Mickey,
Framing my comment here, I suspect you know that I'm largely a "back of the envelope" guy like yourself. Having said that, I'd be more inclined to approach an answer to your question via simulation than finding an appropriate calculation. Here's why:
I interpret the problem you pose as requiring the augmentation of the standard probability math with the question of frequency of 60+ game droughts for "3 of 4" hits.
The underlying math of that augment is anything but straightforward. Rather than dig for it, I'd be inclined to write a simple program that uses the inherent frequencies to produce a string of game outcomes and evaluate that string for 60+ game droughts, evaluating the frequency over a large run.
Once you have that frequency, adjusting the "23.12" is relatively straightforward.
---In vpFREE@yahoogroups.com, <mickeycrimm@...> wrote :
I'm currently analyzing a new video keno game that has banking/progressive features. While you are playing your numbers the machine makes 4 picks. When 3 of those four numbers hit certain events happen. Calculating the 3 out of 4 frequency is pretty easy:
Total combinations
80X79X78X77/4X3X2X1 = 1,581,580
Combinations that make 3 of 4
20X19X18/3X2X1 = 1,140
1,140X60 = 68,400
Frequency of 3 of 4
1,581,580/68,400 = 23.12
But there is rule in the game that if you don't hit the 3 of 4 in 60 games you are automatically awarded the 3 of 4. This is 2.59 cycles. So the frequency in this spot of hitting the 3 of 4 has to be a little lower than 23.12. I've never encountered this kind of math before. Does anyone know how to do it? If so any help is appreciated.
[Non-text portions of this message have been removed]
Posted by: harry.porter@verizon.net
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