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 $$\Pr[Y_1=k_1,\ldots,Y_m=k_m] = \mathbb{E}\Pr[Y_1=k_1,\ldots,Y_m=k_m|x] = \mathbb{E}\prod_{i=1}^m x_{k_i} = \frac{n!}{(n+r)!}\prod_{j=1}^m frac{n!}{(n+r)!}\prod_{j=1}^n r_j!,$$ where $r_j = \#\{i\in U #\{1\le i \le m : k_i=j\}$ and $r=r_1 + \cdots r_m$r_n$. This last expectation can be proved most easily from Lemma 1 in this paper. Post Undeleted by Mark Meckes 2 Rewrote, now hopefully a correct answer. I'll assume that the answer to the question in my comment above is "yes", so You're right that we can assume m=1things change for m>1; I was thinking sloppily. Assume also for simplicity that $U=\{1,\ldots,n\}$ for concreteness. As in Michael's comment, let's denote by$x=(x_1,\ldots, x_n)$a uniform random point in the standard If$n$-simplex, Y_1,\ldots,Y_m$ are chosen independently and denote by $Y$ the number uniformly from $U$ chosen according to U$, then for any$x$. Then k_1,\ldots,k_m\in U$, we of course have the conditional probability $\Pr[Y=i|x$ \Pr[Y_1=k_1,\ldots,Y_m=k_m] = x_i]$\frac{1}{n^m}. $$On the other hand, if x=(x_1,\ldots,x_m) is chosen uniformly from the standard n-simplex and so Y_1,\ldots,Y_m are then chosen independently according to x, then$$ \Pr[Y=i] = Pr[Y_1=k_1,\ldots,Y_m=k_m] = \mathbb{E} \Pr[Y=i|x] mathbb{E}\Pr[Y_1=k_1,\ldots,Y_m=k_m|x] = \mathbb{E} x_i mathbb{E}\prod_{i=1}^m x_{k_i} = \frac{1}{n}\sum_{j=1}^n \mathbb{E} x_j =\frac{1}{n}. frac{n!}{(n+r)!}\prod_{j=1}^m r_j!,$$So where $Y$does indeed have the uniform distribution on r_j = \#\{i\in U : k_i=j\}$ and $U$.r=r_1 + \cdots r_m\$. This last expectation can be proved most easily from Lemma 1 in this paper.