# Probability St Petersburg Game

I am reading a book on probability and there is an interesting chapter on the St Petersburg game - where a coin is flipped until a head is landed and the prize is £2 if there is a head on the first throw, £4 is there is a head on the second flip, £8 for a head on the third etc. The idea of this game is it introduces a paradox since someone should be prepared to pay any amount to play since the expected yield is infinite.

However in the book it introduces the notion of changing the rewards to 2,4,6,8.... etc rather than the doubling in the original game. The book contends that even though the possible payouts increase without bound the calculation of the expected value yields a sensible answer of £4. It mentions that this is simple to calculate so omits the calculations. I have pondering this and cant see how the £4 has been arrived at - a entry price of £4 would be a fair price for the game but why? Any help greatly appreciated. I am sure I'm missing something to do with summing to infinity perhaps

If you are familiar with the series $$\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots,$$ you may recall that it adds up to $$1$$: $$1 = \frac12 + \frac14 + \frac18 + \frac1{16} + \cdots,$$ which is just a number-heavy way to say that the probability is $$1$$ that the coin eventually will come up heads. (Probability $$\frac12$$ on the first throw, $$\frac14$$ that it takes exactly two throws, $$\frac18$$ that it takes exactly three throws, etc.)

If we multiply all the terms by $$2$$ we get twice as much for the sum: \begin{align} 2 &= 1 + \frac12 + \frac14 + \frac18 + \cdots, \\ 4 &= 2 + 1 + \frac12 + \frac14 + \cdots. \end{align}

If we divide all the terms by $$2$$ we get half as much: \begin{align} \frac12 & = \frac14 + \frac18 + \frac1{16} + \frac1{32} + \cdots,\\ \frac14 & = \frac18 + \frac1{16} + \frac1{32} + \frac1{64} + \cdots,\\ \frac18 & = \frac1{16} + \frac1{32} + \frac1{64} + \frac1{128} + \cdots,\\ \end{align} and so forth.

Rearranging terms in a series can alter the sum if some terms are positive and some are negative; but in the series you are looking at, all terms are positive, so we can write: \begin{align} 2\cdot\frac12 + 4\cdot\frac14 &+ 6\cdot\frac18 + 8\cdot\frac1{16} + \cdots\\ &= 1 + 2\cdot\frac12 + 3\cdot\frac14 + 4\cdot\frac18 + \cdots\\ &= 1 + (1+1)\cdot\frac12 + (1+1+1)\cdot\frac14 + (1+1+1+1)\cdot\frac18 + \cdots\\ &= 1 + \frac12 + \frac14 + \frac18 + \cdots\\ &\phantom{{}= 1} + \frac12 + \frac14 + \frac18 + \cdots\\ &\phantom{{}= 1 + \frac12} + \frac14 + \frac18 + \cdots\\ &\phantom{{}= 1 + \frac12 + \frac14} + \frac18 + \cdots\\ &\phantom{{}= 1 + \frac12 + \frac14 + \frac18} + \cdots\\ &= 2 \\ &\phantom{{}= 2} + 1\\ &\phantom{{}= 2 + 1} + \frac12\\ &\phantom{{}= 2 + 1 + \frac12} + \frac14\\ &\phantom{{}= 2 + 1 + \frac12 + \frac14} + \cdots\\ &= 4. \end{align}

• That's a great explanation thank you very much! – Jon Percival Mar 3 at 17:35

Let $$x$$ be the expected payout of the game.

You have a probability of $$1/2$$ of the game ending with a head on the first throw, in which case you receive $$2$$. If it lands tails (also probability $$1/2$$), it is now as if you start a new game where the payouts are $$4,6,8,...$$. This is identical to the original game, but all the payouts are exactly $$2$$ higher than before. So the expected payout of this new game is $$x+2$$.

The formula for the expected winnings is then: $$x = \frac12\cdot 2 + \frac12(x+2)$$ which has the solution $$x=4$$.

The probability that the head shows up first in $$n$$-th throw is: $$p(n)=\frac1{2^n},$$ and the corresponding expected value of the win: $$\sum_{n=1}^\infty 2n p(n)=4.$$

• +1. The OP will perhaps someday learn about infinite series. This is a simple example of one. We do not know what book it is, but there are, indeed, popular books about probability that do not assume math as advanced as infinite series. – GEdgar Mar 3 at 13:21
• Many thanks for your answers. I am intrigued by the infinite series. I know that the sum of the series (1/2^n) is 1; so how does the sum of series 2n give 4? – Jon Percival Mar 3 at 13:56
• @JonPercival This can be proven in many ways. If you are interested ask the question or just search over the site. – user Mar 3 at 15:29

Let $$s$$ be the expected payout. Then $$s=2\cdot\frac12 + 4\cdot\frac14 + 6\cdot\frac18 + 8\cdot\frac1{16}+\dots$$ Subtract $$2 = 2\left(\frac12 + \frac14 + \frac18 + \frac1{16}+\dots\right)$$ $$s-2=2\cdot\frac14 + 4\cdot\frac18 + 6\cdot\frac1{16}+\dots$$ Double $$2(s-2)=2\cdot\frac12 + 4\cdot\frac14 + 6\cdot\frac18+\dots=s$$ That is, $$2s-4=s$$ so $$s=4$$