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HMMT 二月 2010 · 冲刺赛 · 第 22 题

HMMT February 2010 — Guts Round — Problem 22

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 12 ] You are the general of an army. You and the opposing general both have an equal number of troopsto distribute among three battlefields. Whoever has more troops on a battlefield always wins (you winties). An order is an ordered triple of non-negative real numbers ( x, y, z ) such that x + y + z = 1, andcorresponds to sending a fraction x of the troops to the first field, y to the second, and z to the third.
    ( )
    1 1 1
    Suppose that you give the order , , and that the other general issues an order chosen uniformly
    4 4 2
    at random from all possible orders. What is the probability that you win two out of the three battles?
解析

英文解析

  1. [ 12 ] You are the general of an army. You and the opposing general both have an equal number of troopsto distribute among three battlefields. Whoever has more troops on a battlefield always wins (you winties). An order is an ordered triple of non-negative real numbers ( x, y, z ) such that x + y + z = 1, andcorresponds to sending a fraction x of the troops to the first field, y to the second, and z to the third.
    ( )
    1 1 1
    Suppose that you give the order , , and that the other general issues an order chosen uniformly
    4 4 2
    at random from all possible orders. What is the probability that you win two out of the three battles?
    Answer: Let x be the portion of soldiers the opposing general sends to the first battlefield, and 5
    y the portion he sends to the second. Then 1 − x − y is the portion he sends to the third. Then x ≥ 0,8
    y ≥ 0, and x + y ≤ 1. Furthermore, you win if one of the three conditions is satisfied: x ≤ and 1
    1 1 1 1 14
    y ≤ , x ≤ and 1 − x − y ≤ , or y ≤ and 1 − x − y ≤ . This is illustrated in the picture below.
    4 4 2 4 2
    Guts Round
    This triangle is a linear projection of the region of feasible orders, so it preserves area and probabilityratios. The probability that you win, then is given by the portion of the triangle that satisfies one ofthe three above constraints — in other words, the area of the shaded region divided by the area of the
    1655
    entire triangle. We can easily calculate this to be = .
    281