Re: Craps question
I had assumed that since 3 consecutive boxcars were needed that the recurrence relation would have 3 terms.
It appears that the 3 consecutive only influence where the Fibonacci-like sequence starts, since it can only occur starting on trial 3. The sequence itself only depends on the probability of boxcars on one particular throw. BTW, my sentence about there being 36 possible outcomes should read that there are 36 outcomes per surviving sequence on the previous stage. In the coin flip problem there were 2 outcomes per surviving sequence. This number is the only thing that changes the series.
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?
I believe my method generalizes to:
sum[i = n to m]t(i)*p^m
t(i) = (1/p - 1)[t(i-1) + t(i-2)]
BTW, you did post your Markov process matrix solution to the coin flip problem, which can be used here too.
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