Everything we have said thus far assumes only ordinal preferences, but that is not going to be enough to allow us to analyze games that involve chance elements, or what I have somewhat facetiously been calling moves by Lady Luck [think Marlon Brando singing "Luck be a lady tonight" in the movie version of *Guys and Dolls*. ] Suppose that at some point in a game the rules call for a roll of the dice, a flip of a coin, or a spin of a wheel, with some player's options determined by the outcome of the chance event. That is going to create problems for our analysis.

Here is the simplest game I could think of to illustrate this idea. A moves first, and she has a choice. She can choose not to toss a coin, in which case B has to choose between a move that has the payoff (2.4, -2.4) to A and B, and a move that has the payoff (2, -2) to A and B. Pretty obviously, B will choose the latter. OR A can opt to flip a coin. If the coin comes up heads, the game ends with the payoff (1, -1). if the coin comes up tails, the game ends with the payoff (6, -6). Not much of a game, but it will do.

What should A do? If she opts not to toss the coin, she has a sure thing payoff [given that B is rational] of 2. if she opts to flip the coin, she has a one-half chance of a payoff of 1 and a one-half chance of a payoff of 6. Now, if we forget that the numbers in the example are ordinal labels, we might be tempted to suppose that A can solve her problem by engaging in an expected utility calculation. After all, (1/2 x 1) + (1/2 x 6) = .5 + 3 = 3.5 so A should apparently choose to flip the coin. But these are ordinals, not cardinals, and all we really know is that for A, the payoffs are ranked 6 first, 2.4 second, 2 third, and 1 fourth. This ranking is preserved if we re-label the 6 as a 2.5 That still makes it first, which is all the information we have. But now, when we carry out an expected utility calculation on the game, we have 2 versus (1/2 x 1) + (1/2 x 2.5) = 1.75. With these numbers, A should change her strategy choice.

Obviously, we cannot analyze games with chance elements unless we assume that the players have cardinal utility functions with utility assignments that are invariant under an affine [linear] transformation. Therefore, we need now to introduce the formal machinery required to allow us to talk about cardinal utility functions. This is going to get seriously gnarly, I am afraid. The faint of heart may wish to take a vacation for a day or two while I lay all of this out. I choose to go into this for two reasons: First, as my son Tobias, who is following this blog, pointed out to me at dinner several evenings ago, I am really a rather nerdy wonk when it comes to this stuff. I just plain like it. I hadn't realized that, but of course he is right. I think it is nifty. Second, one of the central ideological messages of this blog is that too many intellectuals and academics adopt the jargon and the style of argument of Game Theory without any real grasp of the assumptions that are embedded in what they are saying. They are like party crashers at a Mass who think that the Eucharist is just a light snack, oblivious to its theological meaning. For those who can handle it, I want to take you through the formal unfolding of the concept of a cardinal utility function. For those who cannot handle it, I want to shock and awe you so that in the future, when you idly assume that someone has a cardinal utility function, you will at least know what lies beneath the surface of that assumption. Here goes.

We must begin with the notion of a probability distribution over a set of outcomes. Remember that all of this theory assumes that in a game there are a finite number of possible outcomes [maybe just win or lose, but maybe also lots of different money payouts, or even things like a trip to the zoo, a fur coat, a dinner date with Kevin Bacon, etc.] The convention in probability theory is that probabilities range from 1 to 0 inclusive. If a possible outcome has a probability of 1, that means it is certain to happen. If it has a probability of 0, that means it is certain not to happen. Thus, probabilities are expressed as real numbers between 1 and 0. Most of the time, they are expressed as a decimal point followed by some real number, like .4 or .125, and so forth.

We also assume that the possible outcomes are independent of one another, not part of or nested inside one another. So we cannot have two outcomes one of which is "I lose the game" and another of which is "I lose the game and have to pay fifty dollars to the person who won." If all of this is so, then the probability that either outcome O will happen or outcome P will happen is equal to the probability that O will happen plus the probability that P will happen. So, if the probability of O is .3 and the probability of P is .6, then the probability of either O or P is (.3 + .6) = .9. A little reflection will tell you that if there are three possible outcomes, O, P, and Q, and if it is certain that one or another of them will happen, then the sum of their probabilities must be 1. In other words, "O or P or Q" is certain to happen.

A probability distribution over the set of possible outcomes is a set of numbers, each of which is between 1 and 0 inclusive [some of the outcomes may be certain not to happen, in which case they have probability zero] and all of which add up to 1. Another way to express this [hang on, I am really reaching with my word processing program here] is this: If the probability of outcome i is p_{i }then for all n possible outcomes from 1 to n, ∑p_{i} = 1. [Cripes, That wasn't worth the effort. Oh well.]

Following Luce and Raiffa, I am going to use the term "Lottery" to refer to an experiment that has built into it a probability distribution over the set of n possible outcomes in a game. For example, if you want to construct a lottery that has built into it a .5 chance of outcome O, a .25 chance of outcome P, and a .25 chance of outcome Q, you can make a wooden wheel with an arrow fixed to its center. You can then draw radii on the wheel dividing it into a half segment and two quarter segments. Then you can spin the arrow in such a way that there was an equal chance of the point of the arrow coming to rest anywhere on the wheel [remember to oil the bearings]. That wheel would be a lottery with the desired probabilities.

O.K. We already know that each player has a consistent ordinal preference over the set of possible outcomes of the game. But we also know that that information all by itself is not enough to authorize us to represent that preference order by cardinal numbers. We can certainly use cardinal numbers in payoff matrices to represent a player's preferences -- that is what I have been doing. But we cannot treat them as cardinal numbers. We can only treat them as ordinals. Obviously we need more information about the player's preference structure if we are to define a cardinal preference order for him or her over the possible outcomes.

Before we state the six axioms that von Neuman and Morgenstern proved are sufficient to allow us to impute a cardinal utility function to a player, we need some more definitions and some more notation. [I warned you this would get gnarly.] First of all, we must extend our notion of a Lottery to something called a Compound Lottery. A Simple Lottery is a probability distribution over as set of outcomes, O,P, Q, etc. A Compound Lottery is a Lottery the prizes in which are tickets in other lotteries over O, P, Q, etc. To make this as clear as I can, let me take a very simple case. Imagine a set of three outcomes (O, P, Q).

O = +$5

P = +$8

Q = -$10 [i.e., the player has to play ten dollars]

and a Lottery, L1, with three prizes, namely tickets in Lotteries L11, L12, and L13. L1 is set up so that there is a:

a .5 chance of winning a ticket in L11,

a .25 chance of winning a ticket in L12, and

a .25 chance of winning a ticket in L13.

The prizes in the Lotteries L11, L12, and L13 are the outcomes (O, P, and Q)

Now, L11, L12, and L13 are probability distributions over O, P, Q, etc. So, let us suppose that these three Lotteries are in fact:

L11: .3 chance of O, .4 chance of P, and .3 chance of Q

L12: 1. chance of O, .0 chance of P, .9 chance of Q

L13: .8 chance of O, .1 chance of P, .1 chance of Q

Notice that in each case the probabilities add up to 1.

If our player, A, buys a ticket in L1, what is her chance of ending up with each of the outcomes, O, P, or Q? Well, she has a .5 chance of winning a ticket in L11, and L11 in turn offers a .3 chance of O, so that gives A so far a (.5)(.3) = .15 chance of O.

She also has a .25 chance of a ticket in L12, and L12 offers a .1 chance of O, so that gives her a (.25)(.1) = .025 chance of O.

Finally, she has a .25 chance of a ticket in L13, which offers a .8 chance of O, so that gives her a (.25)((.8) = .2 chance of O.

Adding .15, .025, and .2, we get a .375 chance of O.

If you carry out the same calculations for outcomes P and Q, you will find that A has a .225 chance of getting P and a .4 chance of getting Q, and sure enough, .375 + .225 + .4 = 1.

Now, what is the money value to A of this gamble? It is the Mathematical Expectation, or:

(.375)(5) + (.225)(8) + (.4)(-10) = 1.875 + 1.8 - 4 = -.325 or minus 32.5 cents.

So, if all A cares about is making money, she ought not to buy a lottery ticket in L at any price. In fact, she should not even play if someone gives her a ticket.

This is what is called reducing a Compound Lottery to a Simple Lottery, and it should be obvious that you can do this with any finite number of prizes and any number of levels of lotteries of lotteries of lotteries. Notice one small point that will be important later: If a Lottery offers a chance of p for one outcome, O, and chances of zero for all the other outcomes save a single one, Q, then the probability for Q will be (1-p).

Bob,

ReplyDeleteGreat Blog! I am really enjoying it.

This is 'picky' but since it could be confussing to some folks, you might want fix where you placed the decimal points for the 2.5 in "For example, if you want to construct a lottery that has built into it a .5 chance of outcome O, a .25 chance of outcome P, and a 2.5 chance of outcome Q, you can make a wooden wheel with an arrow fixed to its center." and for the 1. in "L12: 1. chance of O, .0 chance of P, .9 chance of Q".

I have also been enjoying your memiors. Thanks for taking the time to write both of these Blogs.

Both Blogs have become part of my morning routine.

Gary Childers

"An 'older', 'newer' R. P. Wolff Student"

Two minuscule questions: 1) Do you promise that knowing the definition of a compound lottery will be necessary (and sufficient) to understand the six axioms of von Neuman and Morgenstern? 2) Can you possibly mean Frank Sinatra instead of Marlon Brando? (let me assure you that I know nothing of either question) :-)

ReplyDeleteWhoops. Thank you,Gary. Welcome to the blog. I will go in and fix it. This is so hard on line!

ReplyDeleteAnn, yes, I am afraid you have to understand what a compound lottery is. This is very geeky.

Sinatra was in the movie [Nathan Detroit], but Brando [Skye Masterson] is, I think, the character who sings Luck be a Lady Tonight. Google it and check. I did.