Reviewing Movies III

Let $M_{a,b}$ be the expected number of reviews until both $a$ "up"s and $b$ "down"s have been received. Then $\begin{aligned} M_{a,b} & = \; 1 + pM_{a-1,b} + qM_{a,b-1} & \hspace{2cm} a,b \ge 1 \\ M_{a,0} & = \; 1 + pM_{a-1,0} + qM_{a,0} & \hspace{2cm} a \ge 1 \\ M_{0,b} & = \; 1 + pM_{0,b} + qM_{0,b-1} & \hspace{2cm} b \ge 1 \\ M_{0,0} & = \; 0 \end{aligned}$ Thus we have $M_{a,0} \; = \; \tfrac{a}{p} \hspace{2cm} M_{0,b} \; = \; \tfrac{b}{q} \hspace{3cm} a,b \ge 0$ On the other hand, let $N_{a,b}$ be the expected number of reviews until either $a$ "up"s or $b$ "down"s have been received. Then $N_{a,b} \; = \; 1 + pN_{a-1,b} + qN_{a,b-1} \hspace{2cm} a,b \ge 1$ while $N_{a,0} \; = \; N_{0,b} \; = \; 0 \hspace{2cm} a,b \ge 0$ If we let $K_{a,b} = M_{a,b} - N_{a,b}$ , then $\begin{aligned} K_{a,b} & = \; pK_{a-1,b} + qK_{a,b-1} & \hspace{2cm} a,b \ge 1 \\ K_{a,0} & = \; \tfrac{a}{p} & \hspace{2cm} a \ge 1 \\ K_{0,b} & = \; \tfrac{b}{q} & \hspace{2cm} b \ge 1 \\ K_{0,0} & = \; 0 \end{aligned}$ It is easy enough to iterate these equations to obtain $\alpha - \beta \; = \; K_{3,7} \; = \; \frac{3 - 10 p + 420 p^4 - 1512 p^5 + 2520 p^6 - 2400 p^7 + 1350 p^8 - 420 p^9 + 56 p^{10}}{p(1 - p)}$ and this expression is minimized when $p = 0.321752$ , making the answer $\boxed{322}$ .

It is worth noting that we can find the generating function of the double sequence $K_{a,b}$ , with $\sum_{a,b \ge 0} K_{a,b} x^ay^b \; = \; \frac{1}{1 - px - qy}\left[\frac{x(1-px)}{p(1-x)^2} + \frac{y(1-qy)}{q(1-y)^2}\right]$

This is merely a sketch of a solution with some notes.

Roger starts reviewing: one, two, three ... Now jump ahead to the review number $n + m$ . Either exactly $n$ thumbs ups were given (so exactly $m$ thumbs downs as well) or there were $n' > n$ ups (so $m' < m$ downs) or there were $n' < n$ ups (so $m' > m$ downs). It feels right to condition on this outcome.

$\mathbb{E}[R] = \mathbb{E}[R \, | \, n' = n] \cdot \mathbb{P}\{n' = n\} + \mathbb{E}[R \, | \, n' > n] \cdot \mathbb{P}\{n' > n\} + \mathbb{E}[R \, | \, n' < n] \cdot \mathbb{P}\{n' < n\}$

Express expected number of reviews $R$ by summing binomial probabilities. Simplify as much as you can. Eventually a summing term remains that can be summed up in a closed form but as a hypergeometric function. I myself left the term as a sum and evaluated it numerically within the ternary search algorithm for the minimum of $\alpha - \beta$ .