# Reducing an infinitely large matrix to a finite matrix using domination

I'd very much appreciate it if anyone with any familiarity of game theory could help out a newbie. I came across the following problem doing homework in an introductory game theory course:

Consider the following game. Left and right each choose a positive integer. If the integers are equal, there is no payout. If they differ by one, the player with the larger number wins one point. If they differ by at least two, the player with the larger number loses two points.

The prompt implied heavily that we'd be able to reduce this to a finite matrix using domination. To my untrained eye, this appeared to be a general sum game with matrix A being:

$\begin{matrix} 0 & 0 & 0 & 0 & . & . & . & . \\ 1 & 0 & 0 & 0 & . & . & . & . \\ -2 & 1 & 0 & 0 & . & . & . & . \\ -2 & -2 & 1 & 0 & . & . & . & . \\ . & . & . & . & . & . & . & . \\ . & . & . & . & . & . & . & .\end{matrix}$

The only row I see that can be removed by domination is the first one, and I don't see any columns that I can get rid of...so this leaves me with an infinitely large matrix :(

I am VERY new to game theory (we've only had 3 lectures) so my apologies for any errors.

Regardless, can anyone point me in the right direction?

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## 1 Answer

I don't know the answer, but I can tell you where I first saw this game discussed: Herstein and Kaplansky, Matters Mathematical, published by Chelsea in 1978. They call it "the game without a name," and the discussion begins on page 212.

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Thanks for the answer. According to the problem, it was posed by Mendelsohn in 1946. – Arthur Skirvin Jun 27 '11 at 1:59
@Arthur, see Mendelsohn's paper, A psychological game, Amer Math Monthly 53 (1946) 86-88. He generalizes the game, then solves it. – Gerry Myerson Jun 27 '11 at 2:10
@Arthur: Presumably it's the difference in points between Left and Right that determines the final outcome, so from Left's point of view the effective payoff is $0$ when the integers are equal, $1$ when Left's integer is one more than Right's, $-1$ when Left's integer is one less than Right's, $2$ when Left's integer is at least two less than Right's, and $-2$ when Left's integer is at least two more than Right's. This isn't quite the same as Mendelsohn's game, but it's of the same type, and it looks to me as if it reduces to a $3 \times 3$ game. – Brian M. Scott Jun 27 '11 at 2:31
Thanks! (Yikes, though. I can't imagine what we'll be expected to do later on if this is material we're supposed to be able to get for ourselves in the first week.) – Arthur Skirvin Jun 27 '11 at 4:44