Monopoly: Mixed Strategy and Driving Out Your Rivals
Microsoft versus Netscape
Each firm is suffering losses of $2 per period while they are both selling internet browsers. When either firm has no competitor they earn a monopoly profit of $10 in each period. Each firm can choose when to drop out. They can drop out immediately, 1996, and lose nothing. They can wait a year and drop out in 1997, or they can stay until the end in 1998.
Microsoft | ||||
1996 | 1997 | 1998 | ||
Netscape |
1996 | 0, 0 | 0, 10 | 0, 20 |
1997 | 10, 0 | -2, -2 | -2, 8 | |
1998 | 20, 0 | 8, -2 | -4, -4 |
As usual we'll use the 'best response' technique to find the pure strategy Nash equilibria.
Netscape's best response | Microsoft's best response |
bN(1996M) = 1998 | bM(1996N) = 1998 |
bN(1997M) = 1998 | bM(1997N) = 1998 |
bN(1998M) = 1996 | bM(1998N) = 1996 |
Notice that neither Netscape nor Microsoft will play 1997 as a best response. When Netscape conjectures that Microsoft will stay in until 1998 then its best response is to drop out immediately. When Netscape conjectures that Microsoft will drop out immediately then its best response is to stay in until the end. The game is symmetric so you could just switch the names of the firms and the previous two sentences would remain unchanged. Hence, there are two pure strategy Nash equilibria.
Would Netscape ever expect Microsoft to concede? Is it reasonable to expect that Bill Gates would willingly forego an opportunity to earn a profit $20? Your answers to these questions make it inadvisable for either firm to play a pure strategy.
Let's suppose that Microsoft plays 1996 with probability p and 1998 with probability 1-p. Since a mixed strategy will never include in its support a strategy which is not itself a best response we needn't include a probability for Microsoft playing 1997. Then we can calculate Netscape's profits from playing pure strategies against MS's mixed strategy as
Netscape plays a pure strategy against MS's mixed strategy (1996, p) | |
Netscape's Strategy | Netscape's Expected Profit |
Drop out immediately | 0 |
Drop out in 1997 | 10p - 2(1-p) |
Play through 1998 | 20p - 4(1-p) |
You could make a similar expected payoff table for Microsoft just by switching the names. For this purpose we will use q to represent the probability that Netscape plays 1996 in its mixed strategy.
Microsoft plays a pure strategy against Netscape's mixed strategy (1996, q) | |
MS's Strategy | MS's Expected Profit |
Drop out immediately | 0 |
Drop out in 1997 | 10q - 2(1-q) |
Play through 1998 | 20q - 4(1-q) |
We can add the results of these expected value calculations to the original payoff matrix.
Microsoft | |||||
1996 | 1997 | 1998 | Mixed (1996, p) | ||
Netscape |
1996 | 0, 0 | 0, 10 | 0, 20 | 0, 20(1-p) |
1997 | 10, 0 | -2, -2 | -2, 8 | 10p-2(1-p), 8(1-p) | |
1998 | 20, 0 | 8, -2 | -4, -4 | 20p-4(1-p), -4(1-p) | |
Mixed (1996,q) | 20(1-q), 0 | 8(1-q), 10q-2(1-q) | -4(1-q), 20q-4(1-q) |
The new row and column are constructed as follows: If Netscape plays its mixed strategy (1996, q) and Microsoft plays the pure strategy 1996, then Netscape will earn an expected 20(1-q) and Microsoft will earn zero. If Microsoft plays its mixed strategy (1996, p) against Netscape's pure strategy of 1997, then it earns an expected profit of 8(1-p) and Netscape earns an expected profit of 10p-2(1-p).
The last row and last column cell corresponds to the expected payoff when both Microsoft and Netscape are playing mixed strategies. To figure out their expected payoffs we'll use another table. In constructing the following table keep in mind that when they both play a mixed strategy there are four possible outcomes.
Netscape Plays | Microsoft Plays | Probability of Intersection | Payoff |
Payoff x Probability |
|
1996 | 1996 | qp | 0, 0 | 0 | 0 |
1996 | 1998 | q(1-p) | 0, 20 | 0 | 20q(1-p) |
1998 | 1996 | (1-q)p | 20, 0 | 20(1-q)p | 0 |
1998 | 1998 | (1-q)(1-p) | -4, -4 | -4(1-q)(1-p) | -4(1-q)(1-p) |
Netscape's expected payoff = 20(1-q)p - 4(1-q)(1-p)
Microsoft's expected payoff = 20q(1-p) - 4(1-q)(1-p)
Now we are ready to fill in the last cell of the table.
Microsoft | |||||
1996 | 1997 | 1998 | Mixed (1996, p) | ||
N |
1996 | 0, 0 | 0, 10 | 0, 20 | 0, 20(1-p) |
1997 | 10, 0 | -2, -2 | -2, 8 | 10p-2(1-p), 8(1-p) |
|
1998 | 20, 0 | 8, -2 | -4, -4 | 20p-4(1-p), -4(1-p) |
|
Mixed (1996,q) | 20(1-q), 0 | 8(1-q), 10q-2(1-q) |
-4(1-q), 20q-4(1-q) |
20(1-q)p -
4(1-q)(1-p) 20q(1-p) - 4(1-q)(1-p) |
If there is a mixed strategy for Netscape then it must be one that makes Microsoft indifferent between the yields associated with any of the three pure strategies. That is, Microsoft is indifferent about its date of exit. Or, Microsoft's own mixed strategy must be such that 0 = 10p-2(1-p) = 20p-4(1-p). If you solve any pair of equations we see that a mixed strategy makes sense for Netscape if Microsoft's mixed strategy sets p=1/6. By symmetry, we must also see that q = 1/6. This results in the startling outcome that with probability 1/36 both firms abandon a market that would be profitable for one of them, with probability 25/36 both stay the course and lose money, and with probability 10/36 one of them abandons the market and the other firm earns the monopoly profit.
You should be able to demonstrate that if p=1/6 then Netscape is indifferent between a pure strategy and a mixed strategy of its own. Hence, the situation when both Netscape and Microsoft play a mixed strategy is also a Nash Equilibrium.