Reviewing Movies I

Roger reviews movies and gives each movie either a "thumbs up" with probability 1 5 \frac15 or a "thumbs down" with probability 4 5 \frac45 . 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 \mu_k be the expected length of run k k .

Find μ 1 + μ 2 + μ 7 \mu_1 + \mu_2 + \mu_7 in the form a b , \frac ab, where a a and b b are coprime positive integers, and give your answer as a + b . a+b.


The answer is 23.

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2 solutions

Patrick Corn
Oct 31, 2017

Let p = 1 / 5 , q = 4 / 5. p = 1/5, q = 4/5. The expected length of an up run is q ( 1 + 2 p + 3 p 2 + ) = q ( 1 p ) 2 = 1 / q = 5 / 4. q(1+2p+3p^2+\cdots) = \frac{q}{(1-p)^2} = 1/q = 5/4. Similarly, the expected length of a down run is 1 / p = 5. 1/p = 5. So μ 1 + μ 2 = 5 + 5 / 4 = 25 / 4 , \mu_1 + \mu_2 = 5 + 5/4 = 25/4, 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 , p, so μ 7 = p ( 1 / q ) + q ( 1 / p ) = 1 / 4 + 4 = 17 / 4. \mu_7 = p(1/q) + q(1/p) = 1/4+4 = 17/4. The sum is 42 / 4 = 21 / 2 , 42/4 = 21/2, so the answer is 23 . \fbox{23}.

(More generally, the same arguments show that μ n = 17 / 4 \mu_n = 17/4 if n n is odd and 2 2 if n n is even.)

Miles Koumouris
Dec 3, 2017

I did basically the same thing as Patrick; we have μ 1 = k = 1 ( ( 4 5 ) k × 1 5 × k ) + k = 1 ( ( 1 5 ) k × 4 5 × k ) = 17 4 . \mu_1=\sum_{k=1}^{\infty }\left(\left(\dfrac45 \right)^k\times \dfrac15 \times k\right)+\sum_{k=1}^{\infty }\left(\left(\dfrac15 \right)^k\times \dfrac45 \times k\right)=\dfrac{17}{4}. Then μ 2 = 4 5 × ( 1 4 + 1 ) + 1 5 × ( 4 + 1 ) = 2 , \mu_2=\dfrac45\times \left(\dfrac14+1\right)+\dfrac15\times (4+1)=2, and moreover, μ 2 j = 4 5 × ( 1 4 + 1 ) + 1 5 × ( 4 + 1 ) = 2 and μ 2 j 1 = 4 5 × ( 4 + 1 ) + 1 5 × ( 1 4 + 1 ) = 17 4 , j Z + . \mu_{2j}=\dfrac45\times \left(\dfrac14+1\right)+\dfrac15\times (4+1)=2\;\; \text{ and }\;\; \mu_{2j-1}=\dfrac45\times (4+1)+\dfrac15\times \left(\dfrac14+1\right)=\dfrac{17}{4},\;\; j\in \mathbb{Z}^+. Then μ 1 + μ 2 + μ 7 = 17 4 + 2 + 17 4 = 21 2 , \mu_1+\mu_2+\mu_7=\dfrac{17}{4}+2+\dfrac{17}{4}=\dfrac{21}{2}, making the answer 21 + 2 = 23 21+2=\boxed{23} .

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