| lobachevskii |
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-- There isn't a DM alive who is going to let a player get away with having two die to confirm his crit, text or no text.
In the absence of any text, EITHER the DM says "Just use the other one which you already rolled," OR he says "Roll another one." In zero percent of cases is he going to say, "Use the leftover die to confirm, unless it stinks; in which case I'll let you roll another."
According to my back of the envelope calculations these two scenarios give (slightly) different probabilities of a confirmed crit.
1) Perfect Strike as written:
number of ways of getting a confirmed crit: 2*(20-X)+1 (where X is the target to hit, X not equal to 1).
total number of ways to roll 2 D20: 400
P(confirmed crit) = (2*(20-X)+1)/400
Brief explanation: You get a confirmed crit if you roll a 20 and the other die hits. You can roll two 20s (that's the 1) or a 20 and any number less than 20 down to X in one of two ways (that's the 2*(20-X) term).
2) Perfect strike but confirm the critical separately:
number of ways of getting a 20 on 2D20: 2*19+1 = 39.
number of ways of hitting on the confirming die = 21-X (X is target number, X not equal to 1)
P(confirmed crit) = 39*(21-X)/(400*20)
Brief explanation: It doesn't matter what you roll on the first pair of dice as long as one's a 20. You can roll two 20s (that's the 1) or a 20 and any number less than 20, there are nineteen of those, in one of two ways (that's the 2*19 term). Probability of confirming a crit is (21-X)/20. We can multiply the two probabilities since they are independent.
Comparing these two distributions we find that 39*(21-X) > 20*(2*(20-X)+1) for all X > 1. The difference becomes increasingly large as X increases.
TLDR: The statement is necessary. It specifies a game mechanic to use which produces a slightly different probability distribution to that which would arise under the standard rules.