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# Game Theory - Mixed Strategy watch

1. Any suggestions of how to get started with this problem would be great:

.................Rock (R) Paper (P) Scissors (S)
Rock (R)......(0,0)......(-1,1).......(1,-1)
Paper (P)....(1,-1).....(0,0)........(-1,1)
Scissors (S).(-1,1).....(1,-1).......(0,0)

Show that there is no Nash-Equilibrium where player I mixes between Rock and Paper, but puts probability zero on Scissors. Hint: To do this you use the method of proof by contradiction. That is assume the opposite of what you want to show, namely assume that there is such equilibrium and then show that the assumption implies a false statement (for example 5=0).

Thanks a lot
2. what happens if player 1 mixes rock and paper and player 2 mixes scissors and paper?

Edit: let me give you some more detail. A mixed strategy equilibrium must be as good a strategy as each of its component pure strategies. Compare the mixed strategy Rock/Paper to the pure strategy Rock against an opposition's response of Scissors/Paper.
3. (Original post by py0alb)
what happens if player 1 mixes rock and paper and player 2 mixes scissors and paper?

Edit: let me give you some more detail. A mixed strategy equilibrium must be as good a strategy as each of its component pure strategies. Compare the mixed strategy Rock/Paper to the pure strategy Rock against an opposition's response of Scissors/Paper.
Thanks a lot for your reply. I'm still a bit uncertain of what to do. I can understand that I need to mix player I between rock and paper to answer this question. But not sure why I need to mix player II between scissors and paper? Could you help further? Thanks a lot.
4. (Original post by BJP)
Thanks a lot for your reply. I'm still a bit uncertain of what to do. I can understand that I need to mix player I between rock and paper to answer this question. But not sure why I need to mix player II between scissors and paper? Could you help further? Thanks a lot.

A Nash equilibrium means that neither player can improve their strategy. It's like a stable point from which neither player has any immediate incentive to move. Each player is playing his best response to his opponents strategy.

So the way you want to go about trying to find a NE is to ask yourself: if player one plays a mixed strategy or rock and paper, what is player 2's best response? My initial guess was a mixed strategy of scissors and paper - but thinking about it, his best response would be to simply play a pure strategy of paper. That way he will win 50% and draw 50% - an expected payoff of 0.5.

So: can player 1 improve his strategy from one in which he has an expected payoff of -0.5?. If he can, then its not a Nash equilibrium. Simply think of any other strategy that has a higher payoff, and you have shown that this is not a NE.

In fact, the only NE for the rock-paper-scissors game is both players playing a mixed strategy between all three options (1/3, 1/3, 1/3). That way they will both win 50%.Anything else and the opposition can take advantage and they will lose more than they win.
5. (Original post by py0alb)
A Nash equilibrium means that neither player can improve their strategy. It's like a stable point from which neither player has any immediate incentive to move. Each player is playing his best response to his opponents strategy.

So the way you want to go about trying to find a NE is to ask yourself: if player one plays a mixed strategy or rock and paper, what is player 2's best response? My initial guess was a mixed strategy of scissors and paper - but thinking about it, his best response would be to simply play a pure strategy of paper. That way he will win 50% and draw 50% - an expected payoff of 0.5.

So: can player 1 improve his strategy from one in which he has an expected payoff of -0.5?. If he can, then its not a Nash equilibrium. Simply think of any other strategy that has a higher payoff, and you have shown that this is not a NE.

In fact, the only NE for the rock-paper-scissors game is both players playing a mixed strategy between all three options (1/3, 1/3, 1/3). That way they will both win 50%.Anything else and the opposition can take advantage and they will lose more than they win.
Hi,

Thanks a lot for your help. I think I understand what you are describing.

I think in my question though, that I need to show a proof by contradiction? I.e. to play this strategy of mixing between R and P; Expected pay-off R = Expected pay-off from P and Expected pay-offs from R and P must be equal to or greater than the expected pay-off from S. Could you help me with this? Thanks a lot
6. (Original post by BJP)
Hi,

Thanks a lot for your help. I think I understand what you are describing.

I think in my question though, that I need to show a proof by contradiction? I.e. to play this strategy of mixing between R and P; Expected pay-off R = Expected pay-off from P and Expected pay-offs from R and P must be equal to or greater than the expected pay-off from S. Could you help me with this? Thanks a lot
As far as I understand it, the method I suggested above is the way you are expected to do the problem. It is using proof by contradiction, because it is showing that there is a higher payoff to be had by using a strategy other than R/P, and thus R/P is not a NE.
7. (Original post by py0alb)
As far as I understand it, the method I suggested above is the way you are expected to do the problem. It is using proof by contradiction, because it is showing that there is a higher payoff to be had by using a strategy other than R/P, and thus R/P is not a NE.
OK thanks.

When you say, "So: can player 1 improve his strategy from one in which he has an expected payoff of -0.5?. If he can, then its not a Nash equilibrium. Simply think of any other strategy that has a higher payoff, and you have shown that this is not a NE."

Could you give an example? Thanks again
8. (Original post by BJP)
OK thanks.

When you say, "So: can player 1 improve his strategy from one in which he has an expected payoff of -0.5?. If he can, then its not a Nash equilibrium. Simply think of any other strategy that has a higher payoff, and you have shown that this is not a NE."

Could you give an example? Thanks again
Right, so if P1 plays (0.5 0.5 0), then P2's best response is to play (0 1 0), giving P1 an expected payoff of -0.5 and P2 an expected payoff of 0.5. (Remember the expected payoffs have to add up to 0)

To be a Nash Equilibrium, then no change that either player can make can improve his expected payoff. But think about what happens if P1 changes to (0 0 1): his expected payoff becomes 1. As 1 > -0.5, then we were NOT in a Nash Equilibrium. Simple as that.

The fact I chose (0 0 1) isn't important, I could have chosen almost anything.
9. (Original post by py0alb)
Right, so if P1 plays (0.5 0.5 0), then P2's best response is to play (0 1 0), giving P1 an expected payoff of -0.5 and P2 an expected payoff of 0.5. (Remember the expected payoffs have to add up to 0)

To be a Nash Equilibrium, then no change that either player can make can improve his expected payoff. But think about what happens if P1 changes to (0 0 1): his expected payoff becomes 1. As 1 > -0.5, then we were NOT in a Nash Equilibrium. Simple as that.

The fact I chose (0 0 1) isn't important, I could have chosen almost anything.
OK, right, thanks.

One last thing, "Right, so if P1 plays (0.5 0.5 0)" - how do we know that he plays 0.5, 0.5, 0 and not 0.75, 0,25, 0 (or something like that)?

Thanks.
10. (Original post by BJP)
OK, right, thanks.

One last thing, "Right, so if P1 plays (0.5 0.5 0)" - how do we know that he plays 0.5, 0.5, 0 and not 0.75, 0,25, 0 (or something like that)?

Thanks.
We don't, but it doesn't matter, because once P2 makes his best response, then his EP can never be as high as 0, which is what it would need to be to be a NE.

EDIT: I just did a back of an envelope calculation to calculate the subgame NE for player 1 restricted to Rock and Paper, and the NE is (1/3 2/3 0), to which P2 responds (0 2/3 1/3). Player 1's expected payoff is -1/3.

So if you want to be a bit more sophisticated, calculated this subgame NE like I just did, and then demonstrate that is not a total game NE by showing that the BR against player 2's stategy against (0 2/3 1/3) is actually (2/3 0 1/3).
11. (Original post by py0alb)
We don't, but it doesn't matter, because once P2 makes his best response, then his EP can never be as high as 0, which is what it would need to be to be a NE.

EDIT: I just did a back of an envelope calculation to calculate the subgame NE for player 1 restricted to Rock and Paper, and the NE is (1/3 2/3 0), to which P2 responds (0 2/3 1/3). Player 1's expected payoff is -1/3.

So if you want to be a bit more sophisticated, calculated this subgame NE like I just did, and then demonstrate that is not a total game NE by showing that the BR against player 2's stategy against (0 2/3 1/3) is actually (2/3 0 1/3).
Could you set out the steps in your calculations?

When you say subgame, are we still talking about the mixed strategy? Thanks a lot.
12. (Original post by BJP)
Could you set out the steps in your calculations?

When you say subgame, are we still talking about the mixed strategy? Thanks a lot.
Watch lectures 8, 9 and 10 (this will take 4 hours, but its more than worth it):

http://academicearth.org/courses/game-theory

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