2013 Mock AIME I Problems/Problem 9

Problem 9

In a magic circuit, there are six lights in a series, and if one of the lights short circuit, then all lights after it will short circuit as well, without affecting the lights before it. Once a turn, a random light that isn’t already short circuited is short circuited. If $E$ is the expected number of turns it takes to short circuit all of the lights, find $100E$ .

Solution

Let $E_n$ be the expected value of turns it takes to short circuit $n$ lights. Note that $E_1=1$ , $E_2=\frac{1}{2}+\frac{1}{2}(E_1+1)$ , and in general, $E_k=\frac{1}{k}+\frac{1}{k}(E_{k-1}+1)+\frac{1}{k}(E_{k-2}+1)+\dots+\frac{1}{k}(E_1+1)=1+\frac{1}{k}(\sum_{i=1}^{k-1} E_i)$ Doing that calculations gives $E_6=\frac{49}{20}$ , so $100E_6=\boxed{245}$ .

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2013 Mock AIME I Problems/Problem 9

Problem 9

Solution