Monday, May 24, 2010

Installment 7.5

I screwed up. This segment should have been posted BEFORE the previous one. Call it Installment 7.5. My apologies. If you are printing these out, insert this installment between Installments 7 and 8.


You might think that most of life is like this, and especially that all bargaining is, but a little reflection will convince you that that is not so. Think of the situation in which Jones has a house to sell and Smith wants to buy a house. They enter into a negotiation, which we can call a bargaining game. Suppose the lowest price for which Jones will sell the house is $350,000 and the highest price Smith will pay for the house is $375,000. Jones cannot get more than $375,000 for the house, and Smith cannot get the house for less than $350,000. Clearly, within that $25,000 spread, they have strictly opposed preferences. But both of them have an interest in concluding a sale, rather than in having the bargaining break down because they cannot come to an agreement in that "bargaining space." So, simplifying considerably, if we suppose there are three possible outcomes, namely (1) a sale price of 355,000, (2) a sale price of 370,000, and (3) no sale price, then clearly Jones prefers (2) to (1) and (1) to (3). Smith prefers (1) to (2) and (2) to (3). Jones' preference order is 213 and Smith's is 123. These are NOT strictly opposed preference orders [because in both orders alternative 3 is last]. Thus, many real world situations to which we might want to apply Game Theory are not cases of strictly opposed preference orders.

    Now consider a simple example of strictly opposed preference orders. Suppose a married couple, Harry and John, are trying to decide where they will go for their vacation, and suppose that all either of them cares about is the weather. For Harry, the warmer the better; for John, the cooler the better. [So why did they get married?, you ask.] They play a game in which the outcomes are the different places they could go for the vacation. Obviously, if Harry prefers destination 1 to destination 2, because 1 is warmer than 2, then we can be sure that Harry will prefer destination 2 to destination 1. You get the idea. Here is the payoff matrix.










9, -9

-4, 4

2, -2

-1, 1


1, -1

3, -3

-1, 1



6, -6

4, -4

5, -5

3, -3


-3, 3

5, -5

1, -1

-2, 2


    This is a more difficult game to analyze, and not merely because each player has four strategies rather than two. The problem is that neither player has a strictly dominating strategy. Consider each of the eight strategies in turn:

    A1 is not best for A if B should choose B2, B3, or B4

    [Because if B2 anything is better, if B3 A3 is better, if B4 A2 or A3 is better]

    A2 is not best if B should choose B1, B2, B3, or B4

    A3 is not best if B should choose B1, or B2

    A4 is not best if B should choose B1. B3. or B4

    B1 is not best if A should choose A1, A2, or A3

    B2 is not best if A should choose A2, A3, or A4

    B3 is not best if A should choose A1, A3, or A4

    B4 is not best if A should choose A1, A2, or A4

    Before we go on, make sure you understand how I arrived at this series of conclusions. Look just for a moment at strategy B3. B says to himself: "If A should choose A1, I will get -2 with B3. But I would get -9 with B1, 4 with B2, and 1 with B4. So clearly B3 does not do best for me no matter what A does, and that is what 'dominant strategy' means. So B3 is not a dominant strategy." The same reasoning leads A and b to conclude that neither one has a dominant strategy.

    Now let us adopt von Neuman and Morgenstern's proposal that the players seek to maximize their security levels. By the same process we followed a short while ago, we find that the security levels for the strategies available to A and B are:

    A1 -4 A2 -1 A3 3 A4 -3

    B1 -9 B2 -5 B3 -5 B4 -3

    So A3 and B4 are the strategies with the maximum security levels, and following von Neuman and Morgenstern's rule, the players choose the strategy pair (A3 B4) which, according to the payoff matrix yields the payoffs (3. -3). If A holds to A3, B cannot do any better by switching strategies, because the other payoffs to B in that row are -6, -4, and -5. If B holds to strategy B4, A cannot any better by switching strategies, because the other payoffs to A in that column are -1, 0, and -2. A pair of strategies with this property is called an equilibrium pair of strategies.

    The following fact is crucial: A pair of strategies (Ai, Bj) is in equilibrium if and only if the entry Aij is the minimum of its row, Ai, and the maximum of its column, Bj. Here is a proof of that important proposition:

    To say that Ai and Bj are in equilibrium is to say that neither player can improve his or her payoff by a strategy switch so long as the other player holds firm. This means that A's payoff, Aij, is larger than any other payoff in its column, these being the payoffs available to A when B is holding to the strategy Bj. By the same reasoning, Bij is the largest, or most preferred, payoff to B in row Ai, since those are the payoffs available to B so long as A holds fast to Ai. But by hypothesis, A and B have strictly opposed preferences [this is where that crucial assumption comes in], so the outcome at Aij will be the least preferred of all the payoffs in row Ai from A's point of view. Thus, if Ai and Bj are in equilibrium, it follows that Aij will be the maximum of its column and the minimum of its row.

    Conversely, suppose that Aij is the maximum of its column and the minimum of uits row. Since it is the maximum of its column, A can only do worse by switching to a different strategy so long as B holds fast. And since A and B have strictly opposed preferences, payoff Bij must be the most preferred of its row, for Aij is the least preferred of its row. So B can only lose by switching so long as A holds fast. But this is the definition of an equilibrium pair of strategies. Q. E. D.

    We have wandered pretty far into the weeds here, so you should take a moment to go over this argument and make sure you understand it. It is a typical Game Theory argument, and you need to become comfortable with that way of reasoning. Remember, we have already talked about whether identifying security levels and choosing a strategy to maximize your security level is a rational way of proceeding in game that presents you with choice under uncertainty. It is interesting to note, as we will see much later, that Rawls adopts this notion of rationality in A Theory of Justice, where he dramatizes it by saying, in effect [not a quote], Design an institution as though your worst enemy was going to assign you a place in it. In such a case, pretty clearly, maximizing the payoff to the least favored role in the institution makes a good deal of sense.

    Thus far, we have looked at games in which each player's maximum security level shows up in only one of the available strategies, but obviously there might be several strategies with identical security levels, and that security level could perfectly well be the maximum one. In that case, the rule to choose the strategy with the maximum security level does not tell player which strategy to choose. All of the strategies exhibiting the maximum security level are equally good, as far ass the rule is concerned. But if we cannot specify which strategy a player will choose, following the rule, then how can we know what the outcome of the game will be? Fortunately, when players have strictly opposed preferences, it makes no difference. The following is a very important fact:     

    If strategy pairs (Ai, Bj) and (Ap, Br) are both equilibrium pairs of strategies, then so too are the pairs (Ai, Br) and (Ap, Bj). What is more, in that case Aij = Air = Apj = Apr and Bij = Bir = Bpr = Bij. So, no matter how A and b mix and match their strategies with the maximum security levels, the results will be the same. [Note: When I say this, I mean the payoffs to the players will be the same. The actual play of the game may differ according to which strategies A and B choose, but they don't care about that, by hypothesis. They only care about the payoffs. keep that in mind, because down the line, it could be problematic.]    Here is the proof. Let us suppose that we have a payoff matrix that is n rows by m columns, or n x m. I am going to show you a central part of the total matrix that is large enough to include all four payoff pairs: (Aij, Bij), (Air, Bir), (Apj, Bpj), and (Apr, Bpr).














Aij, Bij


Air, Bir






Apj, Bpj


Apr, Bpr







(1) Apr ≥ Air because (Apr, Bpr) is an equilibrium point, and hence Apr is the maximum of its column. [Notice, the symbol ≥ means "equal to or greater than." That is the way my word processing program writes it. ]

(2) Air ≥ Aij because (Aij, Bij) is an equilibrium point, and hence Aij is a minimum of its row.

(3) Aij ≥ Apj same reasoning as (1)

(4) Apj ≥ Apr same reasoning as (2). Hence

(5) Apr ≥ Air ≥ Aij ≥ Apj ≥ Apr Therefore

(6) Apr = Air = Aij = Apj

    The same reasoning establishes that Bij = Bir = Bpr = Bpj and therefore obviously (Air, Bir) and (Apj, Bpj) are also equilibrium pairs.

    Just to review, the key to the proof is the fact that A and B have strictly opposed preference orders. If that is not the case, the argument clearly does not go through.


  1. Why is the assumption of opposed preference orders so prominent? Does it merely simplify the proofs, or does it have a more substantive role in the development of the theory?

  2. Without that assumption, nothing very interesting or powerful can be proved. See the next couple of posts.

  3. Is it ever possible to get to an equilibrium point without strictly opposed preference orders?

  4. "But I would get 9 with B1, 4 with B2, and 1 with B4." I think you mean -9 in the above sentence :)

  5. Whoops. Thank you, Bacon. I will go back into it and correct it.