Suppose I have a family of random variables
$$X_i \sim SomeDistribution_i, \ \ i = 1,..., n$$
and I know how to sample $SomeDistribution_i$ for any $i$.
Suppose I also define a random variable $Y = f(X_1,...,X_n)$ for some deterministic function $f$.
Then I can approximate the probability of some event $\phi(Y)$ by obtaining samples of $Y$, which in turn is computed from samples of $X_1, ..., X_n$, and checking how often a sample of $Y$ satisfies the conditions for the event $\phi(Y)$.
So far, so good. Now suppose I tell you $SomeDistribution_i$ is the same distribution for any $i$. In other words, $X_i$ are independently and identically distributed. It seems to me this is valuable information that ought to be useful in getting better estimates of $P(\phi(Y))$, because how I have less uncertainty involved (from $n$ unknown distributions to a single one).
However, I don't quite know how to make use of that information. It seems I would still need to generate samples of $Y$ in the same manner as before, without ever making use of the fact that $\{SomeDistribution_i\}_i$ are identical.
In other words, the quality of my approximation of $P(\phi(Y))$ would be exactly the same as the one I obtained when I did not know about the identical distributions.
Is there a way to take advantage of that information? Note that I am not making any assumptions about $f$ or $X_i$, not even whether they are discrete or continuous. Is there a way to use this information even without any assumptions?
One intuition I have is that I could, for each sample of $X_1, ..., X_n$, generate $n!$ samples of $Y$ by permuting $X_1, ..., X_n$ in all possible ways. This would generate many more samples of $Y$ for the same amount of samples from $SomeDistribution_i$. Does that hold water? Is it the best approach? Is there a name for this method?
Edit here's an illustration of the permutation method described. Take for example such an $f$ in $f: \{0,1\}^2 \rightarrow \{0, 1, 2, 3\}$ such that $f(X_1,X_2) = 2X_1 + X_2$ (note that $f$ does not treat both inputs in the same manner). If $X_i \sim Uniform(\{0,1\})$, then $Y=f(X_1,X_2)$ is uniformly distributed over $\{0, 1, 2, 3\}$. If we take a single sample for $X_1,X_2$, say, $(1, 0)$, then the estimated distribution for $Y$ is $P(Y=y) = 0$ if $y=2$ and $0$ otherwise. However, because $X_i$ are identically distributed, the sample $(1,0)$ is just as likely as $(0,1)$ and in the limit they would occur an equal number of times. Therefore it makes sense, given sample $(1,0)$, to deterministically introduce sample $(0,1)$ into the sample set. Such sample set would be more representative of what the sample set will look like in the limit, and provide a better approximation for the distribution of $Y$: $P(Y=y)=0.5$ if $y \in \{1, 2\}$ and $0$ otherwise.
