Is it okay to roll multiple attacks that all have advantage in one cluster?Do you have advantage when multiple sources give advantage and only one gives disadvantage?Is giving one side advantage always equivalent to giving disadvantage to the opponent?Do multiple attacks get advantage for both attacks if the creature is under the effects of Faerie Fire?Help me ensure I have a proper grasp of Hiding and Stealth as it applies to the UA RangerWhen making multiple attacks, does advantage from being unseen apply to all attacks?Can you Hide from some but not all enemies?If you have dis/advantage on a roll, do you have dis/advantage on a reroll?Is the ranged attack bonus +27 for a maxed out 20th level archery-styled Hunter with applicable RAW magic items?If I have multiple attacks, does the penalty from Combat Expertise apply to all attacks?Does the Wild Magic sorcerer's Tides of Chaos feature grant advantage on all attacks, or just the first one?
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Is it okay to roll multiple attacks that all have advantage in one cluster?
Do you have advantage when multiple sources give advantage and only one gives disadvantage?Is giving one side advantage always equivalent to giving disadvantage to the opponent?Do multiple attacks get advantage for both attacks if the creature is under the effects of Faerie Fire?Help me ensure I have a proper grasp of Hiding and Stealth as it applies to the UA RangerWhen making multiple attacks, does advantage from being unseen apply to all attacks?Can you Hide from some but not all enemies?If you have dis/advantage on a roll, do you have dis/advantage on a reroll?Is the ranged attack bonus +27 for a maxed out 20th level archery-styled Hunter with applicable RAW magic items?If I have multiple attacks, does the penalty from Combat Expertise apply to all attacks?Does the Wild Magic sorcerer's Tides of Chaos feature grant advantage on all attacks, or just the first one?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am planning to make my first Ranger and going Hunter to eventually get Volley. If I get the preemptive strike from an unseen position, then RAW I get advantage on attack rolls.
Let's say I am moving stealthily through a forest and spot 6 orcs marching down in a caravan, all within 10 ft of each other. I need to make 6 attack rolls, all with advantage.
Because of the advantage do I need to do them one by one (or use pairs of color coded dice)? Or is it considered fair to roll 12 x d20 at once, then take the highest 6 rolls?
dnd-5e statistics advantage-and-disadvantage attack-roll
$endgroup$
add a comment |
$begingroup$
I am planning to make my first Ranger and going Hunter to eventually get Volley. If I get the preemptive strike from an unseen position, then RAW I get advantage on attack rolls.
Let's say I am moving stealthily through a forest and spot 6 orcs marching down in a caravan, all within 10 ft of each other. I need to make 6 attack rolls, all with advantage.
Because of the advantage do I need to do them one by one (or use pairs of color coded dice)? Or is it considered fair to roll 12 x d20 at once, then take the highest 6 rolls?
dnd-5e statistics advantage-and-disadvantage attack-roll
$endgroup$
2
$begingroup$
Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage?
$endgroup$
– kviiri
8 hours ago
add a comment |
$begingroup$
I am planning to make my first Ranger and going Hunter to eventually get Volley. If I get the preemptive strike from an unseen position, then RAW I get advantage on attack rolls.
Let's say I am moving stealthily through a forest and spot 6 orcs marching down in a caravan, all within 10 ft of each other. I need to make 6 attack rolls, all with advantage.
Because of the advantage do I need to do them one by one (or use pairs of color coded dice)? Or is it considered fair to roll 12 x d20 at once, then take the highest 6 rolls?
dnd-5e statistics advantage-and-disadvantage attack-roll
$endgroup$
I am planning to make my first Ranger and going Hunter to eventually get Volley. If I get the preemptive strike from an unseen position, then RAW I get advantage on attack rolls.
Let's say I am moving stealthily through a forest and spot 6 orcs marching down in a caravan, all within 10 ft of each other. I need to make 6 attack rolls, all with advantage.
Because of the advantage do I need to do them one by one (or use pairs of color coded dice)? Or is it considered fair to roll 12 x d20 at once, then take the highest 6 rolls?
dnd-5e statistics advantage-and-disadvantage attack-roll
dnd-5e statistics advantage-and-disadvantage attack-roll
edited 4 hours ago
Derek Stucki
22.5k7 gold badges73 silver badges114 bronze badges
22.5k7 gold badges73 silver badges114 bronze badges
asked 9 hours ago
RyanFromGDSERyanFromGDSE
2,4094 gold badges19 silver badges50 bronze badges
2,4094 gold badges19 silver badges50 bronze badges
2
$begingroup$
Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage?
$endgroup$
– kviiri
8 hours ago
add a comment |
2
$begingroup$
Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage?
$endgroup$
– kviiri
8 hours ago
2
2
$begingroup$
Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage?
$endgroup$
– kviiri
8 hours ago
$begingroup$
Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage?
$endgroup$
– kviiri
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Rolling 12d20 and taking the 6 highest is not equivalent
Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 attacks, you roll double 1s. On the last 3 attacks, you roll double 20s. That means that 3 out of 6 attacks hit. If you instead pooled those 12 rolls all together and took the 6 highest, all 6 attacks would hit, because you rolled 6 20s. With a little work, it's not too hard to show that taking the top 6 dice is always guaranteed to give a result at least as good as rolling with advantage normally, and will usually give a better result (i.e. more hits and/or more crits).
To make it fast and fair: decide before rolling how to pair the dice
Obviously, it would be nice to resolve all 6 attacks at once with a single roll of 12d20, and you can do so in a fair way, even with 12 identical dice, by defining a rule ahead of time that dictates how you will pair up the dice. A simple rule is to pair the dice up from left to right. Roll all the dice, then take the two leftmost dice and pair them up. Then the next two leftmost after that, and so on, until you have 6 pairs. Finally, once you have paired up the dice, start resolving attacks. Make sure to tell the DM what rule you are using so they can check that you are following it instead of cherry-picking pairs.
$endgroup$
add a comment |
$begingroup$
I would say, if I was DM, that you would have to use separate rolls or some differentiation of dice pairs.
For instance, if you rolled at advantage for six attacks, you could get a result on the dice of 20 and 12 on the first attack, 18 and 14 on the second, 1 and 4 on the third, 12 and 7 on the fourth, 20 and 1 on the fifth, and 15 and 17 on the sixth. Considering each pair in order, you'd have a result of 20, 18, 4, 12, 20, and 17. However, if you rolled 12 identical dice at once, you'd get to pick all the highest dice for a corresponding series of results: 20, 20, 18, 17, 15, and 14. This could make for a major difference in the outcome of the battle. Rolling 12 dice at once would allow you to essentially swap a low roll against one Orc for a higher roll against a different Orc.
If you did this roll in a group of undifferentiated dice, you could pick the lower of the two dice intended for your attack roll against Orc A and instead use it on Orc B, when your two attack rolls against B were both lower than that value (and might otherwise have caused you to miss B). This would be essentially cheating, like changing the face-up face of the dice by sleight of hand after the dice stopped rolling.
In other words no, it would not be "fair".
$endgroup$
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
add a comment |
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2 Answers
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$begingroup$
Rolling 12d20 and taking the 6 highest is not equivalent
Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 attacks, you roll double 1s. On the last 3 attacks, you roll double 20s. That means that 3 out of 6 attacks hit. If you instead pooled those 12 rolls all together and took the 6 highest, all 6 attacks would hit, because you rolled 6 20s. With a little work, it's not too hard to show that taking the top 6 dice is always guaranteed to give a result at least as good as rolling with advantage normally, and will usually give a better result (i.e. more hits and/or more crits).
To make it fast and fair: decide before rolling how to pair the dice
Obviously, it would be nice to resolve all 6 attacks at once with a single roll of 12d20, and you can do so in a fair way, even with 12 identical dice, by defining a rule ahead of time that dictates how you will pair up the dice. A simple rule is to pair the dice up from left to right. Roll all the dice, then take the two leftmost dice and pair them up. Then the next two leftmost after that, and so on, until you have 6 pairs. Finally, once you have paired up the dice, start resolving attacks. Make sure to tell the DM what rule you are using so they can check that you are following it instead of cherry-picking pairs.
$endgroup$
add a comment |
$begingroup$
Rolling 12d20 and taking the 6 highest is not equivalent
Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 attacks, you roll double 1s. On the last 3 attacks, you roll double 20s. That means that 3 out of 6 attacks hit. If you instead pooled those 12 rolls all together and took the 6 highest, all 6 attacks would hit, because you rolled 6 20s. With a little work, it's not too hard to show that taking the top 6 dice is always guaranteed to give a result at least as good as rolling with advantage normally, and will usually give a better result (i.e. more hits and/or more crits).
To make it fast and fair: decide before rolling how to pair the dice
Obviously, it would be nice to resolve all 6 attacks at once with a single roll of 12d20, and you can do so in a fair way, even with 12 identical dice, by defining a rule ahead of time that dictates how you will pair up the dice. A simple rule is to pair the dice up from left to right. Roll all the dice, then take the two leftmost dice and pair them up. Then the next two leftmost after that, and so on, until you have 6 pairs. Finally, once you have paired up the dice, start resolving attacks. Make sure to tell the DM what rule you are using so they can check that you are following it instead of cherry-picking pairs.
$endgroup$
add a comment |
$begingroup$
Rolling 12d20 and taking the 6 highest is not equivalent
Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 attacks, you roll double 1s. On the last 3 attacks, you roll double 20s. That means that 3 out of 6 attacks hit. If you instead pooled those 12 rolls all together and took the 6 highest, all 6 attacks would hit, because you rolled 6 20s. With a little work, it's not too hard to show that taking the top 6 dice is always guaranteed to give a result at least as good as rolling with advantage normally, and will usually give a better result (i.e. more hits and/or more crits).
To make it fast and fair: decide before rolling how to pair the dice
Obviously, it would be nice to resolve all 6 attacks at once with a single roll of 12d20, and you can do so in a fair way, even with 12 identical dice, by defining a rule ahead of time that dictates how you will pair up the dice. A simple rule is to pair the dice up from left to right. Roll all the dice, then take the two leftmost dice and pair them up. Then the next two leftmost after that, and so on, until you have 6 pairs. Finally, once you have paired up the dice, start resolving attacks. Make sure to tell the DM what rule you are using so they can check that you are following it instead of cherry-picking pairs.
$endgroup$
Rolling 12d20 and taking the 6 highest is not equivalent
Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 attacks, you roll double 1s. On the last 3 attacks, you roll double 20s. That means that 3 out of 6 attacks hit. If you instead pooled those 12 rolls all together and took the 6 highest, all 6 attacks would hit, because you rolled 6 20s. With a little work, it's not too hard to show that taking the top 6 dice is always guaranteed to give a result at least as good as rolling with advantage normally, and will usually give a better result (i.e. more hits and/or more crits).
To make it fast and fair: decide before rolling how to pair the dice
Obviously, it would be nice to resolve all 6 attacks at once with a single roll of 12d20, and you can do so in a fair way, even with 12 identical dice, by defining a rule ahead of time that dictates how you will pair up the dice. A simple rule is to pair the dice up from left to right. Roll all the dice, then take the two leftmost dice and pair them up. Then the next two leftmost after that, and so on, until you have 6 pairs. Finally, once you have paired up the dice, start resolving attacks. Make sure to tell the DM what rule you are using so they can check that you are following it instead of cherry-picking pairs.
edited 5 hours ago
answered 7 hours ago
Ryan ThompsonRyan Thompson
16k2 gold badges53 silver badges116 bronze badges
16k2 gold badges53 silver badges116 bronze badges
add a comment |
add a comment |
$begingroup$
I would say, if I was DM, that you would have to use separate rolls or some differentiation of dice pairs.
For instance, if you rolled at advantage for six attacks, you could get a result on the dice of 20 and 12 on the first attack, 18 and 14 on the second, 1 and 4 on the third, 12 and 7 on the fourth, 20 and 1 on the fifth, and 15 and 17 on the sixth. Considering each pair in order, you'd have a result of 20, 18, 4, 12, 20, and 17. However, if you rolled 12 identical dice at once, you'd get to pick all the highest dice for a corresponding series of results: 20, 20, 18, 17, 15, and 14. This could make for a major difference in the outcome of the battle. Rolling 12 dice at once would allow you to essentially swap a low roll against one Orc for a higher roll against a different Orc.
If you did this roll in a group of undifferentiated dice, you could pick the lower of the two dice intended for your attack roll against Orc A and instead use it on Orc B, when your two attack rolls against B were both lower than that value (and might otherwise have caused you to miss B). This would be essentially cheating, like changing the face-up face of the dice by sleight of hand after the dice stopped rolling.
In other words no, it would not be "fair".
$endgroup$
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
add a comment |
$begingroup$
I would say, if I was DM, that you would have to use separate rolls or some differentiation of dice pairs.
For instance, if you rolled at advantage for six attacks, you could get a result on the dice of 20 and 12 on the first attack, 18 and 14 on the second, 1 and 4 on the third, 12 and 7 on the fourth, 20 and 1 on the fifth, and 15 and 17 on the sixth. Considering each pair in order, you'd have a result of 20, 18, 4, 12, 20, and 17. However, if you rolled 12 identical dice at once, you'd get to pick all the highest dice for a corresponding series of results: 20, 20, 18, 17, 15, and 14. This could make for a major difference in the outcome of the battle. Rolling 12 dice at once would allow you to essentially swap a low roll against one Orc for a higher roll against a different Orc.
If you did this roll in a group of undifferentiated dice, you could pick the lower of the two dice intended for your attack roll against Orc A and instead use it on Orc B, when your two attack rolls against B were both lower than that value (and might otherwise have caused you to miss B). This would be essentially cheating, like changing the face-up face of the dice by sleight of hand after the dice stopped rolling.
In other words no, it would not be "fair".
$endgroup$
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
add a comment |
$begingroup$
I would say, if I was DM, that you would have to use separate rolls or some differentiation of dice pairs.
For instance, if you rolled at advantage for six attacks, you could get a result on the dice of 20 and 12 on the first attack, 18 and 14 on the second, 1 and 4 on the third, 12 and 7 on the fourth, 20 and 1 on the fifth, and 15 and 17 on the sixth. Considering each pair in order, you'd have a result of 20, 18, 4, 12, 20, and 17. However, if you rolled 12 identical dice at once, you'd get to pick all the highest dice for a corresponding series of results: 20, 20, 18, 17, 15, and 14. This could make for a major difference in the outcome of the battle. Rolling 12 dice at once would allow you to essentially swap a low roll against one Orc for a higher roll against a different Orc.
If you did this roll in a group of undifferentiated dice, you could pick the lower of the two dice intended for your attack roll against Orc A and instead use it on Orc B, when your two attack rolls against B were both lower than that value (and might otherwise have caused you to miss B). This would be essentially cheating, like changing the face-up face of the dice by sleight of hand after the dice stopped rolling.
In other words no, it would not be "fair".
$endgroup$
I would say, if I was DM, that you would have to use separate rolls or some differentiation of dice pairs.
For instance, if you rolled at advantage for six attacks, you could get a result on the dice of 20 and 12 on the first attack, 18 and 14 on the second, 1 and 4 on the third, 12 and 7 on the fourth, 20 and 1 on the fifth, and 15 and 17 on the sixth. Considering each pair in order, you'd have a result of 20, 18, 4, 12, 20, and 17. However, if you rolled 12 identical dice at once, you'd get to pick all the highest dice for a corresponding series of results: 20, 20, 18, 17, 15, and 14. This could make for a major difference in the outcome of the battle. Rolling 12 dice at once would allow you to essentially swap a low roll against one Orc for a higher roll against a different Orc.
If you did this roll in a group of undifferentiated dice, you could pick the lower of the two dice intended for your attack roll against Orc A and instead use it on Orc B, when your two attack rolls against B were both lower than that value (and might otherwise have caused you to miss B). This would be essentially cheating, like changing the face-up face of the dice by sleight of hand after the dice stopped rolling.
In other words no, it would not be "fair".
edited 19 mins ago
V2Blast♦
31.9k5 gold badges116 silver badges194 bronze badges
31.9k5 gold badges116 silver badges194 bronze badges
answered 7 hours ago
KaræthonKaræthon
313 bronze badges
313 bronze badges
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
add a comment |
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
$endgroup$
– V2Blast♦
19 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you.
$endgroup$
– Karæthon
12 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
$begingroup$
No problem, glad to help! :)
$endgroup$
– V2Blast♦
11 mins ago
add a comment |
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$begingroup$
Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage?
$endgroup$
– kviiri
8 hours ago