Economics - Game Theory Question

[tabako]

I am having a hard time intuitively understanding the mixed Nash equilibrium in certain games. I can work out the numbers well enough, but I can't seem to grasp why they are Nash equilibrium.

Take the coordination game:


I can easily identify the 3 Nash equilibrium, but the mixed strategy one does not really make sense to me.

The way I see it, is if I know that the other player is going to random his choices (50%,50%), then any mixed strategy of mine will have the same expected utility. I make the case that any strategy other than (50%,50%) is optimal because then the other player would adjust to a pure strategy, and we would be able to shift to a pure Nash equilibrium, both of which have higher expected utilities than the mixed strategy Nash equilibrium (50% of the time in the mixed equilibrium, we choose opposing strategies and both receive the payoff of 0).

Maybe my problem is that I am thinking of this in terms of playing the game multiple times (am I supposed to? I don't really know). If anybody can add some insight into this, it would be greatly appreciated. Thanks.

[T50_Omaha8]

Unless it is specified you generally don't play a game multiple times. And I believe that is a property of a mixed strategy Nash equilibrium--your expected payoff is constant no matter what the opponent does. So in any of these problems where you find a mixed strategy, it shouldn't matter what your opponent does, you'll still have the same EV. My game theory class was super easy though so I don't know that much more on it.

[venom007]

Right-- thats the limitation of games like this (im guessing your studying Cournot solution as well right?)-- it only assumes its being played once. Also in the matrix you posted there are not 3 nash equlibria-- how do you figure? The 0,0 and 0,0 are not nash equilibria.

[armPitt]

Tabako is saying there are three equilibrium. The two pure equilibrium (100,100) and another equilibrium, in mixed strategy.

[Split Suit]

for most Econ classes u can assume one trial of the game. if ur havin a difficult time with basic game theory, get 'Prisoner's Dilema' and read the intro section. (great read otherwise btw)

[JaredL]

Tabako,

In a two-player game, a Nash Equilibrium is a set of strategies where given what player 1 is doing, player 2 is acting optimally and given what player 2 is doing, player 1 is acting optimally. Make sure that is straight in your head.

It's important to note that a NE is NOT (necesarily) describing what will or should happen in a game. It's just describing a set of strategies that are stable (which is why it's Nash <u>Equilibrium</u>). In your class, in particular when you move to sequential move games, there are often NE that should never happen and rely on players doing or assuming bizarre things. A NE doesn't have to make sense, it just has to be that each player is playing a best response to the others' strategies.

So with that in mind let's look at this game. We each get 1 if we play the same strategy and zero if we don't. There are obviously 2 pure strategy equilibria. Moreover, there are no equilibria where one player plays a pure strategy and the other plays a mixed strategy.

To see this, note that if player i plays strategy 1, the unique best response for j is to also play 1. The same holds for 2. Therefore, if i plays a pure strategy, j cannot in equilibrium play a mixed strategy (for a mixed strategy to exist in equilibrium a player must mix over best responses).

OK, I'll leave the above as it may be helpful to you, but I just reread your post and figured out your problem.

[ QUOTE ]
The way I see it, is if I know that the other player is going to random his choices (50%,50%), then any mixed strategy of mine will have the same expected utility. I make the case that any strategy other than (50%,50%) is optimal because then the other player would adjust to a pure strategy, and we would be able to shift to a pure Nash equilibrium, both of which have higher expected utilities than the mixed strategy Nash equilibrium (50% of the time in the mixed equilibrium, we choose opposing strategies and both receive the payoff of 0).

[/ QUOTE ]

As I mentioned above, a NE is a set of strategies where given what 1 is doing 2 is responding optimally, and vice-versa. It doesn't matter in the slightest whether or not 1's strategy is a best-response if you slightly alter 2's strategy. In other words, at your current point in the class at least, stability doesn't matter. As you say, if 1 plays (1/2,1/2), then anything 2 does is a best response. This makes it clear that each playing 1/2,1/2 is a NE.

Though you are correct when you say that (1/2,1/2) is not a BR to (1/2+epsilon,1/2-epsilon) for any positive epsilon, it's not at all relevant. That (1/2,1/2) is a BR to (1/2,1/2) is all that matters.

As I said above (ok so it was relevant anyway [img]/images/graemlins/wink.gif[/img] ), your point seems to be that ((1/2,1/2),(1/2,1/2)) should never actually happen. That's true, but it doesn't matter.

Notice that in any NE involving mixing, this is the case. If player i mixes over strategies A, B and C that must mean that she is indifferent between playing A, B, or C. Given that, there is no reason for her to actually mix at all, much less using the exact probabilities that an equilibrium may call for. None the less, it is an equilibrium for her to mix as given, and for other players to do whatever they're supposed to do. Does that mean that something like that will happen if you actually play the game? No, but it is an equilibrium.

Especially in zero-sum games (note that the game in the OP isn't), it can make sense that you have certain mixing under the pretense of not being exploited, but in general there is usually no reason to expect the players to actually mix with the necesary frequencies.

BTW your thinking about what would happen would be much better applied to a course in behavioral and/or experimental economics if that is available at your school. In either field (some think of them as the same field), the researcher gets at what will actually happen when you're playing a game and why.

Hope this helps, feel free to ask follow-up questions.

Jared

[tabako]

That post was very helpful Jared.

I was thinking too much along the lines of a NE being something that would actually happen when playing the game, but you cleared that up.

You mentioned mixing as a way to not be exploited in zero sum games, which is something that did make intuitive sense to me (which is why I was careful to only mention the cooperative/coordination game in my OP). My mistake was lumping the mixed NE from both the cooperative and competitive games into the category of strategies that should actually happen while playing the game.

[xorbie]

tabako,

Jared covered a lot of it. If you want to know more about game theory, I would recommend looking into what are called "evolutionarily stable strategies", which describe "better" Nash Equilibriums in some sense.