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;








6












$begingroup$


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?










share|improve this question











$endgroup$







  • 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

















6












$begingroup$


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?










share|improve this question











$endgroup$







  • 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













6












6








6


1



$begingroup$


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?










share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








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












  • 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










2 Answers
2






active

oldest

votes


















17












$begingroup$

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.






share|improve this answer











$endgroup$




















    1












    $begingroup$

    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".






    share|improve this answer











    $endgroup$












    • $begingroup$
      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













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    2 Answers
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    2 Answers
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    active

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    17












    $begingroup$

    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.






    share|improve this answer











    $endgroup$

















      17












      $begingroup$

      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.






      share|improve this answer











      $endgroup$















        17












        17








        17





        $begingroup$

        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.






        share|improve this answer











        $endgroup$



        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.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        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























            1












            $begingroup$

            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".






            share|improve this answer











            $endgroup$












            • $begingroup$
              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















            1












            $begingroup$

            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".






            share|improve this answer











            $endgroup$












            • $begingroup$
              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













            1












            1








            1





            $begingroup$

            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".






            share|improve this answer











            $endgroup$



            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".







            share|improve this answer














            share|improve this answer



            share|improve this answer








            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
















            • $begingroup$
              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

















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