※ 引述《pig030 (猫博3号)》之铭言： : 这一次是一个一阶段的赛局，即两个人分一块钱的故事。 : 规则是这样子的，两个人A、B同时提出要多少钱，如果 : 两个人提出之合超过1元，则两个人一毛钱都拿不到。 : 其中假设两个可以要的钱是，A为a=[0,1]元，B为b=[0,1]元 : 若 a + b > 1 则 两个人的报酬是 0 : 若 a + b <=1 则 两个人的报酬是 a 及 b : 请你找出上述所有pure的Nash 均衡解！ : 另外请你证明 (a=1,b=1)也是一个NE均衡解！ : (最令人觉得奇怪的解，为(a=1,b=1)) (1) Any (a*,b*) with a*+b*<1 is not NE . Because player 1 could choose a'=1-b*>a* to be strictly better off. (2) Any (a*,b*) with a*+b*=1 is NE . Given b* , player 1's utility is U1(a,b*)= 0 if a>a*=1-b* = a* if a=a*=1-b* = a if a<a*=1-b* clearly , player 1 optimally chooses a=a*=1-b* . Likewise , player 2 optimally chooses b=b*=1-a* (3) Any (a*,b*) with a*+b*>1 and min{a*,b*}<1 is not NE Here both players got zero. Min{a*,b*}<1 implies that a*<1 or b*<1. Suppose b*<1. Then player 1 will be strictly better by setting a =1-b*>0,so that it is not NE. Argument is analogous for a*<1. (4) (a*,b*) with a*+b*>1 and min{a*,b*}=1 is NE. That is, a*=b*=1 Given b*=1 , player 1 gets zero utility for any a ,so a*=1 is one best response for plater 1. Silimarly , b*=1 is also one best response for player 2 given a*=1 In sum , NEs are those (a*,b*) st a*+b*=1 or a*=b*=1. --

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