HMMT 二月 2011 · ALGCALC 赛 · 第 5 题
HMMT February 2011 — ALGCALC Round — Problem 5
题目详情
英文原题
- 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
- 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