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Respond to the question: Solution with a saddle point?

 09/12/2000 09:27 AM by Walter;
I didn't say you can't use linear algebra to find a pure strategy NE. I said that it is easier to check for them first (at least for me it is easier to do it when the game is simple), and also it gives me reference of how many NE in

 09/10/2000 07:44 AM by KR Simpson; Solution...
Thanks Walter. What doesn't make sense to me is why I cannot use linear algebra to determine that there is a pure strategy. In other words, why I don't get t=1, and 1-t=0. There are other pure strategy games where I don't get t=7/5 [View full text and thread]

 09/04/2000 09:11 AM by Walter;
Let me repeat the game

COL L R ROW U 3 5 D 1 -2

The numbers are player ROW's payoffs, since COL's payoffs are the negative of those (or some other number such that the sum of the utilities is the same constant).

Well, U is dominant strategy for ROW. COL will choose L (-3>-5). The Nash equilibrium is in pure strategies (U,L).

When you compute equilibrium/a in mixed strategies, you did it right.

Q=7/5 means that there is no NE in mixed in which player ROW mixes between U and D. In fact 7/5 > 1 and that gives you the information that ROW will play U.

Suppose this other game:

COL L R ROW U (3,-3) (-5,5) D (-1,1) (2,-2)

In this case you must find ma mixed strategy equilibrium

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 09/02/2000 01:59 PM by KR Simpson; Solution with a saddle point...
I am trying to solve the zero-sum game: 3, 5 1, -2 Ignoring for the moment that this game has a saddle point, how can I solve using linear algebra? If I assign the column player fractional times Q and (1-Q), I should get Q=1, as [View full text and thread]