The general form of the N-option Mediocrity game can be depicted as follows:

1 | 2 | 3 | 4 | 5 | 6 | ... | 3N-2 | 3N-1 | 3N |

A | B | C | A | B | C | ... | A | B | C |

Let us assume A's optimum strategy is of mixed type:

S_{A} = a_{1}(A←1) + a_{2}(A←4) + ··· + a_{N}(A←3N-2),

with 0 ≤ a_{i} ≤ 1, ∑a_{i} = 1 (a pure strategy is just a particular case where some a_{i} = 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:

_{C}= a

_{1}(C←3N) + a

_{2}(C←3N-3) + ··· + a

_{N}(C←3).

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

_{B}= b

_{1}(B←2) + b

_{2}(B←5) + ··· + b

_{N}(B←3N-1),

with 0 ≤ b_{i} ≤ 1, ∑b_{i} = 1, then B's payoff is

_{B}= ∑b

_{i}(P(A<i)P(i<C) + P(i<A)P(C<i))=

∑b

_{i}((a

_{1}+···+a

_{i})(1−a

_{N−i+2}−···−a

_{N}) + (1−a

_{1}−···−a

_{i})(a

_{N−i+2}+···+a

_{N}

_{})),

where a_{N−i+2}+···+a_{N} is by convention taken to be zero when i = 1. We abbreviate this expression as

U_{B} = ∑b_{i}s_{i}.

Note that s_{i} = s_{N-i+1}.

Case 1: N even. A strategy (a_{1},...,a_{}_{N}) giving s_{i} = 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:

s_{1} = a_{1} = 0,

s_{2} = (a_{1}+a_{2})(1−a_{N}) + (1−a_{1}−a_{2})a_{N} =0,

...

s_{N/2} = (a_{1}+···+a_{N/2})(1−a_{(N/2)+2}−···−a_{N})+(1−a_{1}−···−a_{N/2})(a_{(N/2)+2}+···+a_{N}) = 0.

The first equation implies a_{1} = 0, so that the second equation reduces to a_{2}(1−a_{N})+(1−a_{2})a_{N} = 0, from which it follows that a_{2} = a_{N} = 0. Going down the rest of equations in a similar way we conclude that every a_{i} except a_{(N/2)+1} must be zero. Hence the unique optimum strategy for A+C has

_{(N/2)+1}= 1,

a

_{i}= 0 otherwise,

which translates to

S_{A}= A←(3N/2)+1,

S_{C }= C←3N/2.

As U_{B} = 0, B's strategy is immaterial and can be identified with

S_{B} = (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 U_{A} = U_{C} = 1/2.

Case 2: N odd, N ≥ 3. Let (a_{1},...,a_{N}) be an arbitrary strategy for A and s_{1},...,s_{N} its associated coefficients. We define a derived strategy (a'_{1},...,a'_{N}) in the following manner:

a'_{(N+1)/2} = a_{1} + ··· + a_{(N+1)/2},

a'_{(N+3)/2} = a_{(N+3)/2} + ··· + a_{N},

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}=

= (a_{1}+ ··· + a_{(N+1)/2})^{2}+ (1 − a_{1}− ··· − a_{(N+1)/2})^{2}=

= s_{(N+1)/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 (b_{1},...,b_{N}) we have ∑b_{i}s_{i} ≤ b_{(N+1)/2}s_{(N+1)/2} = ∑b_{i}s'_{i}, the inequality being strict if any of a_{1},..., a_{(N−1)/2},a_{(N+5)/2},...,a_{N} 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

U_{B} = b_{(N+1)/2}s_{(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:

S_{A}= (1/2)(A←(3N−1)/2) + (1/2)(A←(3N+5)/2),

S_{B}= B←(3N+1)/2,

S_{C} = (1/2)(B←(3N−3)/2) + (1/2)(B←(3N+3)/2),

yielding payoffs U_{A} = U_{C} = 1/4, U_{B} = 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.

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