This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

What should you be willing to pay to play a game in which the payoff is

calculated as follows: a coin is flipped until it comes up heads on the

nth toss and the payoff is set at 2^n dollars?

decision/stpetersburg.s

Classical decision theory says that you should be willing to pay any

amount up to the expected value of the wager. Let's calculate the

expected value: The probability of winning at step n is 2^-n, and the

payoff at step n is 2^n, so the sum of the products of the

probabilities and the payoffs is:

E = sum over n (2^-n * 2^n) = sum over n (1) = infinity

So you should be willing to pay any amount to play this game. This is

called the "St. Petersburg Paradox."

The classical solution to this problem was given by Bernoulli. He

noted that people's desire for money is not linear in the amount of

money involved. In other words, people do not desire $2 million twice

as much as they desire $1 million. Suppose, for example, that people's

desire for money is a logarithmic function of the amount of money.

Then the expected VALUE of the game is:

E = sum over n (2^-n * C * log(2^n)) = sum over n (2^-n * C' * n) = C''

Here the C's are constants that depend upon the risk aversion of the

player, but at least the expected value is finite. However, it turns

out that these constants are usually much higher than people are really

willing to pay to play, and in fact it can be shown that any

non-bounded utility function (map from amount of money to value of

money) is prey to a generalization of the St. Petersburg paradox. So

the classical solution of Bernoulli is only part of the story.

The rest of the story lies in the observation that bankrolls are always

finite, and this dramatically reduces the amount you are willing to bet

in the St. Petersburg game.

To figure out what would be a fair value to charge for playing the game

we must know the bank's resources. Assume that the bank has 1 million

dollars (1*K*K = 2^20). I cannot possibly win more than $1 million

whether I toss 20 tails in a row or 2000.

Therefore my expected amount of winning is

E = sum n up to 20 (2^-n * 2^n) = sum n up to 20 (1) = $20

and my expected value of winning is

E = sum n up to 20 (2^-n * C * log(2^n)) = some small number

This is much more in keeping with what people would really pay to

play the game.

Incidentally, T.C. Fry suggested this change to the problem in 1928

(see W.W.R. Ball, Mathematical Recreations and Essays, N.Y.: Macmillan,

1960, pp. 44-45).

The problem remains interesting when modified in this way,

for the following reason. For a particular value of the bank's

resources, let

e denote the expected value of the player's winnings; and let

p denote the probability that the player profits from the game, assuming

the price of getting into the game is 0.8e (20% discount).

Note that the expected value of the player's profit is 0.2e. Now

let's vary the bank's resources and observe how e and p change. It

will be seen that as e (and hence the expected value of the profit)

increases, p diminishes. The more the game is to the player's

advantage in terms of expected value of profit, the less likely it is

that the player will come away with any profit at all. This

is mildly counterintuitive.

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