In game theory, there is something called the Folk Theorem, which basically says that you can create a special strategy for any average of possible payoffs as long as the average payoffs are better than the worst case equilibrium payoff for each player.
In more math terms, let us consider a payoff x and a payoff y. Let U be the mean of an infinite series of terms that can only either be x or y.
1) Is it possible for x and y to be 3 and 4 and somehow have U be pi? This is impossible for a finite series, but as Taylor series show us, the sum of an inifinite number of rational numbers can be irrational.
2) How can you construct the order of this series (i.e. which terms are x versus which terms are y)?
3) Can you generalize this question to determine for what types of values of x and y you can achieve what types of payoffs?
4) In game theory you generally deal with multiple players, so x, v, and U would be vectors, with the mean operator applying element wise. How would this change the results?
In game theory for a symmetric game we can view there being k^n possible strategies for n players with k choices each, each with possibly unique payoffs, which means instead of x and y we would be working with more variables. This Folk Theorem applies when the choices are finite, so these results should hold for more than just the case of either x or y above
The average of infinitely many numbers is not a well defined concept.
That said, it is possible to find an infinite sequence $$ a_1, a_2, \ldots $$ where each $a_i$ is either $3$ or $4$ such that the limit of the average of the partial sums $$ \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n a_i = \pi . $$
To prove that, start with $a_1 = 4$. Then append enough consecutive $3$'s to make the average of the entries so far smaller than $\pi$. That's clearly possible since appending $3$'s can get the average as close to $3$ as you wish.
Now append $4$'s until the average exceeds $\pi$.
Continue this way.
@automaticallyGenerated links in a comment to a sequence at OEIS that does the job. That one starts with $3$, not $4$.