HMMT-Nov 2017 thm, Problem 9

New this year at HMNT: the exciting game of $R N G$ baseball! In RNG baseball, a team of infinitely many people play on a square field, with a base at each vertex; in particular, one of the bases is called the home base. Every turn, a new player stands at home base and chooses a number $n$ uniformly at random from $\{0,1,2,3,4\}$. Then, the following occurs: \begin{itemize} If $n>0$, then the player and everyone else currently on the field moves (counterclockwise) around the square by $n$ bases. However, if in doing so a player returns to or moves past the home base, he/she leaves the field immediately and the team scores one point. If $n=0$ (a strikeout), then the game ends immediately; the team does not score any more points. \end{itemize} What is the expected number of points that a given team will score in this game?

Solution: For $i=0,1,2,3$, let $P_{i}$ be the probability that a player on the $i$-th base scores a point before strikeout (with zeroth base being the home base). We have the following equations: \[ \begin{aligned} P_{0} & =\frac{1}{5}\left(P_{1}+P_{2}+P_{3}+1\right) \\ P_{1} & =\frac{1}{5}\left(P_{2}+P_{3}+1+1\right) \\ P_{2} & =\frac{1}{5}\left(P_{3}+1+1+1\right) \\ P_{3} & =\frac{1}{5}(1+1+1+1) \end{aligned} \] Solving the system of equations gives $P_{3}=\frac{4}{5}, P_{2}=\frac{19}{25}, P_{1}=\frac{89}{125}, P_{0}=\frac{409}{625}$, so the probability that a batter scores a point himself is $\frac{409}{625}$, given that he is able to enter the game before the game is over. Since the probability that the $n$th player will be able to stand on the home base is $\left(\frac{4}{5}\right)^{n-1}$ (none of the previous $n-1$ players receive a strikeout), the expected value is $\frac{409}{625}\left(1+\frac{4}{5}+\left(\frac{4}{5}\right)^{2}+\cdots\right)=\frac{409}{625} \cdot \frac{1}{1-\frac{4}{5}}=\frac{409}{125}$. $\fbox{\frac{409}{125}}$.

Answer: \frac{409}{125}