I worked out the case of a two-element chain and the uniform distribution on [0,1]. This was more complicated than I expected, and although the same sort of approach should allow one to compute chains of length 3, 4, 5, etc., it looks messy. However, maybe someone else will see a clean way to argue the general case.

Let $r_1, r_2, \ldots$ denote the sequence of random numbers generated.
It is pretty clear that the optimal algorithm for labeling a two-element chain is the following:

If $r_1>1/2$ then label the larger element $r_1$; otherwise label the smaller element $r_1$. On each subsequent turn, if labeling the unlabeled element with $r_i$ wins at once, then do so; otherwise, relabel the already-labeled element with $r_i$.

Let $N$ be the number of turns; then $N$ is a random variable, and we want to find the expected value of $N$. So we need to compute $\Pr(N=k)$ for each integer $k$. There are two ways that $N$ can equal $k$: Either
$$1/2 \le r_1 \le r_2 \le \cdots \le r_{k-1} > r_k $$
or
$$1/2 \ge r_1 \ge r_2 \ge \cdots \ge r_{k-1} < r_k.$$
By symmetry we may focus on the latter case and multiply by 2. The calculation splits into two cases depending on whether $r_k > 1/2$ or $r_k \le 1/2$. The case $r_k > 1/2$ has probability
$${(1/2)^{k-1}\over (k-1)!} \cdot {1\over 2}$$
where $(1/2)^{k-1}$ is the probability that $r_1, \ldots, r_{k-1}$ are all less than $1/2$ and $1/(k-1)!$ is the probability that they are in decreasing order and the final $1/2$ is the probability that $r_k >1/2$. The case $r_k \le 1/2$ has probability
$$ {1 \over 2^k} \cdot {k-1\over k!}$$
where $1/2^k$ is the probability that $r_1, \ldots, r_k$ are all less than $1/2$ and $(k-1)/k!$ is the probability that they are in decreasing order *except* that $r_{k-1} > r_k$. So the expected value of $N$ is
$$E[N] = 2 \sum_{k\ge 2} k \cdot \left( {(1/2)^{k-1}\over (k-1)!} \cdot {1\over 2} + {1 \over 2^k} \cdot {k-1\over k!}\right) = 2\exp(1/2) - 1 \approx 2.29744.$$

To analyze the three-element chain, it seems that one will need to consider partially labeled posets. The above analysis easily generalizes to show that the expected number of turns to finish labeling a two-element chain whose smaller element is already labeled with some number $x<1/2$ is
$$\sum_{k\ge 1} k \cdot \left( {x^{k-1}\over (k-1)!} \cdot {(1-x)} + x^k \cdot {k-1\over k!}\right) = \exp(x). $$
However, as I said, extending the analysis to larger chains looks messy.