Difference between revisions of "2013 Mock AIME I Problems/Problem 9"
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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 <math> E </math> is the expected number of turns it takes to short circuit all of the lights, find <math> 100E </math>. | 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 <math> E </math> is the expected number of turns it takes to short circuit all of the lights, find <math> 100E </math>. | ||
Latest revision as of 13:31, 30 July 2024
Problem
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 
is the expected number of turns it takes to short circuit all of the lights, find 
.
Solution
Let 
be the expected value of turns it takes to short circuit 
lights. Note that 
, 
, and in general, ![\[E_k=\frac{1}{k}+\frac{1}{k}(E_{k-1}+1)+\frac{1}{k}(E_{k-2}+1)+\dots+\frac{1}{k}(E_1+1)=1+\frac{1}{k}(\sum_{i=1}^{k-1} E_i)\]](http://latex.artofproblemsolving.com/b/c/8/bc8e35e3de47d4ea6ade02c84b96345668239991.png)
Doing that calculations gives 
, so 
.
