Game holder is always losing money in the St. Petersberg Paradox?

Question

The St. Petersberg Paradox is described as follows:

A gambler pays an entry fee $M$ dollar to play the following game: A fair coin is tossed repeated until the first head occurs and you win $2^{n-1}$ amount of money where $n$ is the total number of tosses. And the question is what is the fair amount of $M$?

By some simple probability knowledge we can get the "Expected Winning Money" is $$E(2^{n-1})=\sum\limits_{k=1}^\infty 2^{k-1}\times\frac1{2^k}=\infty$$

But Bernoulli claims that $M$ is not worth infinity because of utility. And log-utility is considered here, which intuitively means that 1000 dollars are not equally significant to a pauper and a rich man.

But I am confused that if so, then $M$ will be set to some finite amount. But if we only focus on the number itself, since the "Expected Winning Money" is infinity, then the game holder is 100% certain to lose money to hold this game(Or to be more precise, the game holder will definitely lose money when the number of people to play this game is large enough)?

Can anyone help me? Thank you!!

Answer 1

The St. Petersburg "paradox" is essentially a statement about peoples' behavior; it asserts that while the expected value of a particular lottery is infinite, no one would actually be willing to pay an infinite amount to play that lottery. Thus, whatever decision criterion people use to make decisions under uncertainty, that criterion cannot simply be using the expected value.

Frankly, using the expected value is neither "right" nor "wrong." Why should you use the expected value? True, the expected value has a nice interpretation as (approximately) the average payout from a lottery if that lottery is played many times. But in a single play, you need not earn the expected value, and if you are playing the St. Petersburg lottery once, you will earn a finite amount with probability one.

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Expected utility theory formalizes a framework under which people make choices under uncertainty by using a suitable concave Bernoulli utility function as an attempt to resolve such "paradoxes." An expected utility maximizer with Bernoulli utility $u(\cdot)$ would be willing to pay up to their certainty equivalent $C$ for a lottery, which satisfies

$$E[u(X)]=u(C),$$

where $X$ is a random variable signifying the lottery payout. Note how that contrasts with an expected value maximizer, whose certainty equivalent would simply be the expected value of the lottery: $$E[X]=C.$$

For the St. Petersburg lottery you mention, a Bernoulli utility $u(\cdot)$ would imply

$$\sum_{k=1}^\infty \frac{u(2^{k-1})}{2^k}=u(C),$$

and taking the example of log Bernoulli utility $u(x)=\log x$, this gives

$$\sum_{k=1}^\infty \frac{\log (2^{k-1})}{2^k}=\log (C)\\ \implies \log 2\underbrace{\sum_{k=1}^\infty \frac{k-1}{2^k}}_{=1}=\log (C)\\ \implies C=2.$$

This sounds more reasonable a prediction than paying $\infty$, right?

It is also worth noting that there are other models that predict how people make decisions under uncertainty. The Nobel-winning prospect theory is itself a critique of expected utility theory that you may find of interest.