To calculate the chance of them all critting, you take the odds of the event happening (1/20) and raise it to the power of how many options you have. For eight times, that is .05^8 = .0000000000390625 or 1 out of ever 25 billion times.
The odds of getting at least one crit are 1 - [the odds of the event not happening (19/20) raised to the power of how many rolls you are doing]. For eight times, that is 1-.95^8 = 33.658% chance that, if you roll eight times, one will be a crit.
Reckless attack gives you advantage, which means the probability of getting a crit on either dice (or both) is equal to 1 minus the probability of not getting a crit on either. This is 1 - (19/20 x 19/20).
Your probability of getting X crits out of 4 attacks is therefore the probability of getting a crit to the power of the number of X, or (1-(19^20/20^20))^X or 0.0975^X
If you want the probability of getting at least Y crits, then that's 1 - the probability of getting 4-Y non-crits or 1-((19^20)^(4-Y)) or 1-(0.9025^(4-Y))
The odds are: 1 - (odds of not getting a 20 on any of them)
So to not get any nat 20s there are 19 options on each die that aren't a 20. So for an individual die that's 19/20
For eight dice that becomes (19/20)^8
So that means the odds of getting at least one crit are: 1 - (19/20)^8 ~= 33.66%
This is a signature. It was a simple signature. But it has been upgraded.
Belolonandalogalo Malololomologalo Tumagalokumagalo, Sunny
Eggo Lass, Bone and Oblivion
Tendilius Mondhaven Paxaramus, Drakkenheim
Recruiting for Troubles in the Silverwood Forest
Get rickrolled here. Awesome music here. Track 50, 9/23/25, The Mystery of Your Gift
To calculate the chance of them all critting, you take the odds of the event happening (1/20) and raise it to the power of how many options you have. For eight times, that is .05^8 = .0000000000390625 or 1 out of ever 25 billion times.
The odds of getting at least one crit are 1 - [the odds of the event not happening (19/20) raised to the power of how many rolls you are doing]. For eight times, that is 1-.95^8 = 33.658% chance that, if you roll eight times, one will be a crit.
A good site for questions like this is anydice.com.
J
Great Wyrm Moonstone Dungeon Master
The time of the ORC has come. No OGL without irrevocability; no OGL with 'authorized version' language. #openDND
Practice, practice, practice • Respect the rules; don't memorize them • Be merciless, not cruel • Don't let the dice run the game for you
Reckless attack gives you advantage, which means the probability of getting a crit on either dice (or both) is equal to 1 minus the probability of not getting a crit on either. This is 1 - (19/20 x 19/20).
Your probability of getting X crits out of 4 attacks is therefore the probability of getting a crit to the power of the number of X, or (1-(19^20/20^20))^X or 0.0975^X
If you want the probability of getting at least Y crits, then that's 1 - the probability of getting 4-Y non-crits or 1-((19^20)^(4-Y)) or 1-(0.9025^(4-Y))
Find my D&D Beyond articles here