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How do I compare the result of “1d20+x, with advantage” to “1d20+y, without advantage”, assuming x


Ranged attacks from hiding with Advantage?How does replacing Advantage with a flat +2 affect Champion critical hit damage output?How else can I get Advantage on Death Saving Throws?How do these mass advantage/disadvantage rolls work out statistically?How to avoid the determinism of Passive PerceptionIf I have both advantage and disadvantage, and my target is Elusive, do I have disadvantage on the attack?Adding Precision Strike to an attack with advantageShould a character get advantage on a roll for attacking someone with a spell from behind?What's the benefit for Saving Throws of rerolling vs. advantage?What issues could arise with this Advantage/Disadvantage Variant?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








15












$begingroup$


I roll a d20.



If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?



Assume x < y.










share|improve this question











$endgroup$


















    15












    $begingroup$


    I roll a d20.



    If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?



    Assume x < y.










    share|improve this question











    $endgroup$














      15












      15








      15


      2



      $begingroup$


      I roll a d20.



      If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?



      Assume x < y.










      share|improve this question











      $endgroup$




      I roll a d20.



      If I have a choice between d20+x with advantage, or d20+y without advantage, what choice do I make to maximise the result? What are the values of x and y where the choice changes (if it changes at all)?



      Assume x < y.







      dnd-5e statistics advantage-and-disadvantage






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 4 hours ago









      V2Blast

      29k5105177




      29k5105177










      asked 13 hours ago









      EradashEradash

      32213




      32213




















          4 Answers
          4






          active

          oldest

          votes


















          21












          $begingroup$

          Theory



          The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).



          beginarrayr
          textNatural DC & textAdvantage & textNo Advantage & textDifference & textEq. Flat Modifier \ hline
          textDC 1- & text100.000% & text100.000% & text0.000% & 0 \
          textDC 2 & text99.750% & text95.000% & text4.750% & 0.95 (1)\
          textDC 3 & text99.000% & text90.000% & text9.000% & 1.8 (2) \
          textDC 4 & text97.750% & text85.000% & text12.750% & 2.55 (3) \
          textDC 5 & text96.000% & text80.000% & text16.000% & 3.2 (4) \
          textDC 6 & text93.750% & text75.000% & text18.750% & 3.75 (4) \
          textDC 7 & text91.000% & text70.000% & text21.000% & 4.2 (5) \
          textDC 8 & text87.750% & text65.000% & text22.750% & 4.55 (5) \
          textDC 9 & text84.000% & text60.000% & text24.000% & 4.8 (5) \
          textDC 10 & text79.750% & text55.000% & text24.750% & 4.95 (5) \
          textDC 11 & text75.000% & text50.000% & text25.000% & 5 \
          textDC 12 & text69.750% & text45.000% & text24.750% & 4.95 (5) \
          textDC 13 & text64.000% & text40.000% & text24.000% & 4.8 (5) \
          textDC 14 & text57.750% & text35.000% & text22.750% & 4.55 (5) \
          textDC 15 & text51.000% & text30.000% & text21.000% & 4.2 (5) \
          textDC 16 & text43.750% & text25.000% & text18.750% & 3.75 (4) \
          textDC 17 & text36.000% & text20.000% & text16.000% & 3.2 (4) \
          textDC 18 & text27.750% & text15.000% & text12.750% & 2.55 (3) \
          textDC 19 & text19.000% & text10.000% & text9.000% & 1.8 (2) \
          textDC 20 & text9.750% & text5.000% & text4.750% & 0.95 (1) \
          textDC 21+ & text0.000% & text0.000% & text0.000% & 0 \
          endarray



          A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.



          There are a few casual observations we can make:



          • It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.

          • At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.

          • It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.

          • But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).

          • In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.

          Practice



          So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:



          • Calculate the difference between y and x

          • Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"

          • Look at the Eq. Flat Modifier for that row in the table

          • If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.

          Examples



          • DC19, 1d20+5/Adv vs 1d20+7/NoAdv

            • Difference is 7 - 5 == 2

            • DC19 - 5 is DC14

            • DC14 has a Eq. Flat Modifier of 4.55

            • Therefore, the Advantage roll is better than the non-Advantage roll.


          • DC3, 1d20+1/Adv vs 1d20+2/NoAdv

            • Difference is 2 - 1 == 1

            • DC3 - 1 is DC2

            • DC2 has a Eq. Flat Modifier of 0.95

            • Therefore, the non-advantage roll is better than the Advantage roll


          • DC17, 1d20+9/Adv vs 1d20+14/NoAdv

            • We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...

            • Difference is 14 - 9 == 5

            • DC17 - 9 is DC8

            • DC8 has a Eq. Flat Modifier of 4.55

            • Therefore, the non-advantage roll is better than the Advantage roll


          Attack Rolls



          Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.



          Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.






          share|improve this answer











          $endgroup$












          • $begingroup$
            If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
            $endgroup$
            – Yakk
            10 hours ago


















          17












          $begingroup$

          It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.



          Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.



          enter image description here



          (source: Zero Hit Points)






          share|improve this answer











          $endgroup$












          • $begingroup$
            In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
            $endgroup$
            – illustro
            13 hours ago











          • $begingroup$
            @illustro thanks, my experience is limited to 3(.5)E ...
            $endgroup$
            – Glorfindel
            13 hours ago











          • $begingroup$
            (and even there you didn't need to beat a DC, but it's been a while ...)
            $endgroup$
            – Glorfindel
            13 hours ago










          • $begingroup$
            I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            @Carcer: Have some more AnyDice. :)
            $endgroup$
            – Ilmari Karonen
            10 hours ago


















          13












          $begingroup$

          If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.



          Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.






          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            This answer is great! And shows me we didn't need actual numbers. +1
            $endgroup$
            – NautArch
            13 hours ago










          • $begingroup$
            @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
            $endgroup$
            – Yakk
            11 hours ago







          • 1




            $begingroup$
            @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
            $endgroup$
            – Carcer
            11 hours ago











          • $begingroup$
            @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
            $endgroup$
            – Carcer
            11 hours ago


















          6












          $begingroup$

          First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:



          Screenshot



          In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.



          (Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)




          Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...




          • $y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;


          • $y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;


          • $y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;


          • $y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or


          • $y ge x + 5$.

          (As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)






          share|improve this answer









          $endgroup$












          • $begingroup$
            Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
            $endgroup$
            – Captain Man
            9 hours ago











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          4 Answers
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          4 Answers
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          active

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          active

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          21












          $begingroup$

          Theory



          The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).



          beginarrayr
          textNatural DC & textAdvantage & textNo Advantage & textDifference & textEq. Flat Modifier \ hline
          textDC 1- & text100.000% & text100.000% & text0.000% & 0 \
          textDC 2 & text99.750% & text95.000% & text4.750% & 0.95 (1)\
          textDC 3 & text99.000% & text90.000% & text9.000% & 1.8 (2) \
          textDC 4 & text97.750% & text85.000% & text12.750% & 2.55 (3) \
          textDC 5 & text96.000% & text80.000% & text16.000% & 3.2 (4) \
          textDC 6 & text93.750% & text75.000% & text18.750% & 3.75 (4) \
          textDC 7 & text91.000% & text70.000% & text21.000% & 4.2 (5) \
          textDC 8 & text87.750% & text65.000% & text22.750% & 4.55 (5) \
          textDC 9 & text84.000% & text60.000% & text24.000% & 4.8 (5) \
          textDC 10 & text79.750% & text55.000% & text24.750% & 4.95 (5) \
          textDC 11 & text75.000% & text50.000% & text25.000% & 5 \
          textDC 12 & text69.750% & text45.000% & text24.750% & 4.95 (5) \
          textDC 13 & text64.000% & text40.000% & text24.000% & 4.8 (5) \
          textDC 14 & text57.750% & text35.000% & text22.750% & 4.55 (5) \
          textDC 15 & text51.000% & text30.000% & text21.000% & 4.2 (5) \
          textDC 16 & text43.750% & text25.000% & text18.750% & 3.75 (4) \
          textDC 17 & text36.000% & text20.000% & text16.000% & 3.2 (4) \
          textDC 18 & text27.750% & text15.000% & text12.750% & 2.55 (3) \
          textDC 19 & text19.000% & text10.000% & text9.000% & 1.8 (2) \
          textDC 20 & text9.750% & text5.000% & text4.750% & 0.95 (1) \
          textDC 21+ & text0.000% & text0.000% & text0.000% & 0 \
          endarray



          A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.



          There are a few casual observations we can make:



          • It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.

          • At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.

          • It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.

          • But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).

          • In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.

          Practice



          So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:



          • Calculate the difference between y and x

          • Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"

          • Look at the Eq. Flat Modifier for that row in the table

          • If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.

          Examples



          • DC19, 1d20+5/Adv vs 1d20+7/NoAdv

            • Difference is 7 - 5 == 2

            • DC19 - 5 is DC14

            • DC14 has a Eq. Flat Modifier of 4.55

            • Therefore, the Advantage roll is better than the non-Advantage roll.


          • DC3, 1d20+1/Adv vs 1d20+2/NoAdv

            • Difference is 2 - 1 == 1

            • DC3 - 1 is DC2

            • DC2 has a Eq. Flat Modifier of 0.95

            • Therefore, the non-advantage roll is better than the Advantage roll


          • DC17, 1d20+9/Adv vs 1d20+14/NoAdv

            • We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...

            • Difference is 14 - 9 == 5

            • DC17 - 9 is DC8

            • DC8 has a Eq. Flat Modifier of 4.55

            • Therefore, the non-advantage roll is better than the Advantage roll


          Attack Rolls



          Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.



          Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.






          share|improve this answer











          $endgroup$












          • $begingroup$
            If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
            $endgroup$
            – Yakk
            10 hours ago















          21












          $begingroup$

          Theory



          The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).



          beginarrayr
          textNatural DC & textAdvantage & textNo Advantage & textDifference & textEq. Flat Modifier \ hline
          textDC 1- & text100.000% & text100.000% & text0.000% & 0 \
          textDC 2 & text99.750% & text95.000% & text4.750% & 0.95 (1)\
          textDC 3 & text99.000% & text90.000% & text9.000% & 1.8 (2) \
          textDC 4 & text97.750% & text85.000% & text12.750% & 2.55 (3) \
          textDC 5 & text96.000% & text80.000% & text16.000% & 3.2 (4) \
          textDC 6 & text93.750% & text75.000% & text18.750% & 3.75 (4) \
          textDC 7 & text91.000% & text70.000% & text21.000% & 4.2 (5) \
          textDC 8 & text87.750% & text65.000% & text22.750% & 4.55 (5) \
          textDC 9 & text84.000% & text60.000% & text24.000% & 4.8 (5) \
          textDC 10 & text79.750% & text55.000% & text24.750% & 4.95 (5) \
          textDC 11 & text75.000% & text50.000% & text25.000% & 5 \
          textDC 12 & text69.750% & text45.000% & text24.750% & 4.95 (5) \
          textDC 13 & text64.000% & text40.000% & text24.000% & 4.8 (5) \
          textDC 14 & text57.750% & text35.000% & text22.750% & 4.55 (5) \
          textDC 15 & text51.000% & text30.000% & text21.000% & 4.2 (5) \
          textDC 16 & text43.750% & text25.000% & text18.750% & 3.75 (4) \
          textDC 17 & text36.000% & text20.000% & text16.000% & 3.2 (4) \
          textDC 18 & text27.750% & text15.000% & text12.750% & 2.55 (3) \
          textDC 19 & text19.000% & text10.000% & text9.000% & 1.8 (2) \
          textDC 20 & text9.750% & text5.000% & text4.750% & 0.95 (1) \
          textDC 21+ & text0.000% & text0.000% & text0.000% & 0 \
          endarray



          A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.



          There are a few casual observations we can make:



          • It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.

          • At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.

          • It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.

          • But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).

          • In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.

          Practice



          So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:



          • Calculate the difference between y and x

          • Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"

          • Look at the Eq. Flat Modifier for that row in the table

          • If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.

          Examples



          • DC19, 1d20+5/Adv vs 1d20+7/NoAdv

            • Difference is 7 - 5 == 2

            • DC19 - 5 is DC14

            • DC14 has a Eq. Flat Modifier of 4.55

            • Therefore, the Advantage roll is better than the non-Advantage roll.


          • DC3, 1d20+1/Adv vs 1d20+2/NoAdv

            • Difference is 2 - 1 == 1

            • DC3 - 1 is DC2

            • DC2 has a Eq. Flat Modifier of 0.95

            • Therefore, the non-advantage roll is better than the Advantage roll


          • DC17, 1d20+9/Adv vs 1d20+14/NoAdv

            • We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...

            • Difference is 14 - 9 == 5

            • DC17 - 9 is DC8

            • DC8 has a Eq. Flat Modifier of 4.55

            • Therefore, the non-advantage roll is better than the Advantage roll


          Attack Rolls



          Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.



          Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.






          share|improve this answer











          $endgroup$












          • $begingroup$
            If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
            $endgroup$
            – Yakk
            10 hours ago













          21












          21








          21





          $begingroup$

          Theory



          The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).



          beginarrayr
          textNatural DC & textAdvantage & textNo Advantage & textDifference & textEq. Flat Modifier \ hline
          textDC 1- & text100.000% & text100.000% & text0.000% & 0 \
          textDC 2 & text99.750% & text95.000% & text4.750% & 0.95 (1)\
          textDC 3 & text99.000% & text90.000% & text9.000% & 1.8 (2) \
          textDC 4 & text97.750% & text85.000% & text12.750% & 2.55 (3) \
          textDC 5 & text96.000% & text80.000% & text16.000% & 3.2 (4) \
          textDC 6 & text93.750% & text75.000% & text18.750% & 3.75 (4) \
          textDC 7 & text91.000% & text70.000% & text21.000% & 4.2 (5) \
          textDC 8 & text87.750% & text65.000% & text22.750% & 4.55 (5) \
          textDC 9 & text84.000% & text60.000% & text24.000% & 4.8 (5) \
          textDC 10 & text79.750% & text55.000% & text24.750% & 4.95 (5) \
          textDC 11 & text75.000% & text50.000% & text25.000% & 5 \
          textDC 12 & text69.750% & text45.000% & text24.750% & 4.95 (5) \
          textDC 13 & text64.000% & text40.000% & text24.000% & 4.8 (5) \
          textDC 14 & text57.750% & text35.000% & text22.750% & 4.55 (5) \
          textDC 15 & text51.000% & text30.000% & text21.000% & 4.2 (5) \
          textDC 16 & text43.750% & text25.000% & text18.750% & 3.75 (4) \
          textDC 17 & text36.000% & text20.000% & text16.000% & 3.2 (4) \
          textDC 18 & text27.750% & text15.000% & text12.750% & 2.55 (3) \
          textDC 19 & text19.000% & text10.000% & text9.000% & 1.8 (2) \
          textDC 20 & text9.750% & text5.000% & text4.750% & 0.95 (1) \
          textDC 21+ & text0.000% & text0.000% & text0.000% & 0 \
          endarray



          A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.



          There are a few casual observations we can make:



          • It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.

          • At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.

          • It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.

          • But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).

          • In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.

          Practice



          So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:



          • Calculate the difference between y and x

          • Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"

          • Look at the Eq. Flat Modifier for that row in the table

          • If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.

          Examples



          • DC19, 1d20+5/Adv vs 1d20+7/NoAdv

            • Difference is 7 - 5 == 2

            • DC19 - 5 is DC14

            • DC14 has a Eq. Flat Modifier of 4.55

            • Therefore, the Advantage roll is better than the non-Advantage roll.


          • DC3, 1d20+1/Adv vs 1d20+2/NoAdv

            • Difference is 2 - 1 == 1

            • DC3 - 1 is DC2

            • DC2 has a Eq. Flat Modifier of 0.95

            • Therefore, the non-advantage roll is better than the Advantage roll


          • DC17, 1d20+9/Adv vs 1d20+14/NoAdv

            • We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...

            • Difference is 14 - 9 == 5

            • DC17 - 9 is DC8

            • DC8 has a Eq. Flat Modifier of 4.55

            • Therefore, the non-advantage roll is better than the Advantage roll


          Attack Rolls



          Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.



          Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.






          share|improve this answer











          $endgroup$



          Theory



          The first thing we'll look at is a table that represents the odds of rolling at least a given DC, given a d20 with or without advantage (no modifiers yet).



          beginarrayr
          textNatural DC & textAdvantage & textNo Advantage & textDifference & textEq. Flat Modifier \ hline
          textDC 1- & text100.000% & text100.000% & text0.000% & 0 \
          textDC 2 & text99.750% & text95.000% & text4.750% & 0.95 (1)\
          textDC 3 & text99.000% & text90.000% & text9.000% & 1.8 (2) \
          textDC 4 & text97.750% & text85.000% & text12.750% & 2.55 (3) \
          textDC 5 & text96.000% & text80.000% & text16.000% & 3.2 (4) \
          textDC 6 & text93.750% & text75.000% & text18.750% & 3.75 (4) \
          textDC 7 & text91.000% & text70.000% & text21.000% & 4.2 (5) \
          textDC 8 & text87.750% & text65.000% & text22.750% & 4.55 (5) \
          textDC 9 & text84.000% & text60.000% & text24.000% & 4.8 (5) \
          textDC 10 & text79.750% & text55.000% & text24.750% & 4.95 (5) \
          textDC 11 & text75.000% & text50.000% & text25.000% & 5 \
          textDC 12 & text69.750% & text45.000% & text24.750% & 4.95 (5) \
          textDC 13 & text64.000% & text40.000% & text24.000% & 4.8 (5) \
          textDC 14 & text57.750% & text35.000% & text22.750% & 4.55 (5) \
          textDC 15 & text51.000% & text30.000% & text21.000% & 4.2 (5) \
          textDC 16 & text43.750% & text25.000% & text18.750% & 3.75 (4) \
          textDC 17 & text36.000% & text20.000% & text16.000% & 3.2 (4) \
          textDC 18 & text27.750% & text15.000% & text12.750% & 2.55 (3) \
          textDC 19 & text19.000% & text10.000% & text9.000% & 1.8 (2) \
          textDC 20 & text9.750% & text5.000% & text4.750% & 0.95 (1) \
          textDC 21+ & text0.000% & text0.000% & text0.000% & 0 \
          endarray



          A +1 to a non-advantage roll will always improve the odds of rolling a given number by exactly 5 percentage points. Conversely, a +1 to an Advantage roll will increase your odds by an amount equal to moving up one row on that table: a DC7 check made with +1 is equivalent to a DC6 check made with +0. A DC20 check made with advantage and a +1 modifier is equivalent to a DC19 check made with +0, which constitutes a 9.250 percentage point improvement.



          There are a few casual observations we can make:



          • It's not possible to roll a natural d20 lower than a 1, so if if the advantage check requires a natural 1, then there's no benefit to gaining any modifier (or advantage, for that matter): it's a check that is impossible to fail.

          • At DC 2, gaining advantage increases the odds of success by 4.750% (to 99.750%) but gaining a +1 modifier increases the odds of success by 5% (to 100%). So intuitively, If we're comparing 1d20+x/ADV vs 1d20+x+1/NoADV, and the natural number we need to hit is a 2 (for the advantage check), then the +1 modifier is better.

          • It's the same deal at DC20: Gaining Advantage will improve from 5% to 9.750%, but gaining +1 will improve from 5% to 10%. Again, the +1 modifier is better.

          • But the differences get more drastic as we move closer to the mean of the roll: At DC3, advantage improves the odds by 9% (90%→99%) but a +1 modifier only improves the odds by 5% (90%→95%), so here, Advantage is better than a +1 modifier; but it's NOT better than a +2 modifier (90%→100%).

          • In the table, I've added the "Eq. Flat Modifier" column: this describes, for each row, how much of a modifier you would need for the benefit from that modifier to be equivalent to the benefit provided by Advantage. Since 5e doesn't have "half" modifiers or fractional DCs, I've included the (rounded up) proper modifier in parenthesis next to it. In each row, if the modifier difference between the Advantage and Non-Advantage rolls is greater than that number, then the modifier is better; if it's not, then the Advantage roll is better.

          Practice



          So back to the original question: Given two rolls, 1d20+x/Adv, and 1d20+y/NoAdv, which is better? Well, as established, it depends on the DC of the check, but to get the results from this table:



          • Calculate the difference between y and x

          • Subtract x (the modifier for the Advantage roll) from the DC to get the "Natural DC"

          • Look at the Eq. Flat Modifier for that row in the table

          • If the difference between y and x is greater than that value, then you should prefer the 1d20+y/NoAdv roll. If not, then you should prefer the 1d20+x/Adv roll.

          Examples



          • DC19, 1d20+5/Adv vs 1d20+7/NoAdv

            • Difference is 7 - 5 == 2

            • DC19 - 5 is DC14

            • DC14 has a Eq. Flat Modifier of 4.55

            • Therefore, the Advantage roll is better than the non-Advantage roll.


          • DC3, 1d20+1/Adv vs 1d20+2/NoAdv

            • Difference is 2 - 1 == 1

            • DC3 - 1 is DC2

            • DC2 has a Eq. Flat Modifier of 0.95

            • Therefore, the non-advantage roll is better than the Advantage roll


          • DC17, 1d20+9/Adv vs 1d20+14/NoAdv

            • We could skip the steps: none of the rows have a Eq. Flat Modifier greater than 5, meaning the +5 modifier will always be better than (or equivalent to) the Advantage improvement. Nonetheless...

            • Difference is 14 - 9 == 5

            • DC17 - 9 is DC8

            • DC8 has a Eq. Flat Modifier of 4.55

            • Therefore, the non-advantage roll is better than the Advantage roll


          Attack Rolls



          Attack Rolls are a little weird, because you no longer simply care about passing the check; you also care what the natural number was because of Critical Hits and Misses.



          Most of the math still checks out: if all you care about is hitting/missing, then the table above can be used, since the scenarios where a Natural 2 hits and a Natural 19 misses are pretty rare in 5e. If, however, you instead care more about the Crits/Auto-Misses, then you should introduce a "subjectivity factor", which you can define however you like: is it important to you that you get a critical hit (or avoid a critical miss)? Then always go Advantage. If not, then use the table above. I generally stick to the table personally, but "clutch factor" is one of those hazy things that can't be objectively defined, so you'll need to make that call for yourself.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 8 hours ago

























          answered 12 hours ago









          XiremaXirema

          26.1k373152




          26.1k373152











          • $begingroup$
            If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
            $endgroup$
            – Yakk
            10 hours ago
















          • $begingroup$
            If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
            $endgroup$
            – Yakk
            10 hours ago















          $begingroup$
          If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
          $endgroup$
          – Yakk
          10 hours ago




          $begingroup$
          If you provide a distribution of DCs you expect to meet plus a factor of "how much do I care if I pass" (basically a weight), you can work out how much advantage is "worth" in flat +X bonus. (With "bounded accuracy", +X is going to be linear)
          $endgroup$
          – Yakk
          10 hours ago













          17












          $begingroup$

          It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.



          Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.



          enter image description here



          (source: Zero Hit Points)






          share|improve this answer











          $endgroup$












          • $begingroup$
            In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
            $endgroup$
            – illustro
            13 hours ago











          • $begingroup$
            @illustro thanks, my experience is limited to 3(.5)E ...
            $endgroup$
            – Glorfindel
            13 hours ago











          • $begingroup$
            (and even there you didn't need to beat a DC, but it's been a while ...)
            $endgroup$
            – Glorfindel
            13 hours ago










          • $begingroup$
            I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            @Carcer: Have some more AnyDice. :)
            $endgroup$
            – Ilmari Karonen
            10 hours ago















          17












          $begingroup$

          It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.



          Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.



          enter image description here



          (source: Zero Hit Points)






          share|improve this answer











          $endgroup$












          • $begingroup$
            In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
            $endgroup$
            – illustro
            13 hours ago











          • $begingroup$
            @illustro thanks, my experience is limited to 3(.5)E ...
            $endgroup$
            – Glorfindel
            13 hours ago











          • $begingroup$
            (and even there you didn't need to beat a DC, but it's been a while ...)
            $endgroup$
            – Glorfindel
            13 hours ago










          • $begingroup$
            I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            @Carcer: Have some more AnyDice. :)
            $endgroup$
            – Ilmari Karonen
            10 hours ago













          17












          17








          17





          $begingroup$

          It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.



          Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.



          enter image description here



          (source: Zero Hit Points)






          share|improve this answer











          $endgroup$



          It will depend on what you're trying to achieve. For example, if you need to reach DC 25, and $x = 4$ and $y = 5$, the advantage on the roll with $x$ doesn't matter; you'll never roll higher than 24. With the +5, you'll at least have a 5% chance.



          Here (scroll down to "Advantage versus Simple Bonuses") is a table which shows which bonus (difference between x and y) corresponds to having advantage or not.



          enter image description here



          (source: Zero Hit Points)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 13 hours ago

























          answered 13 hours ago









          GlorfindelGlorfindel

          6821614




          6821614











          • $begingroup$
            In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
            $endgroup$
            – illustro
            13 hours ago











          • $begingroup$
            @illustro thanks, my experience is limited to 3(.5)E ...
            $endgroup$
            – Glorfindel
            13 hours ago











          • $begingroup$
            (and even there you didn't need to beat a DC, but it's been a while ...)
            $endgroup$
            – Glorfindel
            13 hours ago










          • $begingroup$
            I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            @Carcer: Have some more AnyDice. :)
            $endgroup$
            – Ilmari Karonen
            10 hours ago
















          • $begingroup$
            In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
            $endgroup$
            – illustro
            13 hours ago











          • $begingroup$
            @illustro thanks, my experience is limited to 3(.5)E ...
            $endgroup$
            – Glorfindel
            13 hours ago











          • $begingroup$
            (and even there you didn't need to beat a DC, but it's been a while ...)
            $endgroup$
            – Glorfindel
            13 hours ago










          • $begingroup$
            I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            @Carcer: Have some more AnyDice. :)
            $endgroup$
            – Ilmari Karonen
            10 hours ago















          $begingroup$
          In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
          $endgroup$
          – illustro
          13 hours ago





          $begingroup$
          In 5e you don't need to "beat" a DC, you need to match it to overcome it. It would be clearer if your first sentence said "If the DC is 25, and x = 4 and y=5 then you will never pass the DC with x = 4, but you have a chance of passing the DC (5%) with y = 5".
          $endgroup$
          – illustro
          13 hours ago













          $begingroup$
          @illustro thanks, my experience is limited to 3(.5)E ...
          $endgroup$
          – Glorfindel
          13 hours ago





          $begingroup$
          @illustro thanks, my experience is limited to 3(.5)E ...
          $endgroup$
          – Glorfindel
          13 hours ago













          $begingroup$
          (and even there you didn't need to beat a DC, but it's been a while ...)
          $endgroup$
          – Glorfindel
          13 hours ago




          $begingroup$
          (and even there you didn't need to beat a DC, but it's been a while ...)
          $endgroup$
          – Glorfindel
          13 hours ago












          $begingroup$
          I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
          $endgroup$
          – Carcer
          12 hours ago




          $begingroup$
          I feel this analysis is improved by MORE ANYDICE. For example, with +2/+5 DC 10 advantage wins, 88% to 80%; but at +2/+5 DC 20, the normal roll is better, with odds of 30% vs. 28%. You can tinker with the variables in the script to investigate the results with different values!
          $endgroup$
          – Carcer
          12 hours ago




          1




          1




          $begingroup$
          @Carcer: Have some more AnyDice. :)
          $endgroup$
          – Ilmari Karonen
          10 hours ago




          $begingroup$
          @Carcer: Have some more AnyDice. :)
          $endgroup$
          – Ilmari Karonen
          10 hours ago











          13












          $begingroup$

          If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.



          Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.






          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            This answer is great! And shows me we didn't need actual numbers. +1
            $endgroup$
            – NautArch
            13 hours ago










          • $begingroup$
            @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
            $endgroup$
            – Yakk
            11 hours ago







          • 1




            $begingroup$
            @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
            $endgroup$
            – Carcer
            11 hours ago











          • $begingroup$
            @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
            $endgroup$
            – Carcer
            11 hours ago















          13












          $begingroup$

          If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.



          Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.






          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            This answer is great! And shows me we didn't need actual numbers. +1
            $endgroup$
            – NautArch
            13 hours ago










          • $begingroup$
            @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
            $endgroup$
            – Yakk
            11 hours ago







          • 1




            $begingroup$
            @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
            $endgroup$
            – Carcer
            11 hours ago











          • $begingroup$
            @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
            $endgroup$
            – Carcer
            11 hours ago













          13












          13








          13





          $begingroup$

          If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.



          Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.






          share|improve this answer











          $endgroup$



          If you only care about maximising the expected result, as opposed to your odds of hitting a specific target number (for instance, you might be making a contested roll against someone else, like in a grapple, or otherwise don't know the target number ahead of time) this is a pretty simple comparison. Having advantage on a d20 roll increases the expected result from an average of 10.5 to 13.82 (illustrated by this anydice program); that's a benefit of +3.32.



          Therefore, in order for a roll without advantage to have a higher expected result than a roll with advantage, the modifier on the normal roll needs to be four or more points better than the modifier on the advantaged roll. +3 with advantage is worse than +7 normally, and so on.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 11 hours ago

























          answered 13 hours ago









          CarcerCarcer

          28.1k583148




          28.1k583148







          • 1




            $begingroup$
            This answer is great! And shows me we didn't need actual numbers. +1
            $endgroup$
            – NautArch
            13 hours ago










          • $begingroup$
            @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
            $endgroup$
            – Yakk
            11 hours ago







          • 1




            $begingroup$
            @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
            $endgroup$
            – Carcer
            11 hours ago











          • $begingroup$
            @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
            $endgroup$
            – Carcer
            11 hours ago












          • 1




            $begingroup$
            This answer is great! And shows me we didn't need actual numbers. +1
            $endgroup$
            – NautArch
            13 hours ago










          • $begingroup$
            @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
            $endgroup$
            – Carcer
            12 hours ago






          • 1




            $begingroup$
            The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
            $endgroup$
            – Yakk
            11 hours ago







          • 1




            $begingroup$
            @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
            $endgroup$
            – Carcer
            11 hours ago











          • $begingroup$
            @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
            $endgroup$
            – Carcer
            11 hours ago







          1




          1




          $begingroup$
          This answer is great! And shows me we didn't need actual numbers. +1
          $endgroup$
          – NautArch
          13 hours ago




          $begingroup$
          This answer is great! And shows me we didn't need actual numbers. +1
          $endgroup$
          – NautArch
          13 hours ago












          $begingroup$
          @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
          $endgroup$
          – Carcer
          12 hours ago




          $begingroup$
          @NautArch It is the kind of question where additional context does inform how you should compare your options, but both approaches are useful in different circumstances.
          $endgroup$
          – Carcer
          12 hours ago




          1




          1




          $begingroup$
          The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
          $endgroup$
          – Yakk
          11 hours ago





          $begingroup$
          The difference in expected value between your rolls does not determine your odds of beating someone else's roll. As an example, suppose I roll 0 25% of the time and 20 75% of the time. My expected value is 15. My enemy rolls 19 100% of the time. Their expected value is 19. I beat them 1/4 of the time. Another enemy rolls 2 100% of the time. I beat them 1/4 of the time as well. In one case, my "average" roll was -4 and in the other it was +13 compared to them.
          $endgroup$
          – Yakk
          11 hours ago





          1




          1




          $begingroup$
          @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
          $endgroup$
          – Carcer
          11 hours ago





          $begingroup$
          @Yakk awkward terminology. It's a comparison of the expected outcomes of your own two different options, not directly against whatever your opposition is. (That said, the question is about 5e D&D, so the distributions you propose are not something you would see in practice.)
          $endgroup$
          – Carcer
          11 hours ago













          $begingroup$
          @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
          $endgroup$
          – Carcer
          11 hours ago




          $begingroup$
          @Yakk rephrased to describe expected results rather than the more ambiguous "beat". Is that more palatable?
          $endgroup$
          – Carcer
          11 hours ago











          6












          $begingroup$

          First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:



          Screenshot



          In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.



          (Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)




          Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...




          • $y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;


          • $y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;


          • $y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;


          • $y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or


          • $y ge x + 5$.

          (As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)






          share|improve this answer









          $endgroup$












          • $begingroup$
            Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
            $endgroup$
            – Captain Man
            9 hours ago















          6












          $begingroup$

          First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:



          Screenshot



          In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.



          (Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)




          Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...




          • $y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;


          • $y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;


          • $y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;


          • $y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or


          • $y ge x + 5$.

          (As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)






          share|improve this answer









          $endgroup$












          • $begingroup$
            Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
            $endgroup$
            – Captain Man
            9 hours ago













          6












          6








          6





          $begingroup$

          First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:



          Screenshot



          In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.



          (Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)




          Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...




          • $y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;


          • $y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;


          • $y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;


          • $y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or


          • $y ge x + 5$.

          (As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)






          share|improve this answer









          $endgroup$



          First, subtract $x$ both from $y$ and from the target number you're rolling against. Then look at this graph:



          Screenshot



          In the graph, find the position on the horizontal axis that matches the target number (minus $x$) that you're trying to meet or exceed, and the colored line that matches the extra bonus $y-x$ to the roll without advantage. If that colored line is higher that the curved black line at that position on the horizontal axis, you should choose the higher bonus over advantage.



          (Specifically, the various lines in the graph show the probability of meeting or exceeding a given target number with various rolls: the black curved line is for d20 with advantage but no bonus, while the five different colored straight lines on top of it are for d20+1 to d20+5.)




          Or, to summarize, you should choose a plain $+y$ bonus over advantage $+x$ when...




          • $y = x + 1$ and the target number is at most $x+2$ or at least $x+20$;


          • $y = x + 2$ and the target number is at most $x+3$ or at least $x+19$;


          • $y = x + 3$ and the target number is at most $x+4$ or at least $x+18$;


          • $y = x + 4$ and the target number is at most $x+6$ or at least $x+16$; or


          • $y ge x + 5$.

          (As noted by Xirema, things can change a little if you're e.g. making an attack roll and care about crits. Rolling with advantage has a 9.75% chance of giving you a natural 20, and only a 0.25% chance of a natural 1, whereas with a normal d20 roll both 1s and 20s show up 5% of the time each. Whether those differences in crit odds are worth trading for a somewhat worse chance to hit depends both on the target DC and on how much you value crits.)







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 9 hours ago









          Ilmari KaronenIlmari Karonen

          11.5k33448




          11.5k33448











          • $begingroup$
            Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
            $endgroup$
            – Captain Man
            9 hours ago
















          • $begingroup$
            Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
            $endgroup$
            – Captain Man
            9 hours ago















          $begingroup$
          Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
          $endgroup$
          – Captain Man
          9 hours ago




          $begingroup$
          Here is what I think to be a simpler graph OP can plug numbers into. Merely change X:1 to what X you want and the same for Y:2. E.g., you want 5 and 10, X:5 and Y:10.
          $endgroup$
          – Captain Man
          9 hours ago

















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