First, let's describe the one-way model of communication, for computing a function $f$ on Alice's input $x$ and Bob's input $y$:
- Alice sends Bob a string $s$ depending on $x$.
- Bob broadcasts the value of $f(x,y)$.
Alice could be a probabilistic algorithm, and in that case, Bob might sometimes output the wrong answer. To make the task nontrivial, for each $x$, Bob should output the correct answer with probability $2/3$. The cost of the protocol is the maximum length of $s$, over all possible $x$ and coin tosses.
There are actually two different variants of this model. The model just described has private coins: Bob only gets to see the string $s$. Another variant has public coins: the string that Alice sends is a function of both $s$ and some randomness $r$ not depending on $x$, and Bob gets to see both Alice's message $s$ and the randomness $r$ used to generate it.
(There's actually one more nicety: is the string $s$ self-delimiting or not? We can convert a protocol of cost $C$ to a self-delimiting protocol of cost $C + O(\log C)$, so the difference is not significant.)
How would a Las Vegas algorithm look like? Bob should now always output the correct answer, and the cost of the protocol is the maximum, over $x$, of the average length of $s$.
Consider first the private coin model. Suppose that cost of such a protocol is $C$. This means that for every input $x$, there is a string $s$ of length at most $C$ which causes Bob to output $f(x,y)$ correctly. Hence we can convert this protocol to a deterministic protocol with the same cost.
Newman's theorem shows that we can convert a protocol in the public coin model to one in the private coin model at an additive cost of $O(\log n)$, thereby converting a public coin protocol of cost $C$ to a deterministic protocol of cost $C + O(\log n)$.
In the presence of interaction, Las Vegas protocols become nontrivial, since Bob can single to Alice that she should "try again". Since no such feedback mechanism is available in the one-way model, Las Vegas algorithms don't really make sense in that model.