Originally Posted by
mickeycrimm
The video keno games I play are either banking or progressive games. This example was neither. I just used it as an example of how to do keno math. The object of doing the math is to determine if the player has an advantage.
Gambling math is pretty simple. It's just addition, subtraction, multiplication and division. There's no "stochastics" or "heuristics" or "geometic distribution." It's just simple math.
Just for fun and if anyone reading this wants another way to solve it which (imo) isn't, "Too fancy," here's how I would do it:
BASE PAYS:
Like Mickey says, just use the WoO Keno Calculator:
2/4: 0.212635465800023
3/4: 0.216239456745786
4/4: 0.199120499753411
BASE GAME TOTAL: 0.627995422299220, which we can just call 62.8%.
GETTING TO THE BONUS GAME:
The only possible mistake someone can make here is trying to come up with a bunch of ridiculous scenarios where a win on the base game impacts the probability of getting the 3/3...which in certain cases you WOULD need to know because there is at least one game where a win on the base game AND hitting the 3/3 combination is required. Fortunately, this is not one of them.
For this, I also use the WoO calculator (without attributing any pays---because just going to the bonus pays nothing) and lift the probability for 3/3, which is:
0.013875365141188
FREE GAMES RETURN:
The next step is to determine the overall return of the Free Games (as a whole) which we will then take and multiply by the probability of going to the Free Games, then having determined that, add to the base return.
Since it's a weird number of draws, we can't cheat with the WoO Calculator.
4/4 = 130
3/4 = 10
2/4 = 2
For this, I like to use an online scientific calculator you can find here:
https://web2.0calc.com/
(nCr(4,2)*nCr(76,22)/nCr(80,24)) * 2 = 0.5374878286270691
(nCr(4,3)*nCr(76,21)/nCr(80,24)) * 10 = 0.7166504381694255
(nCr(4,4)*nCr(76,20)/nCr(80,24)) * 130 = 0.8734177215189873
MickeyCrimm added them up and got 2.1276 units/game on Free Games, and:
0.5374878286270691+0.7166504381694255+0.8734177215 189873 = 2.1275559883154819
I agree with that, except I haven't rounded off yet just because I took the results the way the calculator gave them to me. Now, multiply by the number of Free Games:
2.1275559883154819*12 = 25.5306718597857828
Which is the expected return of all Free Games, so all that remains is to multiply that by the probability of Free Games:
25.5306718597857828*0.013875365141188 = 0.3542473943543810567001571019664
Which gives us our added expected return (per unit bet) that comes from Free Games, which we can call 35.4%, add to 62.8% and get 98.2%. Therefore, my results agree with those of MickeyCrimm with differences due to the way we rounded. It's not positive, anyway, so there would be very limited scenarios where I would care about this game.
Of course, if I had a multiplier day such that it came up to 2% return on the points multiplier and there was no reasonable video poker at this location (or any positive machines) this would be a good machine to know if I needed to earn points anyway...obviously I would not play this for well under a 0.5% edge alone.
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