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%
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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.
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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))
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