craps and math question

What is the formula to calculate the following...

given probability of Event A happening and Event B happening, what is the probability that Event A will happen x number of times before B happens?

I'm curious to know how many times I would have to expect to roll the dice before seeing 10 come up x number of times before a 7 hits.

I have written a program in WinCraps to keep track of that. I have rolled about 1 million times and never seen a 10 roll 10 times before a 7. I have seen a few 10's come up 9 times, and it comes up 7 times before a 7 a handful.

I would like to know the actuall expected probability of this happening - and better yet, how to calculate it (if it isn't too complicated).

The easier question to answer is the probability that Event A will happen x times before event B.

a = prob(A)
b = prob(B)

prob(A before B) = a / (a+b)

prob( A at least X times before B) = (a / a+b)^x
prob(A exactly X times before B) = b*(a / a+b)^x

example:
if A is roll a 10, a = 1/12
B is roll a 7, b = 1/6

prob( roll 10 at least 10 times before a 7) = 1/3^10 = 1/59049 = 0.00001693

As long as 10 comes up once every other time, the 2:1 (almost) payout makes it worth it..
I can't tell you the number of times I've played a session with a $30 buy on the 10, and absolutely cleaned up...

Quote: dwheatley

The easier question to answer is the probability that Event A will happen x times before event B.

a = prob(A)
b = prob(B)

prob(A before B) = a / (a+b)

prob( A at least X times before B) = (a / a+b)^x
prob(A exactly X times before B) = b*(a / a+b)^x

example:
if A is roll a 10, a = 1/12
B is roll a 7, b = 1/6

prob( roll 10 at least 10 times before a 7) = 1/3^10 = 1/59049 = 0.00001693

thanks!