# If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?

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.

• The notation you're using for the crit range is unusual. Are you describing a Champion Fighter? If so, where are you getting the 17/20 range from? – Xirema Apr 10 '19 at 15:33
• What class/sub class is your character? – KorvinStarmast Apr 10 '19 at 15:34
• Possible duplicate of What are my chances of rolling a natural 19/20 critical if I roll 3d20? – Julien Lopez Apr 10 '19 at 17:55
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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 not crit: 17/20 = 0.85
Chance that all three dice will not crit: (17/20) x (17/20) x (17/20) = 0.614125
Chance that at least one die will 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 not crit: 16/20 = 0.8
Chance that all three dice will not crit: (16/20) x (16/20) x (16/20) = 0.512
Chance that at least one die will crit: 1 - (16/20)3 = 0.488

• This is excellent advice. Probability of success is 100% - probability of failure. Many times calculating failure is way easier than calculating success. – Nelson Apr 11 '19 at 1:52

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!)

# 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.

• Could you explain how dropping the higher die drops your chance of getting a critical on one of the three dice? Assuming you wouldn't drop the higher die if it's already a critical... – Ifusaso Apr 11 '19 at 13:49
• It might be worth adding a sentence explaining why you might want to drop the higher die -- there are situations where you might want to land an attack that doesn't crit, and there might be situations where you don't even want to hit (usually as result of roleplay, that one), and advantage is a bad thing in those cases. Dropping highest partially negates advantage. These situations are rare, but they can occur. – Phlarx Apr 11 '19 at 14:47
• @lfusaso Dropping the lower die blindly (as in, without know the results, just that it's lower) makes it more likely that you're dropping a non-critical roll, which then may become a critical roll. That is, if I drop the lower dice, I'm more likely to drop a 1-17 than I am if I drop the higher dice. – Monica Apologists Get Out Apr 11 '19 at 16:35
• @Ifusaso: this whole answer is an exercise in statistics, not plausible situations. Are there really people who would ever reroll the higher die on purpose, not by accident? (e.g. they forgot it's not just triple advantage, and has to replace the original roll) You still want to roll high to make sure you hit, so the only use-case I can think of is attacking an ally for some reason. Maybe if you're Dominated, the DM would let the player choose to use their Elven Accuracy this way hoping to miss, or at least to not crit. Or some other case where you want to attack but not hit or not crit. – Peter Cordes Apr 12 '19 at 7:52
• Thank you some much for the info and how I can get a range of 17-20 it has to do that I'm playing a D&D 5e battle arena game where the max player lv cap is forty and a up to four multiclass options and just a few things have been changed to fit this setup better. – Braymal Gaming Apr 12 '19 at 14:10

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:

$$(\frac{17}{20} + \frac{3}{20} * x)^3$$ which is

$$(\frac{17}{20})^3 + 3(\frac{17}{20})^2\frac{3}{20} * x + 3 \frac{17}{20}(\frac{3}{20})^2 x^2 + (\frac{3}{20})^3x^3$$ 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.