-12 and do not require a specific buy-in.
You play with your own money and obviously keep all you win, and lose what you lose. But let’s not think about the latter as who wants negative vibrations. In fact, we want “Good Vibrations” (YouTube Beach Boys). Keno balls vibrating in the goose then to the selector wheel with our numbers. Thus a ticket can get a big win to place in the tourney and still not cost a lot, while satisfying the requirement of 10 ways to play at a quarter a way.
Try a grouping of 2-2-2-1-1-1 and play just the 7s, 8s, and 9s. There is a 9, three 8s, and six 7s for a total of 10 ways. Now you may ask how do I know this? Knowing combination theory helps but there are other ways to figure it out.
First of all, with six different groups (2-2-2-1-1-1 is six groups) there are 2 to the nth -1 total number of ways or 63 total on this ticket. Now let’s use symmetry. We know there is only one way to make a 9-spot (all nine numbers). Taking away a king, which is a group of one, we can do this three times, taking away Henry the 8th once, Solomon once, and David once.
Using names on keno tickets is based in history when each number had the name of a race horse. That is why games are called races by some. Giving that this gives us three possible 8-spots, which leaves one unused number in each case.
Thus you have four groups of one (three kings). Notice that 8 + 1 = 9, thus the number of groups of eight must equal the number of groups of one. Use the same principle for the 7s. Thus for each 7 chosen you have two numbers (which must be in the same group) left over. The number of possible 7-spots must equal the number of 2-spots.
Same with 6- and 3-spot and 5- and 4-spots. The total is always the total number of spots you are playing. You may ask how this helps. Counting and figuring how many 7-spots may not be easy but figuring out the number of 2-spots is simple.
Since you have three deuces (groups of two) and three kings (groups of one) and are able to leave a king out of each group of two, that gives you three more groups of two, a total of six. Now since you have six possible 2-spots by definition you must also have six 7-spots.
Do the same to find the 6-spots by counting the number of groups of three and find the 5-spots by counting the number of 4-spots as they must be equal to the number of 5-spots. This grouping of 2-2-2-1-1-1 shows a 9, three 8s, six 7s, ten 6s, twelve 5s, twelve 4s, ten 3s, six 2s, and three 1s (kings).
If you are familiar with Pascal’s triangle this may help you in its understanding. Add the math and you have 63 total groups.
You may ask why not also play the 6s and lower? This is a tourney, small wins do not mean a lot but big wins do. Spend your money going for the big win and a share of the cash prize. This way you spend $2.50 per game, relax and enjoy yourself in Paradise (the buffet) or fill out your ribs (Tony Roma’s).
Since you are playing with your own money (as long as it is not the rent money) and must use your slot card for tracking, you will be earning comps as well as playing in the tourney. Call the Fremont keno department for details.
There are the Deuce and Strip Express Buses to downtown if you are staying on the Strip and the Boulder Highway Express bus (BHX) if you are staying on Boulder Highway.
Figuring ways is relatively easy once you know the methodology to do so. Any questions, email me at the address at the end of this column.