HMMT 十一月 2017 · THM 赛 · 第 9 题
HMMT November 2017 — THM Round — Problem 9
题目详情
- N ew this year at HMNT: the exciting game of RNG 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: • 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. What is the expected number of points that a given team will score in this game? √ 1+ 5
解析
- N ew this year at HMNT: the exciting game of RNG 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: • 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. What is the expected number of points that a given team will score in this game? Proposed by: Yuan Yao 409 Answer: 125 For i = 0 , 1 , 2 , 3, let P be the probability that a player on the i -th base scores a point before strikeout i (with zeroth base being the home base). We have the following equations: 1 P = ( P + P + P + 1) 0 1 2 3 5 1 P = ( P + P + 1 + 1) 1 2 3 5 1 P = ( P + 1 + 1 + 1) 2 3 5 1 P = (1 + 1 + 1 + 1) 3 5 4 19 89 409 Solving the system of equations gives P = , P = , P = , P = , so the probability that a 3 2 1 0 5 25 125 625 409 batter scores a point himself is , given that he is able to enter the game before the game is over. 625 4 n − 1 Since the probability that the n th player will be able to stand on the home base is ( ) (none of the 5 409 4 4 409 1 409 2 previous n − 1 players receive a strikeout), the expected value is (1+ +( ) + · · · ) = · = . 4 625 5 5 625 125 1 − 5 √ 1+ 5