The St. Petersburg Paradox
Kapitoly: The Three Door Problem, The Probabilistic Liar's Paradox, The question mark paradox, Simpson's Paradox, The medical paradox, The St. Petersburg Paradox, Non-transitive cubes
The St. Petersburg Paradox mixes statistics, decision making, and probability together. In St. Petersburg, we have a casino that offers us a game in which we can win a certain amount of money. Our task is to figure out what would be a fair entry fee for this game.
The rules of the game
If we enter the game, the operator starts to flip a coin. If the first toss comes up heads, the game ends and we win one euro. If tails is rolled, the game continues. In the second round, the coin is flipped again. If heads comes up, the game ends and we have won twice the previous possible prize, i.e. two euros. If tails comes up, we continue. If heads comes up on the third round, we win four euros. If it lands on the fourth round, we win eight euros.
Generalizing - if the head falls in the k-th round, then we have won 2k − 1 euros. The question now is, what would be a fair entry fee for this game?
The expected value
The fair price should be based on the expected (mean) value. For example, if on average we can win one hundred euros, the entry price should be one hundred euros, or a little more, to make the casino a profit. But what is the expected value in our game? Let's break down the details with which we can get each win. In the first column are the prizes, in the second column is our chance of getting that prize. For example, the chance of getting heads on the first roll is $\frac12$. The chance of getting tails first and then heads is $\frac14$. etc.
$$\begin{array}{cc} 1&1/2\\ 2&1/4\\ 4&1/8\\ 8&1/16\\ 16&1/32\\ …&… \end{array}$$
Now how do we calculate the mean? We multiply the amount we can get by the probability of getting that amount and add it all up. We get:
$$\begin{eqnarray} E&=&\frac12\cdot1+\frac14\cdot2+\frac18\cdot4+\frac{1}{16}\cdot8+\ldots\\ &=&\frac12+\frac12+\frac12+\frac12+\ldots\\ &=&\sum_{k=1}^{\infty}\frac12=\infty \end{eqnarray}$$
The expected value is infinity. The mean value of the prize is thus, in the idealised case, infinitely many euros. That doesn't look bad. The problem is that at the same time the entry should be equal to infinity. This is, of course, nonsense.
Critics of this paradox argue, of course, that you can't play for infinitely long, you can't win infinitely much money, and even if you "reduce" the number of maximum coin flips from infinity to some finite number n, then you'll still get to amounts that nobody in the world has pretty soon. For example, after 41 flips you would have already won 240 euros, which is about a trillion euros (a thousand billion euros). After another ten rolls you would have won a thousand times more money.
The final entry fee
Of course, no one will give you an infinite number of euros in entry fees. However, the paradox can be partially demonstrated with a finite amount of money. For each whole euro amount, there is a maximum number of coin flips for which the average value comes out to be what we need. For example, if we want to have an input of 1,000 euros, we say that the maximum number of coin tosses is 2,000. Then we calculate such an amount:
$$\sum_{k=1}^{2000}\frac12=2000\cdot\frac12=1000$$
The mean value will then be 1000 euros. If we want to have an entry fee of D euros, then we say that the maximum number of coin flips is 2D.
But of course no sensible person will pay an entry fee of, say, a thousand euros if they have absolutely minimal chance of winning more than a thousand euros.