I want now to take some time to make sure that everyone understands just how strong these assumptions are, and also exactly how to interpret them. The first point to understand is in a way the hardest. You might think that our subject, A, decides how she feels about all of these simple and compound lotteries by carrying out expected utility calculations and then saying to herself, "Well, since this one has a greater mathematical expectation than that one, I prefer this one to that one." You might think that, because, good heavens, how else could she possibly decide which she prefers to which? But if you thought that [which of course none of you does], you would be **WRONG, WRONG, WRONG! TOTALLY WRONG, WRONG, WRONG!** That would be, to use correctly a phrase that these days is almost always used incorrectly, **begging the question**. It would be assuming what is to be proved, and thus arguing in a circle. What von Neuman actually supposes is that our subject, A, looks at the outcomes O1, O2, etc and decides how she feels about them. She ranks them in order of her preference. She then looks at the infinitude of simple lotteries and compound lotteries and decides how she feels about them as well. She merges this all in her mind into a single complete, transitive ordering of all of those outcomes and simple lotteries and compound lotteries. **Then** von Neuman posits that her preferences, thus arrived at, in fact obey the six Axioms. **If that is so**, then, von Neuman shows, her preferences can be represented **AS THOUGH** she were carrying out expected utility calculations in her head in accordance with the axioms.

We are talking here about an enormously powerful set of idealizing and simplifying assumptions, as powerful in their way as the assumptions economists have to make before they can talk about continuously twice differentiable production functions [which they need in order to prove their nifty equilibrium theorems.] Let me draw on something I said earlier to show you just how powerful these Axioms are. Look at Axiom V, the transitivity axiom, and let us recall the eye doctor example. Suppose that the lotteries A is comparing are big Amusement Park wheels, on which are marked off different sized wedges [each defined by two radii], each one of which is associated with one of the outcomes in the set, O. It would be no problem at all to construct a whole series of wheels, each of which is such a tiny bit different from the one next to it that when A is shown the two wheels together, she looks at them and says, I am indifferent between those two lotteries." But suitably arranged, the series of wheels might very slowly, indiscernibly, alter the size of the wedges associated with two prizes or outcomes, Oi and Oj, until, if we were to show A the first and the last in the series, she would look at them and say, flatly, I prefer the one on the left to the one on the right. Whoops. No transitivity! Axiom V rules out any such state of affairs.

Well, you can think about each one of the Axioms and see whether you can imagine a situation in which the assumption of that Axiom clearly requires something very strong and even counterintuitive. But rather than go on about that, I am going to take the next step.

We are now ready to extend our notion of strictly opposed preference orders. Recall that we describe the preference orders of A and B over a set of outcomes, O, as "strictly opposed" when A prefers Oi to Oj or is indifferent between them if and only if B prefers Oj to Oi or is indifferent between them. We will describe the preference orders of A and B over the infinite set of lotteries, simple and compound, over the set of outcomes, O, as "strictly competitive" when A prefers Lottery L1 to Lottery L2 or is indifferent between them if and only if B prefers L2 to L1 or is indifferent between them. This means that A and B not only rank all of the outcomes in exactly opposite ways. They also rank all of the lotteries, simple or compound, over those outcomes in exactly opposite ways.

In this very specific set of circumstances [where all six axioms apply to both A's preferences and B's preferences, and A and B have strictly competitive preferences], we can normalize the utility functions of A and B so that for any lottery, L, simple or compound, over the set of outcomes, O, the sum of the utility index assigned to L by A's utility function and the utility index assigned to L by B's utility function is a constant. This is what is meant by saying that a game played by A and B is a **constant sum game**.

Rather than grind out an algebraic proof, I will offer a simple, intuitive proof that should be easy to grasp. We shall use u(L) to mean the utility that A's utility function assigns to L, and u'(L) to mean the utility that B's utility function assigns to L. Now, we are permitted arbitrarily to let A's most preferred outcome, O1, have a utility of 1, and A's least preferred outcome have a utility of 0. Since A and B have strictly opposed preferences for outcomes, B's most preferred outcome is On and his least preferred outcome is O1. We are permitted to set B's utility for On equal to 1 and for O1 equal to 0. So the utility assignments of both A and B can be portrayed as lying along a line that runs between 1 and 0.

No matter what lottery, L, we have chosen, we know from the Axioms that it is equivalent, for A, to some lottery over just O1 and On whose probability weights are u and (1-u) for some u. Think of that as a point somewhere on the line running between 1 and 0. [Remember that for the best and worst alternatives, O1 and On, the point is an endpoint of the line.] The same thing is true for B. We are now going to prove that the point on the line representing A's utility for L and the point on the line representing B's utility for L are the same point. To prove this, we will assume the contrary and derive a contradiction with our assumption that A and B have strictly opposed preferences. So, let us choose a point representing u(L) and a different point representing u'(L), and then choose some point that lies between those two points, which we shall call S. Here is a picture of the situation. The line runs from 1 to 0 for A, and from 0 to 1 for B:

1 0

|-----------------u(L)---------------S-------------u'(L)-------------------------|

0 1

The point S represents a lottery, Ls, with weights S for On and (1-S) for O1. Now, just from looking at the diagram, we can see the following:

(i) A prefers L to Ls, because L puts greater weight on O1 than Ls does. [u(L) is closer to the 1 than S is].

(ii) B prefers L to Ls, because L puts greater weight on On [his favorite] than Ls does. [u'(L) is closer to *his* 1 than S is.]

But this means that A and B do not have strictly opposed preferences, since they both prefer L to Ls. And this contradicts the assumption. So no matter which lottery L we choose, there cannot be a point S between u(L) and u'(L), which means they are the same point.

But if they are the same point, then A's utility is u and B's utility is u' = (1-u), regardless of which lottery, L, we choose. and:

u + u' = u + (1-u) = 1

Now, B's utility function is invariant under an affine (linear) transformation. So let us introduce the following affine transformation:

u'' = u' - 1

What this does is to re-label B's utility assignments so that instead of running from 1 to 0, the run from 0 to -1. This means that A's and B's utilities for any arbitrary lottery L are no longer u and (1-u). Instead, they are now u and -u. And the sum of u and -u is zero.

**THIS, AND ONLY THIS, IS WHAT IS MEANT BY SAYING THAT A GAME PLAYED BY A AND B IS A ZERO-SUM GAME**.

Is it possible for a game with more than two players to be a zero sum game? If we have a game with players A, B, and C, and if B and C's utility goes down by the amount which A's goes up, etc, then do we have a multiple person zero sum game?

ReplyDeleteNo. Since in general it is impossible to add up the utility indices of different players [because each of them is invariant under an affine transformation, so the sum is meaningless], it is only under the very special case of a two person game with strictly competitive preferences that we can represent their separate utility indices in a manner that has a constant sum. Once we establish that the sum is constant, a trivial transformation can adjust the sum to be zero.

ReplyDeleteNote that in the absence of this set of circumstances [and of course for a game with more than two players], we do NOT have a "Variable sum game." We have a game in which the concept of a sum of utilities cannot be defined.

It is of course always possible to have a situation in which the money won and lost by players sums to zero. That is always true in a friendly poker game without a house taking a cut. But that tells us nothing at all about the utility indices that players attach to those sums of money.