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When I do the basic calculations for mixed probability, I get that the Column player always plays B. However, I am getting a negative probability for the row. Any help is appreciated.

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    $\begingroup$ Perhaps there is something wrong with the underlying assumption you are making when you are calculating the mixed Nash equilibrium? $\endgroup$
    – Giskard
    Jun 21 '18 at 6:44
  • $\begingroup$ You are correct that Player 2 always chooses $B$ as it is a dominant strategy. Maybe if you show us the calculations that you made, we can spot the mistake. $\endgroup$
    – tdm
    May 12 at 11:52
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I assume that you are calculating a mixed strategy for the row player so that the column player is indifferent between choosing $A$ and $B$. And the reason why you get a negative probability is that the row player cannot make the column player indifferent by choosing, say strategy $C$, with a positive probability $\in [0,1]$. This is because the column player (always) strictly prefers $B$.

Side-note: To avoid such confusion in the future, try to see whether you can apply math to calculate a mixed strategy to a game. Here you clearly cannot (because of the above argument), and that is why math is giving you a ''weird'' answer. Namely, math yields an answer to a question: What would be a hypothetical probability with which the row player chooses, say, strategy $C$ to make the column player indifferent?

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Hints:

  • As you say, B strictly dominates A for the second player. You can draw a conclusion in terms of Nash equilibria about the second player

  • If the second player plays B then C and D produce the same result for the first player. You can draw a conclusion in terms of Nash equilibria about the first player

If you draw the sensible conclusions, you may be able to see why your calculations for a unique mixed strategy Nash equilibrium is unlikely to work

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