A paradoxical three-person game: numerical solution

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 = a₁(A←1) + a₂(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:

S_C = a₁(C←3N) + a₂(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

S_B = b₁(B←2) + b₂(B←5) + ··· + b_N(B←3N-1),

with 0 ≤ b_i ≤ 1, ∑b_i = 1, then B's payoff is

U_B = ∑b_i(P(A<i)P(i<C) + P(i<A)P(C<i))=
∑b_i((a₁+···+a_i)(1−a_N−i+2−···−a_N) + (1−a₁−···−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_is_i.

Note that s_i = s_N-i+1.

Case 1: N even. A strategy (a₁,...,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:

s₁ = a₁ = 0,
s₂ = (a₁+a₂)(1−a_N) + (1−a₁−a₂)a_N =0,
...
s_N/2 = (a₁+···+a_N/2)(1−a_(N/2)+2−···−a_N)+(1−a₁−···−a_N/2)(a_(N/2)+2+···+a_N) = 0.

The first equation implies a₁ = 0, so that the second equation reduces to a₂(1−a_N)+(1−a₂)a_N = 0, from which it follows that a₂ = 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

a_(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 (a₁,...,a_N) be an arbitrary strategy for A and s₁,...,s_N 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 + ··· + 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 =
  = (a₁ + ··· + a_(N+1)/2)²+ (1 − a₁ − ··· − 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 (b₁,...,b_N) we have ∑b_is_i ≤ b_(N+1)/2s_(N+1)/2 = ∑b_is'_i, the inequality being strict if any of a₁,..., 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)/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:

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.