How to calculate the probability of rolling 6 at least 5 times in a row, out of 50 tries? If I roll the dice 50 times, how do I calculate the chance that I will roll 6 at least 5 times in a row?
Why this problem is hard

*

*With 5 tries this would be easy: take $(1/6)$ to fifth power.

*With 6 tries this is manageable; take the probability of rolling 6 the first five times, add the probability of rolling 6 the last five times, then subtract the overlap (all six results are 6).

*Given two overlapping sets of 5 rolls, the probability that one is all 6's is not independent from the probability that the other is all 6's.

*In principle this could be continued, but inclusion-exclusion gets out of hand. There has to be a better way; what is it?


This is a hard kind of question to answer, because you might roll 5 6's in a row more than once with separation in between, and/or you may roll more than 5 6's in a row. There are 46 places where a run of 5 6's might start, and although these events can overlap, if we consider them all separate then we get that the expected number of times you get 5 6's in a row is less than $46/6^5 = 46/7776$ which is pretty small. By the Markov inequality, the probability that you get 5 6's in a row at least once is thus less than $46/7776$.
