Roger reviews movies and gives each movie either a "thumbs up" with probability 5 1 or a "thumbs down" with probability 5 4 . If his scoring starts as "up, down, down, down, up, up, down...," there is an up-run of length one at the beginning, followed by a down-run of length three, an up-run of length two, etc. Let μ k be the expected length of run k .
Find μ 1 + μ 2 + μ 7 in the form b a , where a and b are coprime positive integers, and give your answer as a + b .
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I did basically the same thing as Patrick; we have μ 1 = k = 1 ∑ ∞ ( ( 5 4 ) k × 5 1 × k ) + k = 1 ∑ ∞ ( ( 5 1 ) k × 5 4 × k ) = 4 1 7 . Then μ 2 = 5 4 × ( 4 1 + 1 ) + 5 1 × ( 4 + 1 ) = 2 , and moreover, μ 2 j = 5 4 × ( 4 1 + 1 ) + 5 1 × ( 4 + 1 ) = 2 and μ 2 j − 1 = 5 4 × ( 4 + 1 ) + 5 1 × ( 4 1 + 1 ) = 4 1 7 , j ∈ Z + . Then μ 1 + μ 2 + μ 7 = 4 1 7 + 2 + 4 1 7 = 2 2 1 , making the answer 2 1 + 2 = 2 3 .
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Let p = 1 / 5 , q = 4 / 5 . The expected length of an up run is q ( 1 + 2 p + 3 p 2 + ⋯ ) = ( 1 − p ) 2 q = 1 / q = 5 / 4 . Similarly, the expected length of a down run is 1 / p = 5 . So μ 1 + μ 2 = 5 + 5 / 4 = 2 5 / 4 , because the first two runs consist of an up run and a down run in some order.
The probability that the seventh run is an up run is the probability that the first run is an up run, which is p , so μ 7 = p ( 1 / q ) + q ( 1 / p ) = 1 / 4 + 4 = 1 7 / 4 . The sum is 4 2 / 4 = 2 1 / 2 , so the answer is 2 3 .
(More generally, the same arguments show that μ n = 1 7 / 4 if n is odd and 2 if n is even.)