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What are my chances of rolling a natural 19/20 critical if I roll 3d20?
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The Oath of Vengeance paladin gains the Channel Divinity option "Vow of Enmity" at 3rd level (PHB, p. 88):
As a bonus action, you can utter a vow of enmity against a creature you can see within 10 feet of you. You gain advantage on attack rolls against the creature for 1 minute or until it drops to 0 hit points or falls unconscious.
Part of the description of the Elven Accuracy feat (Xanathar's Guide to Everything, p. 74) states:
Whenever you have advantage on an attack roll using Dexterity, Intelligence, Wisdom, or Charisma, you can reroll one of the dice once.
The Hexblade warlock patron grants the Hexblade's Curse feature at 1st level (Xanathar's Guide to Everything, p. 55), which allows you to mark and curse a creature. One benefit of this curse is:
- Any attack roll you make against the cursed target is a critical hit on a roll of 19 or 20 on the d20.
Combining all of these things, what are my chances of rolling a natural 19/20 if I roll 3d20?
dnd-5e statistics critical-hit advantage attack-roll
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
add a comment |Â
up vote
22
down vote
favorite
The Oath of Vengeance paladin gains the Channel Divinity option "Vow of Enmity" at 3rd level (PHB, p. 88):
As a bonus action, you can utter a vow of enmity against a creature you can see within 10 feet of you. You gain advantage on attack rolls against the creature for 1 minute or until it drops to 0 hit points or falls unconscious.
Part of the description of the Elven Accuracy feat (Xanathar's Guide to Everything, p. 74) states:
Whenever you have advantage on an attack roll using Dexterity, Intelligence, Wisdom, or Charisma, you can reroll one of the dice once.
The Hexblade warlock patron grants the Hexblade's Curse feature at 1st level (Xanathar's Guide to Everything, p. 55), which allows you to mark and curse a creature. One benefit of this curse is:
- Any attack roll you make against the cursed target is a critical hit on a roll of 19 or 20 on the d20.
Combining all of these things, what are my chances of rolling a natural 19/20 if I roll 3d20?
dnd-5e statistics critical-hit advantage attack-roll
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
Comments are not for extended discussion; this conversation has been moved to chat.
– doppelgreener♦
Aug 30 at 21:25
add a comment |Â
up vote
22
down vote
favorite
up vote
22
down vote
favorite
The Oath of Vengeance paladin gains the Channel Divinity option "Vow of Enmity" at 3rd level (PHB, p. 88):
As a bonus action, you can utter a vow of enmity against a creature you can see within 10 feet of you. You gain advantage on attack rolls against the creature for 1 minute or until it drops to 0 hit points or falls unconscious.
Part of the description of the Elven Accuracy feat (Xanathar's Guide to Everything, p. 74) states:
Whenever you have advantage on an attack roll using Dexterity, Intelligence, Wisdom, or Charisma, you can reroll one of the dice once.
The Hexblade warlock patron grants the Hexblade's Curse feature at 1st level (Xanathar's Guide to Everything, p. 55), which allows you to mark and curse a creature. One benefit of this curse is:
- Any attack roll you make against the cursed target is a critical hit on a roll of 19 or 20 on the d20.
Combining all of these things, what are my chances of rolling a natural 19/20 if I roll 3d20?
dnd-5e statistics critical-hit advantage attack-roll
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
The Oath of Vengeance paladin gains the Channel Divinity option "Vow of Enmity" at 3rd level (PHB, p. 88):
As a bonus action, you can utter a vow of enmity against a creature you can see within 10 feet of you. You gain advantage on attack rolls against the creature for 1 minute or until it drops to 0 hit points or falls unconscious.
Part of the description of the Elven Accuracy feat (Xanathar's Guide to Everything, p. 74) states:
Whenever you have advantage on an attack roll using Dexterity, Intelligence, Wisdom, or Charisma, you can reroll one of the dice once.
The Hexblade warlock patron grants the Hexblade's Curse feature at 1st level (Xanathar's Guide to Everything, p. 55), which allows you to mark and curse a creature. One benefit of this curse is:
- Any attack roll you make against the cursed target is a critical hit on a roll of 19 or 20 on the d20.
Combining all of these things, what are my chances of rolling a natural 19/20 if I roll 3d20?
dnd-5e statistics critical-hit advantage attack-roll
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
edited Aug 29 at 22:57
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
asked Aug 29 at 20:40
Daniel Zastoupil
4,2011051
4,2011051
Comments are not for extended discussion; this conversation has been moved to chat.
– doppelgreener♦
Aug 30 at 21:25
add a comment |Â
Comments are not for extended discussion; this conversation has been moved to chat.
– doppelgreener♦
Aug 30 at 21:25
Comments are not for extended discussion; this conversation has been moved to chat.
– doppelgreener♦
Aug 30 at 21:25
Comments are not for extended discussion; this conversation has been moved to chat.
– doppelgreener♦
Aug 30 at 21:25
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
48
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accepted
About 27%, or a little over 1/4
This is a pretty easy calculation to run in anydice:
As you can see, the percentage chance of rolling at least a 19 one of the d20 rolls is 27.10% So you can expect, on average, to crit a little over once in every four attacks made in this manner.
How do you calculate that without a simulation?
The probability of getting at least one 19 or 20 among the three dice is equal to 1 minus the probability of not getting any 19s or 20s among the three dice.
Since there is a 2 out of 20 chance to get a 19 or a 20 on each die, we know that there is an 18 out of 20 chance to not get any of those values. 18/20 = 0.9, so there is a 90% chance we won't get any 19s or 20s on any given die.
To find out the odds of not getting a 19 or a 20 on any of the three dice together, we just have to multiply the probabilities for each individual die together.
So, the probability of getting at least one 19 or 20 among the three dice would be $$1 - (0.9 times 0.9 times 0.9) = 1 - 0.729 = 0.271$$
Multiplying that probability by 100 gives us a 27.1% chance to crit on any given attack.
answered Aug 29 at 20:48
Adam
18.1k373123
2
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
48
down vote
accepted
About 27%, or a little over 1/4
This is a pretty easy calculation to run in anydice:
As you can see, the percentage chance of rolling at least a 19 one of the d20 rolls is 27.10% So you can expect, on average, to crit a little over once in every four attacks made in this manner.
How do you calculate that without a simulation?
The probability of getting at least one 19 or 20 among the three dice is equal to 1 minus the probability of not getting any 19s or 20s among the three dice.
Since there is a 2 out of 20 chance to get a 19 or a 20 on each die, we know that there is an 18 out of 20 chance to not get any of those values. 18/20 = 0.9, so there is a 90% chance we won't get any 19s or 20s on any given die.
To find out the odds of not getting a 19 or a 20 on any of the three dice together, we just have to multiply the probabilities for each individual die together.
So, the probability of getting at least one 19 or 20 among the three dice would be $$1 - (0.9 times 0.9 times 0.9) = 1 - 0.729 = 0.271$$
Multiplying that probability by 100 gives us a 27.1% chance to crit on any given attack.
answered Aug 29 at 20:48
Adam
18.1k373123
2
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
add a comment |Â
up vote
48
down vote
accepted
About 27%, or a little over 1/4
This is a pretty easy calculation to run in anydice:
As you can see, the percentage chance of rolling at least a 19 one of the d20 rolls is 27.10% So you can expect, on average, to crit a little over once in every four attacks made in this manner.
How do you calculate that without a simulation?
The probability of getting at least one 19 or 20 among the three dice is equal to 1 minus the probability of not getting any 19s or 20s among the three dice.
Since there is a 2 out of 20 chance to get a 19 or a 20 on each die, we know that there is an 18 out of 20 chance to not get any of those values. 18/20 = 0.9, so there is a 90% chance we won't get any 19s or 20s on any given die.
To find out the odds of not getting a 19 or a 20 on any of the three dice together, we just have to multiply the probabilities for each individual die together.
So, the probability of getting at least one 19 or 20 among the three dice would be $$1 - (0.9 times 0.9 times 0.9) = 1 - 0.729 = 0.271$$
Multiplying that probability by 100 gives us a 27.1% chance to crit on any given attack.
answered Aug 29 at 20:48
Adam
18.1k373123
2
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
add a comment |Â
up vote
48
down vote
accepted
up vote
48
down vote
accepted
About 27%, or a little over 1/4
This is a pretty easy calculation to run in anydice:
As you can see, the percentage chance of rolling at least a 19 one of the d20 rolls is 27.10% So you can expect, on average, to crit a little over once in every four attacks made in this manner.
How do you calculate that without a simulation?
The probability of getting at least one 19 or 20 among the three dice is equal to 1 minus the probability of not getting any 19s or 20s among the three dice.
Since there is a 2 out of 20 chance to get a 19 or a 20 on each die, we know that there is an 18 out of 20 chance to not get any of those values. 18/20 = 0.9, so there is a 90% chance we won't get any 19s or 20s on any given die.
To find out the odds of not getting a 19 or a 20 on any of the three dice together, we just have to multiply the probabilities for each individual die together.
So, the probability of getting at least one 19 or 20 among the three dice would be $$1 - (0.9 times 0.9 times 0.9) = 1 - 0.729 = 0.271$$
Multiplying that probability by 100 gives us a 27.1% chance to crit on any given attack.
answered Aug 29 at 20:48
Adam
18.1k373123
About 27%, or a little over 1/4
This is a pretty easy calculation to run in anydice:
As you can see, the percentage chance of rolling at least a 19 one of the d20 rolls is 27.10% So you can expect, on average, to crit a little over once in every four attacks made in this manner.
How do you calculate that without a simulation?
The probability of getting at least one 19 or 20 among the three dice is equal to 1 minus the probability of not getting any 19s or 20s among the three dice.
Since there is a 2 out of 20 chance to get a 19 or a 20 on each die, we know that there is an 18 out of 20 chance to not get any of those values. 18/20 = 0.9, so there is a 90% chance we won't get any 19s or 20s on any given die.
To find out the odds of not getting a 19 or a 20 on any of the three dice together, we just have to multiply the probabilities for each individual die together.
So, the probability of getting at least one 19 or 20 among the three dice would be $$1 - (0.9 times 0.9 times 0.9) = 1 - 0.729 = 0.271$$
Multiplying that probability by 100 gives us a 27.1% chance to crit on any given attack.
answered Aug 29 at 20:48
Adam
18.1k373123
edited Aug 30 at 13:05
answered Aug 29 at 20:48
Adam
18.1k373123
answered Aug 29 at 20:48
Adam
18.1k373123
answered Aug 29 at 20:48
Adam
18.1k373123
18.1k373123
2
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
add a comment |Â
2
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
2
2
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
giantitp.com/forums/… This answer is being referenced as basically a doomsday device against DM's. Nicely done!
– Daniel Zastoupil
Aug 29 at 21:11
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
I think it's worth noting that this is a better chance of critting than your chance of critting normally with at least one of three normal attacks with advantage. (1- (.95 x .95)^3 = .2649)
– Gandalfmeansme
Aug 30 at 16:40
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
@Gandalfmeansme That's always true though. If all you're interested in is at least one success, then increasing your single chance probability by n times will always net better results than increasing the number of attempts by n times.
– Ordous
Aug 30 at 17:49
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
An awkward situation I could see is when a Rogue gets a hit, can apply sneak attack, but has to decide not to in order to fish for a crit. Kinda funny, thinking about it. "Nah, I'm gonna decide not to stab him really hard to hopefully stab him harder later".
– Daniel Zastoupil
Aug 30 at 18:32
add a comment |Â
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Comments are not for extended discussion; this conversation has been moved to chat.
– doppelgreener♦
Aug 30 at 21:25