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I have Elven Accuracy so with advantage on an attack roll using Dexterity, Intelligence, Wisdom, or Charisma, you can reroll one of the dice once.
So what are my crit chance with a crit range of 18/20 with three rolls also what would be the crit chance with a crit range of 17/20 with three rolls.
For "at least one" probability problems, it's usually easier to start by calculating the chance that none of the dice crit, as that saves you the hassle of combining the probabilities of getting 1/2/3 crits.
Chance that a single die will not crit: 17/20 = 0.85
Chance that all three dice will not crit: (17/20) × (17/20) × (17/20) = 0.614125
Chance that at least one die will crit: 1 - (17/20)3 = 0.385875
Chance that a single die will not crit: 16/20 = 0.8
Chance that all three dice will not crit: (16/20) × (16/20) × (16/20) = 0.512
Chance that at least one die will crit: 1 - (16/20)3 = 0.488
The other answers do a good job of answering the question, but I'll point out how you can answer questions like this in the future:
https://anydice.com/ is a very powerful (if slightly complicated) calculator for these sorts of questions. In your case, you'd enter the query:
output [highest 1 of 3d20]
And then select "At Least" from options below to get this table, which shows the odds of getting at least each number:
The result, is 38.59% for a crit range of 18, and 48.8% for a crit range of 17
A general way to solve this kind of problem is with counting polynomials.
$$\frac{17}{20} + \frac{3}{20} x$$ this polynomial represents rolling 1d20 and having a 3/20 chance of getting a crit. The crit chance is the coefficient to the \$x^1\$ term.
For 3d20 it looks like:
$$\left(\frac{17}{20} + \frac{3}{20} x\right)^3$$ which is
$$\left(\frac{17}{20}\right)^3 + 3\left(\frac{17}{20}\right)^2\frac{3}{20} x + 3 \frac{17}{20}\left(\frac{3}{20}\right)^2 x^2 + \left(\frac{3}{20}\right)^3x^3$$ where the \$x^1\$ through \$x^3\$ represent the 1 through 3 of the dice landing on 18 19 or 20.
We could add up the coefficients of the \$x^1\$ through \$x^3\$ cases, but we also know that the coefficients of all 4 terms add up to 1 – so we can just take the \$x^0\$ coefficient and subtract 1.
$$1-\left(\frac{17}{20}\right)^3$$ or
$$\frac{8000-4913}{8000}$$ aka about $$38.6\%$$
Now this is a bit complicated; but we can use it to analyze more complicated cases.
Imagine a rule that states that crits from elven accuracy deal an extra 50 damage, but only if you had already critted. We can distinguish the elven accuracy crit from the others:
$$\left(\frac{17}{20} + \frac{3}{20} x\right)^2 \left( \frac{17}{20} + \frac{3}{20} y \right)$$ by using a different variable (\$y\$ instead of \$x\$).
We can then expand
$$\left(\frac{17}{20}\right)^2 + 2\frac{3\cdot17}{20^2}x + \left(\frac{3}{20} x\right)^2 \left(\frac{17}{20} + \frac{3}{20} y \right)$$ or $$\frac{17^3}{20^3} + \frac{3\cdot17^2}{20^3}y + 2\frac{3\cdot17^2}{20^3}x + 2\frac{3^2\cdot17}{20^3}xy + \frac{3^2\cdot17}{20^3} x^2 + \frac{3^3}{20^3} x^2y$$ then isolate the cases that have both an \$x\$ and a \$y\$, from those with only \$x\$s or only a \$y\$.
For the most part, this technique really gets useful when you can feed the polynomials to a program that can do the number crunching for you.
If \$p\$ is the probability of a crit on a single roll, then \$P=1-(1-p)^N\$ is the probability of at least one crit on \$N\$ rolls.
so for \$N=3\$ and \$p=3/20, P=38.6\%\$
For a critical hit range of \$17-20, P=48.8\%\$ (WOW!)
There's two ways that players are legally allowed to use the Elven Accuracy Die:
In the former case, this roll is mathematically equivalent to rolling 3 dice and taking the highest. In the latter case, it's more like rolling two dice, taking the lower, and then taking the higher of that result and a third die.
\begin{array}{r|l|l} \text{Outcomes} & \text{18-20 A} & \text{18-20 B} \\ \hline \text{Non-Crit} & \text{61.413%} & \text{83.088%} \\ \text{Crit} & \text{38.588%} & \text{16.913%} \\ \end{array} \begin{array}{r|l|l} & \text{17-20 A} & \text{17-20 B} \\ \hline \text{Non-Crit} & \text{51.200%} & \text{76.800%} \\ \text{Crit} & \text{48.800%} & \text{23.200%} \\ \end{array}
Note: as far as I'm aware, in 5th Edition D&D, it is not possible to get a Critical hit range that includes 17. It's possible I'm unaware of a specific class feature or magic item that is expanding the range beyond what can be attained by a Champion Fighter at level 15. But as a result, the second table (the 17-20 range) does not have practical use in this game.
