Consider the following variation on the classic St Petersburg Game:
The Alternating St Petersburg Game: A fair coin is tossed until it lands heads for the first time. If it lands heads on the first toss, you are paid \$2; if it lands heads on the second toss, you must pay \$4; if it lands heads on the third toss, you are paid \$8; and so on.
The payoff table for this game is as follows:
| Outcome | Probability | Payoff |
|---|---|---|
| H | $1/2$ | \$$2$ |
| TH | $1/4$ | \$$-4$ |
| TTH | $1/8$ | \$$8$ |
| TTTH | $1/16$ | \$$-16$ |
| $\vdots$ | $\vdots$ | $\vdots$ |
The expected value of the Alternating St Petersburg Game turns out to be undefined. If we try to calculate it, we get "Grandi's series" which doesn't converge:
\begin{equation} \sum_{i = 1}^\infty \frac{1}{2^i} (-1)^{i-1}2^i = 1 - 1 + 1 -1 + \cdots \end{equation}
Still, we might wonder whether we should play the Alternating St Petersburg Game, if given the chance. Imagine we kept playing the game many times while keeping track of the cumulative payoff (which could be negative). We might then wonder about the chance that the cumulative payoff is positive in the long run.
Let $S_n$ stand for the cumulative payoff after $n$ rounds. We'd like to examine the behavior of the following probability, as $n$ increases.
\begin{equation} P(S_n > \$0) \end{equation}
Does this probability have upper/lower bounds? Does it have a limit?
We do not know the answers to these questions. However, we have calculated the values of $P(S_n > 0)$ for $n = 1, 2, 3$. The probabilities in those cases are $2/3$, $8/15$, and $293/630$ respectively. Calculations of the first two probabilities follow below.
We also performed a simulation to examine the behavior of this probability experimentally. We played the Alternating St Petersburg Game over 100 rounds, with 25,000 iterations. The following data plot shows the frequency with which the game showed a positive cumulative payoff, for each number of rounds ($n = 1,2,3,\ldots,100$).
Graph: https://i.stack.imgur.com/b2PLf.jpg
Preliminary calculations
$n = 1$
To determine $P(S_1 > \$0)$, all we need to do is add up the probabilities of all outcomes that yield a positive payoff. Looking at the payoff table, this yields:
\begin{equation} P(S_1 > \$0) = P(\text{H}) + P(\text{TTH}) + P(\text{TTTTH}) + \cdots = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots = \frac{2}{3} \end{equation}
So, there is a 2/3 probability that a single round of play yields a positive payoff.
$n = 2$
After two rounds of play, things already get a bit more complicated. Some pairs of worlds yield a positive cumulative payoff, whereas others yield a negative cumulative payoff. For example, the pair (TH,TTH) yields a cumulative payoff of \$$-4 + \$8 = \$4$, whereas (H,TH) yields a cumulative payoff of $\$2 - \$4 =$ \$$-2$. Our first task is to determine what pairs of outcomes yield a positive cumulative payoff. To this end, let us divide the pairs of outcomes into four categories ($o_i$ is the $i$th row in the payoff table):
$C_1$: $\{(o_i,o_j): o_i \textrm{ yields a positive payoff}, o_j \textrm{ yields a positive payoff}\}$
$C_2$: $\{(o_i,o_j): o_i \textrm{ yields a positive payoff}, o_j \textrm{ yields a negative payoff}\}$
$C_3$: $\{(o_i,o_j): o_i \textrm{ yields a negative payoff}, o_j \textrm{ yields a positive payoff}\}$
$C_4$: $\{(o_i,o_j): o_i \textrm{ yields a negative payoff}, o_j \textrm{ yields a negative payoff}\}$
We can then consider which pairs in each category yield a positive cumulative payoff. Obviously all pairs in $C_1$ yield a positive cumulative payoff, and all pairs in $C_4$ yield a negative cumulative payoff. The question is what pairs in $C_2$ and $C_3$ yield a positive cumulative payoff. Here is a table that includes the payoffs of both rounds of play:
| Outcome | Probability | 1st round payoff | 2nd round payoff |
|---|---|---|---|
| H | $1/2$ | \$$2$ | \$$2$ |
| TH | $1/4$ | \$$-4$ | \$$-4$ |
| TTH | $1/8$ | \$$8$ | \$$8$ |
| TTTH | $1/16$ | \$$-16$ | \$$-16$ |
| TTTTH | $1/32$ | \$$32$ | \$$32$ |
| $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |
Looking at the table, we notice a pattern: when the first payoff is positive and the second payoff is negative, the cumulative payoff after both rounds of play is positive iff the second payoff is numerically smaller than the first. So the pairs in $C_2$ that yield a positive cumulative payoff, call them $C_2^+$, are:
\begin{equation} C_2^+ = \bigcup_{i=1}^\infty \bigcup_{j=1}^{i-1} (o_{2i-1},o_{2j}) \end{equation}
A similar pattern holds for $C_3$: when the first payoff is negative and the second payoff is positive, the cumulative payoff after both rounds of play is positive iff the first payoff is numerically smaller than the second:
\begin{equation} C_3^+ = \bigcup_{i=1}^\infty \bigcup_{j=1}^{i-1} (o_{2i-1},o_{2j}) \end{equation}
Now we can calculate $P(S_2 > 0)$ by adding the probabilities of each pair in $C_1 \cup C_2^+ \cup C_3^+$:
\begin{equation} \begin{split} P(S_2 > 0) = \sum_{i=1}^\infty \sum_{j=1}^\infty P(o_{2i-1})P(o_{2j-1}) + \sum_{i=1}^\infty \sum_{j=1}^{i-1} P(o_{2i-1})P(o_{2j}) \\ + \sum_{i=1}^\infty \sum_{j=i+1}^\infty P(o_{2i})P(o_{2j-1}) \end{split} \end{equation}
If we plug in $P(o_i) = 1/2^i$ and do the calculations, we get:
\begin{equation} P(S_2 > 0) = \frac{4}{9} + \frac{2}{45} + \frac{2}{45} = \frac{8}{15} \end{equation}
So, there is a 8/15 probability of getting a positive cumulative outcome after two rounds of play.