He is wrong, you were right in the first place. The only relevant numbers are the odds of getting the 1st 4 rolls without a 1, 2 or 3.
So, I think this is one of the cases where the maths is super important, so it's worth the time/effort to discuss it until we all agree..
I think my initial calculation was wrong, because the existence of a vig changes the odds of not having any {1,2,3} in the first 4 rolls. Sure, the vig could have been the fifth roll, but that's just the problem, it now needs to be the fifth roll. That decreases the probability.
I don't think Egork's math is correct either, I just said 'you are right' to simplify things and the discussion, because the result I calculated was very similar:
I think what we really need to calculate is this:
P(A given B) where
A := We have a setup of: {C, RC, W, RB, RB}
B := We have exactly 1 Cop, exactly 1 Watcher, exactly 1 Rolecop, either 1 or 2 Roleblockers, and no other PRs
In other words, A is the setup if both claims are true, and B is every setup that's still possible. We want to know how likely A is given that B is true.
P(A | B) = P(A ∩ B) / P(B) = P(A) / P(B) ≈ 23%