w****a 发帖数: 155 | 1 Two players, A and B.
The strategy space for A is { a1, a2};
The strategy space for B is {b1, b2, b3, b4}.
The expected payoff for A to employ strategy a1 is
Ea[a1] = P1*U(a1,b1)+P2*U(a1,b2)+P3*U(a1,b3)+P4*U(a1,b4)
Ea[a2] = P1*U(a2,b1)+P2*U(a2,b2)+P3*U(a2,b3)+P4*U(a2,b4)
The ensure A is indifferent to a1 and a2, we have
Ea[a1] = Ea[a2]
Another trivial equation is p1+p2+p3+p4 = 1.
我的问题是我们有四个变量: p1, p2, p3 and p4, 但只有两个方程。 在这样的情况下
,如何求mixed strategy呢? | U*****e 发帖数: 2882 | 2 Player B's optimization will give you the rest equations. | w****a 发帖数: 155 | 3 谢谢回复。
但我不太明白的是, 考虑 player B, 我们有:
Eb[b1] = L1*U(a1,b1)+L2*U(a2,b1)
Eb[b2] = L1*U(a1,b2)+L2*U(a2,b2)
Eb[b3] = L1*U(a1,b3)+L2*U(a2,b3)
Eb[b4] = L1*U(a1,b4)+L2*U(a2,b4)
The ensure B is indifferent to b1, b2, b3 and b4, we have
Eb[b1] = Eb[b2]=Eb[b3]=Eb[b4]
这并不能帮助解出 p1,p2, p3 和 p4 呀?
【在 U*****e 的大作中提到】 : Player B's optimization will give you the rest equations.
| U*****e 发帖数: 2882 | 4 Yeah, you are right. It seems the two players questions are independent.
Then with 4 variables and two equations, you must have multiple solutions.
Now you see for Player B there are 4 equations (3 from what you wrote, 1
from the definition of probability) and 2 variables. It implies that Player
B might not mix all the actions. If he mixes only 3 actions, then one of P1-
P4 is zero. You might have to discuss all the situations.
Therefore it is necessary to solve both player's problem simultaneou
【在 w****a 的大作中提到】 : 谢谢回复。 : 但我不太明白的是, 考虑 player B, 我们有: : Eb[b1] = L1*U(a1,b1)+L2*U(a2,b1) : Eb[b2] = L1*U(a1,b2)+L2*U(a2,b2) : Eb[b3] = L1*U(a1,b3)+L2*U(a2,b3) : Eb[b4] = L1*U(a1,b4)+L2*U(a2,b4) : The ensure B is indifferent to b1, b2, b3 and b4, we have : Eb[b1] = Eb[b2]=Eb[b3]=Eb[b4] : 这并不能帮助解出 p1,p2, p3 和 p4 呀?
| w****a 发帖数: 155 | 5 谢谢。
我是不是可以这样理解你的解释, 因为B有4个action, 我可以每次选出两个action,
和A的两个action合起来得到一个mixed strategy。 然后再选出B的另外两个action的
组合,在和A的两个action合起来得到另外一个mixed strategy。B有4个action, 所以
我们能得到6个不同的mixed strategy。 然后再从这六个mixed strategies 中根据特
定的指标选出一个。我这样的理解对吗?
另外的问题就是,为了达到equilibrium我们强制不mix 某一个action , 这从物理上应
该怎么解释?
假如B有很多 actions,那求解量将会非常大,请问有好的software package 解决这类
问题吗?
Player
P1-
【在 U*****e 的大作中提到】 : Yeah, you are right. It seems the two players questions are independent. : Then with 4 variables and two equations, you must have multiple solutions. : Now you see for Player B there are 4 equations (3 from what you wrote, 1 : from the definition of probability) and 2 variables. It implies that Player : B might not mix all the actions. If he mixes only 3 actions, then one of P1- : P4 is zero. You might have to discuss all the situations. : Therefore it is necessary to solve both player's problem simultaneou
| U*****e 发帖数: 2882 | 6 四选2,四选三,还有四个都选,都应当考虑.其实本质是四个都选,其他都是利用
game的特殊性质让求解更方便.
第二个问题.为什么要强制不mix某些action?我觉得是因为Player B的某些action作
为纯策略有可能被其他策略dominated,因而要使所有payoff都相等的方程组无解.
我做过的题目大多数是2*2。求解程序不太了解.你可以看看下面这个网页
http://gambit.sourceforge.net/doc/gambit-0.2007.01.30/gambit/c622.html#UTILS.NASH
【在 w****a 的大作中提到】 : 谢谢。 : 我是不是可以这样理解你的解释, 因为B有4个action, 我可以每次选出两个action, : 和A的两个action合起来得到一个mixed strategy。 然后再选出B的另外两个action的 : 组合,在和A的两个action合起来得到另外一个mixed strategy。B有4个action, 所以 : 我们能得到6个不同的mixed strategy。 然后再从这六个mixed strategies 中根据特 : 定的指标选出一个。我这样的理解对吗? : 另外的问题就是,为了达到equilibrium我们强制不mix 某一个action , 这从物理上应 : 该怎么解释? : 假如B有很多 actions,那求解量将会非常大,请问有好的software package 解决这类 : 问题吗?
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