A fair die is rolled repeatedly, and a running total is kept (which is, at each time, the total of all the rolls up until that time). Let
be the probability that the running total is ever exactly n (assume the die will always be rolled enough times so that the running total will eventually exceed n, but it may or may not ever equal n).

(a) Write down a recursive equation for (relating to earlier terms in a simple way). Your equation should be true for all positive integers n, so give a definition of and for k < 0 so that the recursive equation is true for small values of n.

(b) Find .

(c) Give an intuitive explanation for the fact that as .

(a) Write down a recursive equation for (relating to earlier terms in a simple way). Your equation should be true for all positive integers n, so give a definition of and for k < 0 so that the recursive equation is true for small values of n.

(b) Find .

(c) Give an intuitive explanation for the fact that as .

Solution:
(a) We will find something to condition on to reduce the case of interest to earlier, simpler cases. This is achieved by the useful strategy of first step analysis: pn = (1/6)(pn−1 + pn−2 + pn−3 + pn−4 + pn−5 + pn−6), where p0 = 1 and pk = 0 for k < 0.
(b) Use the recursive equation from part a to get that p7 = (1/6)(p1 + p2 + p3 + p4 + p5 + p6) = (1/6)((1 + 1/6)^6 - 1) ~ 0.2536.
(c) An intuitive explanation is as follows. The average number thrown by the die is (total of dots)/6, which is 21/6 = 7/2, so that every throw adds on an average of 7/2. We can therefore expect to land on 2 out of every 7 numbers, and the probability of landing on any particular number is 2/7. Consult iTunes course for a proof involving taking the limit as n goes to infinity.

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