Re: Craps question
I was very surprised to see the Fibonacci like formula
t(n) = 35[t(n-1) + t(n-2)].
I had assumed that since 3 consecutive boxcars were needed that the recurrence relation would have 3 terms.
Have you solve the general problem:
What is the probability of n consecutive successes in a list of m tries if the probability of success for any particular try is p?
Using your approximate method we would get something like
1-(1-p^n)^k
where
k = m/( q + p q + p q^2 + ... + p q^(n-1) ).
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