请教一道数学题 - V2EX
WayTooExplore

请教一道数学题

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  •   WayTooExplore Dec 2, 2020 3032 views
    This topic created in 1994 days ago, the information mentioned may be changed or developed.

    你参与某多人游戏,规则是:

    • 游戏按照轮次、重复循环
    • 每一轮,所有参与者从 [红] 和 [蓝] 中任选 1 个颜色
    • 每一轮你最后一个选择,且能看到其它参与者选择的颜色
    • 所有参与者选完颜色后,主办方随机抽 1 名选手:
      => 如果被抽中的选手选了 [红] ,则:
      ① 他要支付给随机的另一位选 [红] 的选手 100 元
      ② 主办方提供 500 元的奖金,平分给所有选 [蓝] 的选手
      => 如果被抽中的选手选了 [蓝] ,则:
      ① 他要支付给随机的另一位选 [蓝] 的选手 50 元
      ② 主办方提供 250 元的奖金,平分给所有选 [红] 的选手

    请问:你按照怎样的策略选择颜色,能收益最大化?
    (例如设参与者有 X 人,其中 Y 人选了 [红] ,则 X 与 Y 满足怎样的条件,你会选择 [红] 以期望最大?)

    2 replies    2020-12-02 09:18:05 +08:00
    binux
        1
    binux  
       Dec 2, 2020 via Android
    选蓝的人数大于√2 倍选红的
    starzh
        2
    starzh  
       Dec 2, 2020
    试着答一下
    现在你还没选
    红:Y 蓝:X-Y-1
    你选红,那么
    红:Y+1 蓝:X-Y-1
    E(红)=(Y+1)/X*(1/(Y+1)*-100+Y/(Y+1)*1/Y*100)+(X-Y-1)/X*(250/(Y+1))
    你选蓝,那么
    红:Y 蓝:X-Y
    E(蓝)=Y/X*(500/(X-Y))+(X-Y)/X*(1/(X-Y)*-50+(X-Y-1)/(X-Y)*1/(X-Y-1)*50)
    E(红)>E(蓝),X>Y>0
    结果是 X>Y+1
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