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Average number of rolls to roll 66

Posted By: Bob Koca
Date: Tuesday, 6 September 2011, at 11:06 a.m.

In Response To: Average number of rolls to roll 66 (Raj Jansari)

Here is another way.

Consider what happens on the first roll. If the event occurs then you are done after one trial. If it does not then you have used up one trial and then it is like starting over again. So if E is the expected number of tosses needed and p is the probability of success we have that

E = (p)(1) + (1-p) (1+E)

Solving for E gives E = 1/p.

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