Is there a Nash equilibrium with only one entrant?

Is there a Nash equilibrium with only one entrant?

Is there a Nash equilibrium with only one entrant?
Is there a Nash equilibrium with only one entrant?

Is there a Nash equilibrium with only one entrant? Once again, the an- swer is “no,” since the gross profit for the second entrant is 400 and that exceeds all companies’ entry costs. Thus, if we consider a strategy profile in which one company enters and the other four do not, any of those other four companies could increase its payoff by entering. Hence, it is not an equilibrium for only one of the firms to enter.

■ Is there a Nash equilibrium with two entrants? Again, the answer is “no,” by the same argument (although we’re getting “warm”). Consider a strat- egy profile in which two companies choose enter and the other three choose do not enter. Then each of the latter earns zero. Now examine col- umn 2 in Table 5.6. All of the payoffs are positive, so, regardless of which three companies chose do not enter, a company can earn a positive pay- off from also entering.

■ Is there a Nash equilibrium with three entrants? There is, indeed. In fact, there are six of them: (1) companies 1, 2, and 3 enter; (2) companies 1, 2, and 4 enter; (3) companies 1, 2, and 5 enter; (4) companies 1, 3, and 4 enter; (5) companies 1, 3, and 5; and (6) companies 1, 4, and 5 enter. Consider the first strategy profile. Each of those entrants earns a positive payoff from entering; company 1 earns 150, company 2 earns 90, and company 3 earns 70. Entry is then optimal. As for companies 4 and 5— which have chosen do not enter—if you examine column 3 in Table 5.6, you can see that entry would result in a negative payoff: for company 4 and for company 5. Thus, the strategy profile in which companies 1, 2, and 3 enter and companies 4 and 5 stay out is a Nash equilibrium. One can similarly confirm each of the other five strategy profiles, as well as confirm that any other strategy profile that has three companies enter (such as companies 3, 4, and 5) is not a Nash equilibrium.

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