I could sample a set of m elements from the uniform distribution over a universe $U$ of n >> m elements. Alternately, I could select a random probability distribution $\mathcal{D}$, and sample $m$ elements from $\mathcal{D}$. 


EDIT (per Michael Lugo): When I say "select a random probability distribution", I mean select a point uniformly at random from the standard $n$-simplex: {$\{(x\_1,\ldots,x\_n) : x\_i \geq 0, x\_1+\ldots+x\_n = 1\}$}. 


Do these two methods lead to the same distribution over my sample? If not, how do they differ? If some event (my sample lies in some set of samples of size m) occurs with probability p using the second method, what can I say about its probability using the first method?



Sample from uniform distribution vs. Sample from random distribution
You're right that things change for m>1; I was thinking sloppily.


Assume $U=\{1,\ldots,n\}$ for concreteness. If $Y\_1,\ldots,Y\_m$ are chosen independently and uniformly from $U$, then for any $k\_1,\ldots,k\_m\in U$, we of course have
$$
\Pr[Y\_1=k\_1,\ldots,Y\_m=k\_m] = \frac{1}{n^m}.
$$


On the other hand, if $x=(x\_1,\ldots,x\_m)$ is chosen uniformly from the standard $n$-simplex and $Y\_1,\ldots,Y\_m$ are then chosen independently according to $x$, then
$$
\Pr[Y\_1=k\_1,\ldots,Y\_m=k\_m] = \mathbb{E}\Pr[Y\_1=k\_1,\ldots,Y\_m=k\_m|x]
= \mathbb{E}\prod\_{i=1}^m x\_{k\_i} = \frac{n!}{(n+r)!}\prod\_{j=1}^n r\_j!,
$$
where $r\_j = \#\{1\le i \le m : k\_i=j\}$ and $r=r\_1 + \cdots r\_n$. This last expectation can be proved most easily from Lemma 1 in [this paper](http://www.jstor.org/stable/2048262).


