s******4 发帖数: 329 | 1 There are four sealed boxed, with $100 in one of them, and others empty. A
player can pay to open a box (price initially set to X, if you choose to
continue playing after two incorrect guesses the price changes to Z) and
take the contents as many times as they like. Assuming it is a fair game,
what is the value fo X and Z?
多谢! | l********y 发帖数: 185 | 2 不知道理解的对不对:
第一次付$X选一次,
选错了,可以付$X再选一次,
如果又选错了,可以付$Z,然后剩下的两个盒子都可以打开(即肯定能拿到$100)。
如果是这样的话,计算如下:
1.第一次选中的概率P1=1/4,玩家赚到100-X。
2.第一次没选中,不想继续选,概率P2=3/4*1/2(有一半的机会会继续选下去),玩家
陪X(即赚到-X)。
3.第一次没选中,继续选第二次而且选中,概率P3=3/4*1/2*1/3,玩家赚到100-2X。
4.两次都没选中,概率P4=3/4*1/2*2/3。这时赔了2X,如果再付Z,可以保证拿到100,
这样赔Y=2X+Z-100。
最公平的方式是Y=X,即玩家继续玩儿使损失减半。所以,2X+Z-100=X => Z=100-X。
上述四种情况,玩家赚钱的平均值:
P1*(100-X)+P2*(-X)+P3*(100-2X)+P4*(-X) = 100*3/8 - 9/8*X
公平起见,上面的平均值=0, 即X=100/3,Z=200/3。
【在 s******4 的大作中提到】 : There are four sealed boxed, with $100 in one of them, and others empty. A : player can pay to open a box (price initially set to X, if you choose to : continue playing after two incorrect guesses the price changes to Z) and : take the contents as many times as they like. Assuming it is a fair game, : what is the value fo X and Z? : 多谢!
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