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I'm given the following incomplete Normal-Form Game:

 | L | R
+--------
O|1,?|7,?
U|?,2|?,1

First I was asked to fill out the missing pieces to obtain a game with exactly two Nash-Equilibria in pure strategies. I came up with:

 | L | R
+--------
O|1,0|7,7
U|2,2|0,1

The next question is to fill out the table to obtain a game with exactly one mixed strategy and I'm stuck at this point on how to approach this.

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  • $\begingroup$ Why is this question downvoted? I'd be happy to include more detail if necessary. $\endgroup$ Commented Jul 30, 2019 at 17:57
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    $\begingroup$ I think the down votes to this question was a little harsh. I guess people were hoping to see more effort shown, for example, what you have tried. Do you know what a MSNE is? If so, have you tried to apply it to this context? At which step were you stuck? The more specific the question, the more likely there will be a useful response. $\endgroup$
    – Herr K.
    Commented Jul 30, 2019 at 18:17
  • $\begingroup$ @HerrK. Do you mean Mixed-Strategy-Nash-Equilibrium? I also know how to calculate them for a given game. I could try adding 4 variables at the places of the 4 question marks and start the calculation, but I was hoping for an easier way to get to the result. $\endgroup$ Commented Jul 30, 2019 at 18:42
  • $\begingroup$ You can always try "guess-and-verify" but this depends on how lucky you are with your first few guesses. If you don't want to leave anything to chances, you should do what you said, by replacing the question marks with variables and solve for the unique MSNE. In general, any 2-by-2 game with no pure strategy NE will have a unique MSNE. (But I think this is the point of this exercise, so you shouldn't quote this result.) $\endgroup$
    – Herr K.
    Commented Jul 30, 2019 at 18:57
  • $\begingroup$ @HerrK. Thank you for the approach, is there an easy proof for the general result? How is this obtained? $\endgroup$ Commented Jul 30, 2019 at 19:16

1 Answer 1

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To ensure that the game has a single, completely-mixed, Nash equilibrium there needs to be "cyclic unique best responses", i.e., either

BR(L) = O, BR(O) = R, BR(R) = U, BR(U) = L,

or

BR(L) = U, BR(U) = R, BR(R) = O, BR(O) = L.

The strict cyclic preferences both rule out pure equilibria and allow a (unique) completely mixed equilibrium.

There are infinitely many ways to achieve either of the two cases above. Aiming for the first case with BR(L) = O, given that the payoff to the row player for (O,L) is 1, to ensure that BR(L) = O, i.e., the unique best response against L is O, one can set the payoff to the row player for (U,L) as any number strictly less than 1, i.e. so that against L, the row player strictly prefers O with payoff 1 to U with payoff strictly less than 1. The rest of the payoffs can be filled in similarly.

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  • $\begingroup$ I feel like this is more of a comment as it suggests what to check for to verify but doesn’t necessarily give a solution or how to go about finding one $\endgroup$
    – Brennan
    Commented Jul 31, 2019 at 5:18

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