The numerical solution to the N-option Mediocrity game disproves in a sense the validity of a kind of circular reasoning that we posed in our presentation of the game:

"I must choose among 1,4,...,3N-2. Clearly 1 can never be in the middle, so I discard it. Player B, however, knows I won't play 1, so she won't choose 2, and thus player C, who can also deduce this, won't play 3. Hence 4 is not an option for me either, and this implies B won't play 5,..."

This was the argument of A; B and C can reason in a similar fashion so as to come to the joint conclusion that no valid choice can be made. How come our numerical solution points to the existence of valid strategies after all?

When N is even, we have proved that B cannot win, so any choice is as good as any other for her. This invalidates the reasoning made by A that goes "Player B, however, knows I won't play 1, so she won't choose 2..." This "wavefront" of position discarding advances only as long as the three players assume there is some strategy with non-null payoff.

When N is odd, the alluded wavefront somehow ripples back when the central positions are reached. Consider the game for N=3:

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

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

Let's unfold a plausible, if admittedly not an ineluctable, reasoning for B:

"A cannot choose the losing position 1, and similarly C cannot choose 9, so this discards my positions at 2 and 8 as valid moves. Now, my having discarded 2 and 8 causes A and C to drop 7 and 3, respectively, and by an iteration of the argument 4 and 6 must also go. But now A and C are without any valid choice to make: going one step back results in A choosing 4 and C playing 6, which is a sure loosing strategy for them; one further step back, my opponents can come up with positions beating my choosing 5."

Again, the circularity breaks down once the chain of reasoning leads to a dead end.

N odd, N ≥ 3: implicit coalitions and self-cheating. Let's examine the optimum strategies for A and C when N=3 (the situation is equivalent for other odd values of N greater than 3): Although the rules we established for the game forbid coalitions, an implicit one is formed between A and C so as to erode B's central position at 5. The possible outcomes of the game are then (bolds mark the winner):

A←4,B←5,C←3

A←4,B←5,C←6

A←7,B←5,C←3

A←7,B←5,C←6

so that A and C manage to take a payoff of 1/4 each away from B. The implicit coalition emerges from the fact that A and C are in entirely symmetrical positions and both are perfectly rational, and know the same is true of the other: this allows them to conclude that their strategies must coincide (doing the appropriate reversions). In a way, symmetry makes up for the lack of communication between players.

But, upon tossing the coin to decide whether to play 4 or 7, A could have a final moment of reflection: "Wait, if I don't choose at random and go instead always for 4, my chances of winning rise from 25% to 33%!" What's wrong with defeating at this point? Well, if C does the analogous reasoning and plays 6 unconditionally, then the sure winner is B --a classical Prisoner's dilemma here between A and C. The twist is that we assumed that the optimum strategies for A and C must be symmetrical, so A's defection is tantamount to self-cheating. I don't know whether this conflict is amenable to a rational analysis after all. Maybe Hofstadter's concept of superrationality could be invoked here to escape the riddle, although some, I among them, may be inclined to regard superrationality as a mere sleight of hand.

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