0
$\begingroup$

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.

Question

$\endgroup$
  • 2
    $\begingroup$ 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
1
$\begingroup$

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?

| improve this answer | |
$\endgroup$
0
$\begingroup$

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

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.