St Petersburg paradox

The St Petersburg paradox mixes together statistics, decision-making and probability. In St Petersburg there is a casino that offers us a game in which we can win a certain amount of money. Our task is to find out what a fair price to enter this game would be.

The rules of the game

If we enter the game, the dealer starts tossing a coin. If heads comes up on the first toss, the game ends and we have won £1. If tails comes up, the game continues. In the second round the coin is tossed again. If heads comes up, the game ends and we have won double the previous possible prize, that is, £2. If tails comes up, we carry on. If heads comes up in the third round, we win £4. If it comes up in the fourth round, we win £8.

Generalising – if heads came up in the kth round, we won £2^{k − 1}. The question now is: what would be a fair price to enter this game?

Expected value

The fair price should be based on the expected value (the mean). For example, if we can win £100 on average, the price of entry should be £100, or slightly more so that the casino makes a profit. But what is the expected value in our game? Let’s write out the probabilities with which we can win each prize. The first column shows the prizes and the second shows our chance of winning that prize. For example, the chance that heads comes up on the first toss is $\frac{1}{2}$. The chance that tails comes up first and then heads is $\frac{1}{4}$. And so on.

$$\begin{array}{cc}1& 1/2\\ 2& 1/4\\ 4& 1/8\\ 8& 1/16\\ 16& 1/32\\ \dots & \dots \end{array}$$

How do we work out the expected value now? We multiply each amount we can win by the probability of winning that amount, and add everything up. We get:

$$\begin{array}{rcl}E& =& \frac{1}{2}\cdot 1+\frac{1}{4}\cdot 2+\frac{1}{8}\cdot 4+\frac{1}{16}\cdot 8+\dots \\ & =& \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots \\ & =& \sum _{k=1}^{\mathrm{\infty }}\frac{1}{2}=\mathrm{\infty }\end{array}$$

The expected value is infinity. So in this idealised case the expected prize is an infinite amount of pounds. That does not sound bad. The problem is that the price of entry should then also be infinite. That is obviously nonsense.

Critics of this paradox naturally object that you cannot play for an infinite amount of time, that we cannot win an infinite amount of money, and that even if you “reduce” the maximum number of coin tosses from infinity to some finite number n, you still fairly quickly reach amounts of money that nobody in the world has. For example, after 41 tosses you would have won £2⁴⁰, which is approximately a trillion pounds (a thousand billion pounds). After another ten tosses you would have won a thousand times more.

A finite price of entry

Of course, nobody will actually pay an infinite amount of money to enter. Nevertheless, the paradox can be partly demonstrated with a finite amount of money. For every whole number of pounds there is a maximum number of coin tosses for which the expected value comes out as the amount we need. For example, if we want the price of entry to be £1,000, we say that the maximum number of tosses is 2,000. Then we calculate this sum:

$$\sum _{k=1}^{2000}\frac{1}{2}=2000\cdot \frac{1}{2}=1000$$

The expected value is then £1000. If we want the price of entry to be £D, we say that the maximum number of tosses is 2D.

Of course, no sensible person will pay an entry price of, say, £1,000 if they have a very slim chance of winning more than £1,000.

Further sources

• English Wikipedia