Consider the following game of chance. A fair coin is tossed until the first tails appears. You place an initial bet of k. If the 1st tails appears on the nth toss, you receive a total of $2^n$ (2 to the power of n) in return for your initial bet. How large should k be in order for your expected winnings to be zero (note, expected winnings of zero is sometimes called a “fair” game)?

I did the question and the answer comes to infinite. Is that correct? If not, what did I do wrong?

In other words, we can rephrase the question as: what is the expectation value of $2^n$ given $p(n)=1/2^{(n+1)}$?

  • $\begingroup$ I believe you have mixed up a $k$ and an $n$ somewhere. As is, the value of $k$ won't change anything. $\endgroup$ – mb7744 Feb 10 '15 at 5:05
  • $\begingroup$ No I am pretty sure I didn't. Yeah as is, the value of $k$ needs to be infinite for it to be a fair game. That's exactly what I got. $\endgroup$ – Quantum_Boson Feb 10 '15 at 5:12
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    $\begingroup$ You can look for a discussion under the St. Petersburg Paradox. $\endgroup$ – André Nicolas Feb 10 '15 at 5:18
  • $\begingroup$ Your description of the wagering and award scheme is unclear. Is the cost of each toss $k$? What do you lose if you do not win? $\endgroup$ – heropup Feb 10 '15 at 5:19
  • $\begingroup$ it should not be dependent upon $k$, as $k$ is the expectation value of $2^n$. So we are finding k. In other words, we can rephrase the questions as: what is the expectation value of $2^n$ given $p(n)=1/2^{(n+1)}$? $\endgroup$ – Quantum_Boson Feb 10 '15 at 5:20

The answer is indefinite expected return.

$$\begin{align} \mathsf P(N=n) & = 2^{-n} \\[2ex] \mathsf E(g(N)) & = \sum_{n=1}^\infty g(n)\times \mathsf P(X=n) \\[2ex] \mathsf E(2^N) & = \sum_{n=1}^\infty ((2^n) \times 2^{-n}) \end{align}$$

While you intuitively expect the game to end at some finite result, however at any point before the end of the game the probability of doubling your return on the next toss is always 1/2.

Another way: The game will either end on the first turn, or continue recursively with double the stake. So using the Law of Iterated Expectations:

$$\begin{align} \mathsf E(2^N) & = \mathsf E(2^N \mid N=1)\mathsf P(N=1) + \mathsf E(2^N\mid N\gt 1)\mathsf P(N\gt 1) \\ & = 2^1 \times \tfrac 1 2 + \mathsf E(2^{N+1})\times \tfrac 1 2 \\ & = 1 + \mathsf E(2^N) \end{align}$$ Which is paradoxical for finite numbers, so the expectation must be indefinite.

So there is no finite value of a stake $k$ which would equal the expected return.

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    $\begingroup$ the question is finding the value $k$ of which $E[2^n-k]=0$, not multiply by $k$, but essentially you are correct. Just edit it so that it would be correct. $\endgroup$ – Quantum_Boson Feb 10 '15 at 5:39

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