Re: [心得] 賽局分一塊錢的故事
※ 引述《pig030 (貓博3號)》之銘言：
: 這一次是一個一階段的賽局，即兩個人分一塊錢的故事。
: 規則是這樣子的，兩個人A、B同時提出要多少錢，如果
: 兩個人提出之合超過1元，則兩個人一毛錢都拿不到。
: 其中假設兩個可以要的錢是，A為a=[0,1]元，B為b=[0,1]元
: 若 a + b > 1 則 兩個人的報酬是 0
: 若 a + b <=1 則 兩個人的報酬是 a 及 b
: 請你找出上述所有pure的Nash 均衡解！
: 另外請你證明 (a=1,b=1)也是一個NE均衡解！
: (最令人覺得奇怪的解，為(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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