6-Spot Keno Math Question

On a 6-spot keno progressive game the meter caps at $800 but goes positive well before that. I can't just make a straight up calculation to get the true advantage because of the $800 cap. I need some insight into the odds. So I'm hoping that some of you strong math guys can help me on this one. The frequency is 7753.

What are the odds I will hit the solid six in one cycle?
What are the odds it will take me 2 cycles?
What are the odds it will take me 3 cycles?
What are the odds it will take me 4 cycles?
What are the odds it will take me 5 cycles?

Can anyone show me how to make these calculations? Any help is greatly appreciated.

When a cycle length is "k" plays, and the number of cycles played is "n" the probability of NOT hitting it within "n" cycles is just:

[(k-1)/k]^(n*k)

And when k is large like it is here the answer can be well approximated by 1/e^n where e is approximately equal to 2.718281828459045. So with this in mind, the probability of going so long without hitting is as follows:

1 cycle ~ 1/e ~ 0.36788
2 cycles ~ 1/e^2 ~ 0.13534
3 cycles ~ 1/e^3 ~ 0.049787
4 cycles ~ 1/e^4 ~ 0.018316
5 cycles ~ 1/e^5 ~ 0.0067379

Fyi, that approximation for "e" was from memory. I was taught a good mnemonic for the decimal places:

Andrew Jackson was the 7th President of the US. He was elected twice starting in 1828. As for the last six digits, those are the angles of an isosceles right triangle.

Wow. That's a pretty impressive mnemonic. Unfortunately, I'll probably use it so rarely that when I DO need it, all I can come up with is 'Yeah! It's something to do with Michael Jackson!'...

Quote: tringlomane

When a cycle length is "k" plays, and the number of cycles played is "n" the probability of NOT hitting it within "n" cycles is just:

[(k-1)/k]^(n*k)

And when k is large like it is here the answer can be well approximated by 1/e^n where e is approximately equal to 2.718281828459045. So with this in mind, the probability of going so long without hitting is as follows:

1 cycle ~ 1/e ~ 0.36788
2 cycles ~ 1/e^2 ~ 0.13534
3 cycles ~ 1/e^3 ~ 0.049787
4 cycles ~ 1/e^4 ~ 0.018316
5 cycles ~ 1/e^5 ~ 0.0067379

I totally knew that!

Okay, no I didn't, this is how I would have done it.

The probability of it not happening on any given play is 1-0.000128984939112 = 0.99987101506

A cycle rounds up to 7753 plays, thus:

(0.99987101506)^7753 = 0.3678482687 (One cycle, no hit)
(0.99987101506)^15506 = 0.13531234878 (Two cycles, no hit)
(0.99987101506)^23259 = 0.04977441323 (Three cycles, no hit)
(0.99987101506)^31012 = 0.01830943173 (Four cycles, no hit)
(0.99987101506)^38765 = 0.00673509276 (Five cycles, no hit)

The method Tringlomane used clearly rocks, the difference between that and doing it the long way is little more than a rounding error, in terms of expectation.

Quote: Venthus

Wow. That's a pretty impressive mnemonic. Unfortunately, I'll probably use it so rarely that when I DO need it, all I can come up with is 'Yeah! It's something to do with Michael Jackson!'...

LOL, yeah "e" isn't used by tons of people. For gambling purposes, it's more useful than "pi" though. :)

Quote: Mission146

The method Tringlomane used clearly rocks, the difference between that and doing it the long way is little more than a rounding error, in terms of expectation.

Thanks. Part of the reason I did it that way is I am posting on my phone right now. I didn't want to keep typing in 7753...lol Thanks for showing the "exact" way. Knowing this way is important too if the cycle is quite small.

I don't blame you, I almost never post any math stuff from my phone. I should also mention that my cycles are still slightly inexact as I rounded up to the 7753 for the cycle length, so the actual probability of going those cycles without a six-hit will be infinitesimally more likely.

Tring, mission, thanks for the math lessons, guys. It's just what the doctor ordered.