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.

Best of three rolls with 18-20 crit range: ~39% chance to crit

Chance that a single die will não crit: 17/20 = 0.85
Chance that all three dice will não crit: (17/20) x (17/20) x (17/20) = 0.614125
Chance that finalmente one die vontade crit: 1 - (17/20)³ = 0.385875

Best of three rolls with 17-20 crit range: ~49% chance to crit

Chance that a single die will não crit: 16/20 = 0.8
Chance that all three dice will não crit: (16/20) x (16/20) x (16/20) = 0.512
Chance that finalmente one die vontade 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

If p is the probability of a crit on a single roll, then 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!)

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:

$$(\frac{17}{20} + \frac{3}{20} * x)^3$$ que é

$$(\frac{17}{20})^3 + 3(\frac{17}{20})^2\frac{3}{20} * x + 3 \frac{17}{20}(\frac{3}{20})^2 x^2 + (\frac{3}{20})^3x^3$$ 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 off all 4 terms add up to 1 -- so we can just take the $x^0$ coefficient and subtract 1.

$$1-(\frac{17}{20})^3$$ 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:

$$(\frac{17}{20} + \frac{3}{20} * x)^2 ( \frac{17}{20} + \frac{3}{20} * y )$$ by using a different variable (y instead of x).

We can then expand

$$(\frac{17}{20})^2 + 2\frac{3*17}{20^2}x + (\frac{3}{20} * x)^2 (\frac{17}{20} + \frac{3}{20} * y )$$ or $$\frac{17^3}{20^3} + \frac{3*17^2}{20^3}y + 2\frac{3*17^2}{20^3}x + 2\frac{3^2*17}{20^3}xy + \frac{3^2*17}{20^3} x^2 + \frac{3^3}{20^3} x^2y$$ then isolate the cases that have both an x and a y, from those with only xs 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.

It depends on how you use the Elven Accuracy Die

There's two ways that players are legally allowed to use the Elven Accuracy Die:

• Replace the lower of the two advantage dice (Type A)
• Replace the higher of the two advantage dice (Type B)

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.