# The St. Petersburg Paradox and the definition of the Exptected Value

As far as I've learned, the expected value is only defined when the series of probabilities is absolutely convergent.

However, the St. Petersburg Paradox presents an expected value which is $\infty$

Thus, renedering the expected value as undefined, which means it does not make sense to work with it.

My question is, why is the conclusion arising from the paradox, is that it is recommended to participate in the game. (Why does it make sense to talk about the expected value if it is undefined?)

Here are the rules, by the way:

• You throw a coin until it lands tails.
• You then get paid $2^{n}$ dollars, where $n$ is the amount of heads you got.

Am I missing something?

Thanks in advance!

• What is your question? – Ittay Weiss Feb 13 '17 at 22:51
• Clarified my question mate :) – Chen Mor Feb 14 '17 at 6:43
• Your objection (undefined expectation) goes away if the house rules cut you off after 2 billion tosses whether the coin has landed tails or not. And the paradox arises if the entry fee is less than a billion dollars. So you would agree with the recommendation to play in this modified version of the game, but not if the restriction is removed? – bof Feb 14 '17 at 7:58

## 1 Answer

With your rules, I would always play -- assuming no entry fee --since since the worst I can do is collect nothing with a 50% chance and collect something with a 50% chance. If I have to pay $4$ dollars to play then there is a only a probability of $1/8$ that I do not lose money or break even.

Would you play if the entry fee were $16$ dollars?

The infinite expectation is an artifact of the payoff growing as $2^n$ with associated probability decaying as $1/2^{n+1}.$ Even moderately large payoffs are extremely unlikely events. Furthermore, the expected gain of $\infty$ is meaningless if you are risk averse since, as the entry fee becomes higher, the probability of not winning money tends to $1$. There would be a limit to how much you would be willing to pay to play in spite of an infinite expectation.

What you would pay to play would then depend on your utility function which would consider the entire return distribution and probability of loss. With log utility as suggested by Bernoulli, the expected utility is finite:

$$\sum_{k=1}^\infty \frac{\log 2^k}{2^{k+1}} = \log 2 \approx 0.7$$

With such a risk averse utility function you would pay no more than a $1$ dollar entry fee.

Note also that the expected duration of the game is finite:

$$\sum_{k=1}^\infty \frac{k}{2^k} = 2.$$

On average the game lasts for $2$ throws where you win $2$ dollars. Now consider that if I ask you to pay $16$ dollars to play.