Thanks for your time. So is it correct to say that it would happen 1 in 12 times so therefore a 8.33% chance of it happening?
NO, the probability 1/12 refers to throwing a 4 on any particular roll. I expected that to be something a craps player would understand inherently.
The probability of NOT throwing a 4 is 11/12,
So the probability of N throws
before the first 4 is
(11/12)*(11/12)*. . . *(1/12)= (11/12)^N * (1/12)
That tells you how long a run you can expect
without a 4 being thrown. [Note that I have defined N as the number of not-4s, while the form used by
pka is N= the first throw that
is a 4.]
Code:
N P(N) P(6|N) P(7|N) P(8|N) . . .
0 0.0833
1 0.0764
2 0.0700
. . .
6 0.0494 0.0000214
7 0.0453 0.0001125 0.0000036
8 0.0415 0.0003376 0.0000214 0.0000006
. . .
For example, the probability is 0.0415 that there will be exactly 8 throws before the first 4. That is one chance in 24. If that is the value of N, then the probability that 6, 7, or 8 of those are 7s is 0.0003596. There is a chance of 1/67000 that there will be 8 throws preceding the 4, AND that at least 6 of the 8 are 7s.
To get the total probability, you have to add up ALL values of N. The best way to do that would be a spreadsheet.
If you don't understand probability, you are the kind of guy the house loves to see at the craps table. Most "systems" work most of the time, IF you have a big enough bankroll to cover adverse runs. You wind up risking a LOT of money to expect a small positive return. You can continue to make small gains, right up to the day your bank goes bust. It is likely your bankroll is not as large as the house's.