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Mixed Strategies

 

example3516

This game has two pure strategy equilibria, namely one of the two firms enters the market niche and the other stays out. But, unlike the games we have encountered thus far, neither player has a dominant strategy. When a player has no dominant strategy, she should consider playing a mixed strategy. In a mixed strategy, each of the various pure strategies is played with some probability, say tex2html_wrap_inline9024 for Strategy 1, tex2html_wrap_inline9026 for Strategy 2, etc with tex2html_wrap_inline9028 . What would be the best mixed strategies for Firms A and B? Denote by tex2html_wrap_inline9024 the probability that Firm A enters the market niche. Therefore tex2html_wrap_inline9032 is the probability that Firm A stays out. Similarly, Firm B enters the niche with probability tex2html_wrap_inline9034 and stays out with probability tex2html_wrap_inline9036 . The key insight to a mixed strategy equilibrium is the following. Every pure strategy that is played as part of a mixed strategy equilibrium has the same expected value. If one pure strategy is expected to pay less than another, then it should not be played at all. The pure strategies that are not excluded should be expected to pay the same. We now apply this principle. The expected value of the ``Enter'' strategy for Firm A, when Firm B plays its mixed strategy, is

displaymath9038

The expected value of the ``Stay out'' strategy for Firm A is tex2html_wrap_inline9040 . Setting tex2html_wrap_inline9042 we get

displaymath9044

Using tex2html_wrap_inline9046 , we obtain

displaymath9048

Similarly,

displaymath9050

As you can see, the payoffs of this mixed strategy equilibrium, namely (0,0), are inefficient. One of these firms could make a lot of money by entering the market niche, if it was sure that the other would not enter the same niche. This assurance is precisely what is missing. Each firm has exactly the same right to enter the market niche. The only way for both firms to exercise this right is to play the inefficient, but symmetrical, mixed strategy equilibrium. In many industrial markets, there is only room for a few firms - a situation known as natural oligopoly. Chance plays a major role in the identity of the firms that ultimately enter such markets. If too many firms enter, there are losses all around and eventually some firms must exit. From the mixed strategy equilibrium, we can actually predict how often two firms enter a market niche when there is only room for one: with the above data, the probability of entry by either firm is 2/3, so the probability that both firms enter is tex2html_wrap_inline9054 . That is a little over 44% of the time! This is the source of the inefficiency. The efficient solution has total payoff of 100, but is not symmetrical. The fair solution pays each player the same but is inefficient. These two principles, efficiency and fairness, cannot be reconciled in a game like Market Niche. Once firms recognize this, they can try to find mecanisms to reach the efficient solution. For example, they may consider side payments. Or firms might simply attempt to scare off competitors by announcing their intention of moving into the market niche before they actually do so.

An important case where mixed strategies at equilibrium are always efficient is the case of constant-sum games. Interestingly, the optimal mixed strategies can be computed using linear programming, one linear program for each of the two players. We illustrate these ideas on an example.

example3554

The payoff matrix is as follows.

tex2html_wrap_inline9056

Suppose Player 1 chooses the mixed strategy tex2html_wrap_inline6914 , where

eqnarray3581

The expected payoff of Player 1 is tex2html_wrap_inline9060 when Player 2 plays Stone, tex2html_wrap_inline9062 when Player 2 plays Paper and tex2html_wrap_inline9064 when Player 2 plays Scissors. The best Player 2 can do is to achieve the minimum of these three values, namely tex2html_wrap_inline9066 . Player 1, on the other hand, should try to make the quantity v as large as possible. So, Player 1 should use probabilities tex2html_wrap_inline5252 , tex2html_wrap_inline5256 and tex2html_wrap_inline5270 that maximize tex2html_wrap_inline9066 . This can be written as the following linear program.

eqnarray3587

Similarly, we obtain Player 2's optimal mixed strategy tex2html_wrap_inline9078 , where

eqnarray3590

by solving the linear program

eqnarray3595

For the Stone, Paper, Scissors game, the optimal solutions are tex2html_wrap_inline9080 and tex2html_wrap_inline9082 with tex2html_wrap_inline9084 .

In general, for any 2-person zero-sum game, let tex2html_wrap_inline9086 and tex2html_wrap_inline9088 be the optima of the linear programs for Players 1 and 2 respectively. Because of the special form of these linear programs, it can be shown that tex2html_wrap_inline9090 . So, the payoff at equilibrium is tex2html_wrap_inline9092 . The fact that, for each player, the equilibrium solutions are the optimal solutions of a linear has another consequence: All equilibrium solutions have the same payoff.


next up previous contents
Next: n-Person Games Up: Two-Person Games Previous: Pure Strategies

Michael A. Trick
Mon Aug 24 16:30:59 EDT 1998