The Arroyo Game versus The Pasadena Game This choice problem builds on St. Petersburg’s Paradox. Part 7 of this SEP article indicates the problem.
Question
For all integers $n\geq1$, the Arroyo game pays $X$ where 
$$\text{P}\left(X=(-1)^{n+1}(n+1)\right)=\frac{1}{n^2+n}$$
and the Pasadena game pays $Y$ where
$$\text{P}\left(Y=\frac{(-1)^{n+1} 2^n}{n}\right)=2^{-n}$$
The tables below illustrate the probabilities and payouts of both games for $1\leq n\leq 5$
$$\begin{array}{ccccccccc}
 x & 2 & -3 & 4 & -5 & 6 & \cdots  & (-1)^{n+1} (n+1) & \cdots  \\
 \text{P} (X=x) & \frac{1}{2} & \frac{1}{6} & \frac{1}{12} & \
\frac{1}{20} & \frac{1}{30} & \cdots  & \frac{1}{n^2+n} & \cdots  \\
 \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \
& \text{} & \text{} \\
 y & 2 & -2 & \frac{8}{3} & -4 & \frac{32}{5} & \cdots  & \
\frac{(-1)^{n+1} 2^n}{n} & \cdots  \\
 \text{P} (Y=y) & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \
\frac{1}{16} & \frac{1}{32} & \cdots  & 2^{-n} & \cdots  \\
\end{array}$$
For both games, negative payouts indicate losses. If our only end is increasing our wealth, which game is the best means to our end and why?
My Work
Expected Values
The standard solution is choosing the game with the higher expected value. However, the expected value of these games is the same conditionally convergent sum:
$$\text{E}[X]=\sum_n^{\infty}\frac{(-1)^{n+1}(n+1)}{n^2+n}=\sum_n^{\infty}\frac{(-1)^{n+1}}{n}$$
$$\text{E}[Y]=\sum_n^{\infty}\frac{2^{-n}(-1)^{n+1}2^n}{n}=\sum_n^{\infty}\frac{(-1)^{n+1}}{n}$$
This result is problematic for two reasons. Firstly, the expected value is independent of the order in which we add the terms. However, different orderings of terms here give different expected values (due to the Riemann series theorem). Secondly, even if we arbitrarily choose some order, the mean is equal for both games and hence is useless for choosing between them. Overall, the expected values are not helpful.
Relative Probabilities
Another approach is finding $\text{P}(X>Y)$ and $\text{P}(X<Y)$ and choosing the game that is more likely to pay more. A brute force calculation shows that $\text{P}(X>Y)\approx0.370308$, $\text{P}(X<Y)\approx0.379692$, and $\text{P} (X=Y)=\frac{1}{4}$ (the rewards are equal only for $n=1$). This approach suggests that the Pasadena game is slightly more valuable. However, the relative probabilities method is also flawed. Consider games G1 and G2. G1 pays $5$ with probability $0.6$, and $0$ with probability $0.4$. G2 pays $4$ with probability $1$. G1’s expected value is $3$, and G2’s is $4$. However, the probability method suggests playing the former. It is absurd to play G1 since there is a $0.4$ chance of earning $0$, and G2 only pays one unit less. Similar conditions may hold for Arroyo and Pasadena.

If we show that Arroyo vs. Pasadena does not suffer from the G1 vs. G2 problem, then Relative Probabilities may serve as a solution.
Further Particulars


*

*Approaches other than those discussed here are welcome.

*Proofs showing that both Arroyo and Pasadena are equally favorable are welcome.

*Avoid answers that change the question, for example, using diminishing utility or probability weighting.

 A: The optimality criterion depends on what you want to do.  
1) Loan shark: 
Suppose a loan shark asks you to pay 10000 dollars in 5 minutes or you will be killed.   In this 5 minutes you can either bet on $X$ or $Y$.  Then, you only care about the bet earning at least 10000 dollars so you simply compare: 
$$P[X\geq 10000] \mbox{ versus } P[Y \geq 10000]$$
which means you are maximizing the probability of living. 
2) A critique of the model:  
The above scenario means that you will be killed if the bet earns less than 10000, in which case there is no way you can actually pay the "negative profit" situation if it just-so-happens that your $X$ or $Y$ bet earns a profit that is negative.  In this case, the random variables $X$ or $Y$ do not really model the true payoffs, as there is no way you can actually pay the negative profit case. 
Similarly, even without the threat of being killed, this scenario is not realistic because there is no true bet that will ever have either an arbitrarily large payoff and/or an arbitrarily large loss. The PMF of $X$ suggests that you may have a negative payoff less than negative 10 trillion, but, everyone knows you will actually not pay that amount off.  So the true profit is not the one that is given in the PMF for $X$.  You may as well truncate the profit to some interval $[-a, b]$, where $-a$ is the true amount you are able to pay back, and $b$ is the true amount the person running the bet is willing to pay out.
3) Theory of repeatable experiments:  
Arguably, the reason expectations are important is that, by the law of large numbers, they are close to the averaged value you will earn if you independently repeat the experiment many times.  The law of large numbers does not apply in this situation where the expectations of $X$ and $Y$ are not well defined. 
So, if you want to understand what happens when you independently and repeatedly play this game, you would need to use a different theory.  For example, even though the expectations are not defined, it is meaningful to track the probability mass functions (PMF) of profit given you independently play $n$ games in a row: 
$$ P[X_1 + X_2 + ... + X_n = x] , P[Y_1 + Y_2 + ... + Y_n = y]$$ 
and you could, for example, find the number of trials $n$ (possibly 0) to maximize the probability of earning more than 10000 subject to the constraint that the probability of loosing more than $500$ is less than $0.01$. 
