The general form of the N-option Mediocrity game can be depicted as follows:
Let us assume A's optimum strategy is of mixed type:
SA = a1(A←1) + a2(A←4) + ··· + aN(A←3N-2),
with 0 ≤ ai ≤ 1, ∑ai = 1 (a pure strategy is just a particular case where some ai = 1). The critical point in our analysis is the obervation that the roles of A and C are perfectly symmetrical, and thus C's optimum strategy must be the following:
This interdependency of A's and C's strategies allows us to regard the game as a two-person one: B against the "combined player" A+C. If the strategy adopted by B is written down as
with 0 ≤ bi ≤ 1, ∑bi = 1, then B's payoff is
∑bi((a1+···+ai)(1−aN−i+2−···−aN) + (1−a1−···−ai)(aN−i+2+···+aN)),
where aN−i+2+···+aN is by convention taken to be zero when i = 1. We abbreviate this expression as
UB = ∑bisi.
Note that si = sN-i+1.
Case 1: N even. A strategy (a1,...,aN) giving si = 0 for i=1,...,N is clearly optimum (for A+C). We show that such strategy exists and is unique by solving the set of N/2 equations:
s1 = a1 = 0,
s2 = (a1+a2)(1−aN) + (1−a1−a2)aN =0,
sN/2 = (a1+···+aN/2)(1−a(N/2)+2−···−aN)+(1−a1−···−aN/2)(a(N/2)+2+···+aN) = 0.
The first equation implies a1 = 0, so that the second equation reduces to a2(1−aN)+(1−a2)aN = 0, from which it follows that a2 = aN = 0. Going down the rest of equations in a similar way we conclude that every ai except a(N/2)+1 must be zero. Hence the unique optimum strategy for A+C has
ai = 0 otherwise,
which translates to
SC = C←3N/2.
As UB = 0, B's strategy is immaterial and can be identified with
SB = (1/N)((B←2) + ··· + (B←3N-1)).
The winner is A or C depending on which side of A and C consecutive positions B plays. Their payoffs are UA = UC = 1/2.
Case 2: N odd, N ≥ 3. Let (a1,...,aN) be an arbitrary strategy for A and s1,...,sN its associated coefficients. We define a derived strategy (a'1,...,a'N) in the following manner:
a'(N+1)/2 = a1 + ··· + a(N+1)/2,
a'(N+3)/2 = a(N+3)/2 + ··· + aN,
a'i = 0 otherwise;
Then the associated coefficients s'i verify the following:
- s'(N+1)/2 = (a'1 + ··· + a'(N+1)/2)2+ (1 − a'1 − ··· − a'(N+1)/2)2 =
= (a'(N+1)/2)2+ (1 − a'(N+1)/2)2 =
= (a1 + ··· + a(N+1)/2)2+ (1 − a1 − ··· − a(N+1)/2)2 =
- For i < (N+1)/2,
s'i = (a'1 + ··· + a'i)(1 − a'N−i+2 − ··· − a'N) +
+ (1 − a'1 − ··· − a'i)(a'N−i+2 + ··· + a'N) =
= 0(1-0) + (1-0)0 = 0, as all the a'i coefficients involved are other than a'(N+1)/2 and a'(N+3)/2.
- For i > (N+1)/2, s'i = s'N−i+1 = 0, since N−i+1 < (N+1)/2.
So, for any strategy for B of the form (b1,...,bN) we have ∑bisi ≤ b(N+1)/2s(N+1)/2 = ∑bis'i, the inequality being strict if any of a1,..., a(N−1)/2,a(N+5)/2,...,aN is different from zero, hence we conclude that an optimum strategy for A must have all its coefficients set to zero except those at positions (N+1)/2 and (N+3)/2. We are then left with the task of finding positive values of a(N+1)/2 and a(N+3)/2 adding to 1 and minimizing the maximum of the expression
UB = b(N+1)/2s(N+1)/2 = b(N+1)/2((a(N+1)/2)2 + (1 − a(N+1)/2)2),
which is obviously reached at b(N+1)/2 = 1. The solution to this trivial problem is a(N+1)/2 = a(N+3)/2 = 1/2. So, the optimum strategies for each player are:
SA= (1/2)(A←(3N−1)/2) + (1/2)(A←(3N+5)/2),
SC = (1/2)(B←(3N−3)/2) + (1/2)(B←(3N+3)/2),
yielding payoffs UA = UC = 1/4, UB = 1/2.
Although the numerical solution of the N-option Mediocrity game seems entirely satisfactory, the argument stating that optimum strategies for A and C must be symmetrical induces, for N odd, N ≥ 3, some psychological difficulties akin to those arising in the analysis of the Prisoner's dilemma. We will examine this issue in a later entry.