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?

Finding overlapping polygons in two shapefiles and deleting them in R?

Would a carnivorous diet be able to support a giant worm?

Is it possible to complete a PhD in CS in 3 years?

What exactly is a "murder hobo"?

Is it okay to roll multiple attacks that all have advantage in one cluster?

No Torah = Revert to Nothingness?

Is it better in terms of durability to remove card+battery or to connect to charger/computer via USB-C?

My previous employer committed a severe violation of the law and is also being sued by me. How do I explain the situation to future employers?

Reference request: quantifier elimination test

Password Hashing Security Using Scrypt & Argon2

Intern not wearing safety equipment; how could I have handled this differently?

Would a Nikon FG 20 film SLR camera take pictures without batteries?

Four ships at the ocean with the same distance

Object's height not a multiple of layer height

When did "&" stop being taught alongside the alphabet?

Was it ever illegal to name a pig "Napoleon" in France?

Why AI became applicable only after Nvidia's chips were available?

Non-Chromatic Orchestral Instruments?

What was the profession 芸者 (female entertainer) called in Russia?

An integral that needs subtitution to be solved.

How to find the positions of replaced elements in a list

Found and corrected a mistake on someone's else paper -- praxis?

What is the problem here?(all integers are irrational proof...i think so)

Need a non-volatile memory IC with near unlimited read/write operations capability



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
    9 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
    9 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
    9 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
    9 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
9 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
9 hours ago










2 Answers
2






active

oldest

votes


















18












$begingroup$

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











$endgroup$




















    2












    $begingroup$

    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











    $endgroup$








    • 1




      $begingroup$
      I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
      $endgroup$
      – V2Blast
      35 mins ago






    • 1




      $begingroup$
      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
      29 mins ago






    • 1




      $begingroup$
      No problem, glad to help! :)
      $endgroup$
      – V2Blast
      28 mins ago










    • $begingroup$
      Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
      $endgroup$
      – KorvinStarmast
      5 mins ago














    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "122"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2frpg.stackexchange.com%2fquestions%2f151289%2fis-it-okay-to-roll-multiple-attacks-that-all-have-advantage-in-one-cluster%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    18












    $begingroup$

    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











    $endgroup$

















      18












      $begingroup$

      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











      $endgroup$















        18












        18








        18





        $begingroup$

        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 6 hours ago

























        answered 7 hours ago









        Ryan ThompsonRyan Thompson

        16k2 gold badges53 silver badges116 bronze badges




        16k2 gold badges53 silver badges116 bronze badges























            2












            $begingroup$

            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











            $endgroup$








            • 1




              $begingroup$
              I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
              $endgroup$
              – V2Blast
              35 mins ago






            • 1




              $begingroup$
              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
              29 mins ago






            • 1




              $begingroup$
              No problem, glad to help! :)
              $endgroup$
              – V2Blast
              28 mins ago










            • $begingroup$
              Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
              $endgroup$
              – KorvinStarmast
              5 mins ago
















            2












            $begingroup$

            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











            $endgroup$








            • 1




              $begingroup$
              I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
              $endgroup$
              – V2Blast
              35 mins ago






            • 1




              $begingroup$
              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
              29 mins ago






            • 1




              $begingroup$
              No problem, glad to help! :)
              $endgroup$
              – V2Blast
              28 mins ago










            • $begingroup$
              Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
              $endgroup$
              – KorvinStarmast
              5 mins ago














            2












            2








            2





            $begingroup$

            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 36 mins ago









            V2Blast

            31.9k5 gold badges116 silver badges194 bronze badges




            31.9k5 gold badges116 silver badges194 bronze badges










            answered 8 hours ago









            KaræthonKaræthon

            413 bronze badges




            413 bronze badges







            • 1




              $begingroup$
              I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
              $endgroup$
              – V2Blast
              35 mins ago






            • 1




              $begingroup$
              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
              29 mins ago






            • 1




              $begingroup$
              No problem, glad to help! :)
              $endgroup$
              – V2Blast
              28 mins ago










            • $begingroup$
              Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
              $endgroup$
              – KorvinStarmast
              5 mins ago













            • 1




              $begingroup$
              I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
              $endgroup$
              – V2Blast
              35 mins ago






            • 1




              $begingroup$
              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
              29 mins ago






            • 1




              $begingroup$
              No problem, glad to help! :)
              $endgroup$
              – V2Blast
              28 mins ago










            • $begingroup$
              Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
              $endgroup$
              – KorvinStarmast
              5 mins ago








            1




            1




            $begingroup$
            I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say.
            $endgroup$
            – V2Blast
            35 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
            35 mins ago




            1




            1




            $begingroup$
            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
            29 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
            29 mins ago




            1




            1




            $begingroup$
            No problem, glad to help! :)
            $endgroup$
            – V2Blast
            28 mins ago




            $begingroup$
            No problem, glad to help! :)
            $endgroup$
            – V2Blast
            28 mins ago












            $begingroup$
            Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
            $endgroup$
            – KorvinStarmast
            5 mins ago





            $begingroup$
            Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification.
            $endgroup$
            – KorvinStarmast
            5 mins ago


















            draft saved

            draft discarded
















































            Thanks for contributing an answer to Role-playing Games Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2frpg.stackexchange.com%2fquestions%2f151289%2fis-it-okay-to-roll-multiple-attacks-that-all-have-advantage-in-one-cluster%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            ParseJSON using SSJSUsing AMPscript with SSJS ActivitiesHow to resubscribe a user in Marketing cloud using SSJS?Pulling Subscriber Status from Lists using SSJSRetrieving Emails using SSJSProblem in updating DE using SSJSUsing SSJS to send single email in Marketing CloudError adding EmailSendDefinition using SSJS

            Кампала Садржај Географија Географија Историја Становништво Привреда Партнерски градови Референце Спољашње везе Мени за навигацију0°11′ СГШ; 32°20′ ИГД / 0.18° СГШ; 32.34° ИГД / 0.18; 32.340°11′ СГШ; 32°20′ ИГД / 0.18° СГШ; 32.34° ИГД / 0.18; 32.34МедијиПодациЗванични веб-сајту

            19. јануар Садржај Догађаји Рођења Смрти Празници и дани сећања Види још Референце Мени за навигацијуу