What is a stable configuration in this society?
What is a stable configuration in this society? That is, when are all sneetches content with their bellies? To answer this question requires deriving Nash equilibria. We can think of each sneetch’s strategy set as including star and nothing, and the objective is to find a strategy profile whereby each and every sneetch cannot do better by doing something different.
Accordingly, consider a strategy profile whereby so that fewer than half of the sneetches choose to be star bellied. For those sneetches who choose to have a star, each is clearly acting optimally, since its payoff is 1 and that’s the highest payoff there is. But what about the sneetches who have chosen not to have a star? Well, they each have a zero payoff. To deter- mine whether they can do better, there are two cases to consider.
In the first case, the number of sneetches with stars is not only less than n/2, but also less than That is, or, equivalently,
An example is the top row in FIGURE 5.3 where and A sneetch who originally planned not to have a star would increase
the number of star-bellied sneetches from m to if, instead, she got a star. This situation is shown in the bottom row in Figure 5.3. Since we have supposed that it is still the case that star-bellied sneetches are in the minority. (Again, see Figure 5.3.) This means that a sneetch without a star can go from being one of the majority without a star (and earning a pay- off of 0) to being one of the minority with a star (and earning a payoff of 1).
m � 1 6 n/2,
m � 1 m � 2.
n � 7m � 1 6 n/2. m 6 (n/2) � 1,(n/2) � 1.
n � m
m 6 n/2,
5.2 Symmetric Games 121
Because any of the starless sneetches would then prefer to have a star, this case is not a Nash equilibrium. In sum, if, at the end of the day, the number of star-bellied sneetches is less than then this strategy profile is not a Nash equilibrium.
In the second case, we continue to suppose that so that star-bellied sneetches are rarer, but now assume that their number exceeds that is, Since we are then assuming both and
these inequalities can be combined to yield But this just means that we are supposing that the number m of star-bellied sneetches equals (Recall that n is odd.) That is, m is the highest in- teger less than half of n. An example is shown in the top row of FIGURE 5.4, where and, therefore, So, is this case an equilibrium? We already know that the sneetches with stars can’t do any better. A sneetch without a star is getting a zero payoff, since more than half of the sneetches have chosen not to have stars. But because if one of those starless sneetches now chooses to have a star, then there would be or sneetches with stars. With one sneetch having changed her mind, there would be more sneetches with stars than without, which would mean the star-bellied sneetches now have a zero payoff. This becomes clear if we move from the top to the bottom row in Figure 5.4. Since having a star doesn’t raise the sneetch’s payoff, the sneetch is content not to have a star. In other words, when (n � 1)/2
(n � 1)/2,((n � 1)/2) � 1, m � (n � 1)/2,
m � 3.n � 7
(n � 1)/2.
(n/2) � 1 6 m 6 n/2.m 6 n/2, (n/2) � 1 6 m(n/2) � 1 6 m.
(n/2) � 1; m 6 n/2,
(n/2) � 1,
FIGURE 5.3 A Starless Sneetch Gets a Star (n � 7, m � 2)
FIGURE 5.4 A Starless Sneetch Gets a Star (n � 7, m � 3)
122 CHAPTER 5: STABLE PLAY: NASH EQUILIBRIA IN DISCRETE n-PLAYER GAMES
sneetches have a star and don’t, a starless sneetch is “damned if she does and damned if she does- n’t”: the sneetch will be in the majority, regardless of what she does. A starless sneetch can then do no better than to remain starless.
In conclusion, there is a Nash equilibrium in which sneetches have stars. By an analogous argu-
ment, there is also a Nash equilibrium in which sneetches have stars. (Convince yourself.) We
then have two stable configurations for a society of sneetches: either have stars (and it is the star- bellied sneetches who are esteemed), or have stars (and it is the starless sneetches who are esteemed).