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HMMT 二月 2011 · ALGCALC 赛 · 第 5 题

HMMT February 2011 — ALGCALC Round — Problem 5

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. If
    Nathaniel goes first, determine the probability that he ends up winning.
解析

英文解析

  1. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep arunning tally of the sum of the results of all rolls made. A player wins if, after he rolls, the numberon the running tally is a multiple of 7. Play continues until either player wins, or else indefinitely. If
    Nathaniel goes first, determine the probability that he ends up winning.
    Answer: For 1 ≤ k ≤ 6, let x be the probability that the current player, say A , will win when 5
    11 kthe number on the tally at the beginning of his turn is k modulo 7. The probability that the total
    1 1
    is l modulo 7 after his roll is for each l 6 ≡ k (mod 7); in particular, there is a chance he wins
    6 6
    immediately. The chance that A will win if he leaves l on the board after his turn is 1 − x . Hence forl
    1 ≤ k ≤ 6,
    ∑
    1 1
    x = (1 − x ) + .
    k l
    6 6
    1 ≤ l ≤ 6 , l 6 = k
    ∑
    x − s 5 x s 6
    k k
    Letting s = x , this becomes x = + 1 or = − + 1. Hence x = · · · = x , and 6 x = sl k 1 6 kl =1
    6 6 6
    11 xk 6
    for every k . Plugging this in gives = 1, or x = .
    6 11 k
    Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7.
    6 5
    Hence Obediah’s chance of winning is and Nathaniel’s chance of winning is .
    11 11