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HMMT 二月 2005 · 冲刺赛 · 第 36 题

HMMT February 2005 — Guts Round — Problem 36

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [12] One hundred people are in line to see a movie. Each person wants to sit inthe front row, which contains one hundred seats, and each has a favorite seat, chosenrandomly. They enter the row one at a time from the far right. As they walk, if theyreach their favorite seat, they sit, but to avoid stepping over people, if they encountera person already seated, they sit to that person’s right. If the seat furthest to the rightis already taken, they sit in a different row. What is the most likely number of peoplethat will get to sit in the first row?
    HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND
解析

英文解析

  1. One hundred people are in line to see a movie. Each person wants to sit in the frontrow, which contains one hundred seats, and each has a favorite seat, chosen randomlyand independently. They enter the row one at a time from the far right. As they walk,
    if they reach their favorite seat, they sit, but to avoid stepping over people, if theyencounter a person already seated, they sit to that person’s right. If the seat furthestto the right is already taken, they sit in a different row. What is the most likely numberof people that will get to sit in the first row?
    Solution: 10
    Let S ( i ) be the favorite seat of the i th person, counting from the right. Let P ( n ) bethe probability that at least n people get to sit. At least n people sit if and only if
    S (1) ≥ n , S (2) ≥ n − 1, . . . , S ( n ) ≥ 1. This has probability:
    100 − ( n − 1) 100 − ( n − 2) 100 100!
    P ( n ) = · · · · = .
    100 100 100 (100 − n )! · 100 n
    The probability, Q ( n ), that exactly n people sit is
    100! 100! 100! · n
    P ( n ) − P ( n + 1) = − = .
    n n +1 n +1
    (100 − n )! · 100 (99 − n )! · 100 (100 − n )! · 100
    Now,
    n 2
    Q ( n ) 100! · n (101 − n )! · 100 n (101 − n ) 101 n − n = · = = ,
    n +1
    Q ( n − 1) (100 − n )! · 100 100! · ( n − 1) 100( n − 1) 100 n − 100
    which is greater than 1 exactly when n − n − 100 < 0, that is, for n ≤ 10. Therefore,2
    the maximum value of Q ( n ) occurs for n = 10.