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Does this AnyDice function accurately calculate the number of ogres you make unconcious with three 4th-level castings of Sleep?


Help Calculating Dice Odds for a Unique d6 Dice PoolHow can I execute code depending on a die value in AnyDice?How do I find the highest number rolled in a pool if dice, and the number of times it is rolled using AnyDice?Help Writing an AnyDice Function for a Weird Dice MechanicAnyDice function to use a die as high/low flip and criticalImplementing multiattack in AnyDiceHow do I model the fighter's Great Weapon Fighting fighting style in Anydice?How can I write and compare these formulas using Anydice?Exploding “D9” success-counting pools in anydice?AnyDice formula for rolling 2d20 and dropping farthest value from 10













6












$begingroup$


Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.



I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?



output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]



If not, how would I properly create the function?










share|improve this question











$endgroup$
















    6












    $begingroup$


    Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.



    I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?



    output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]



    If not, how would I properly create the function?










    share|improve this question











    $endgroup$














      6












      6








      6





      $begingroup$


      Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.



      I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?



      output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]



      If not, how would I properly create the function?










      share|improve this question











      $endgroup$




      Say we have three ogres, and want to put them to sleep as a handful of level 7 casters with the sleep spell. We want to know how many of them, on average, will get put to sleep.



      I tried to make an AnyDice function that searches a listing of all of the results of 11d8 for the HP totals that would put to sleep one of the ogres (using the average of 59 HP). Is this AnyDice function accurate to the goal?



      output [count 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88 in 3d(11d8)]



      If not, how would I properly create the function?







      dnd-5e anydice






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 3 hours ago









      V2Blast

      24.6k383155




      24.6k383155










      asked 6 hours ago









      David CoffronDavid Coffron

      37.6k3129264




      37.6k3129264




















          2 Answers
          2






          active

          oldest

          votes


















          11












          $begingroup$

          The function is correct but can be made simpler



          The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.



          However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.



          A simpler and faster way to do the same function would be this anydice function:



          output 3d(11d8 > 58)


          Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.



          Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.






          share|improve this answer











          $endgroup$








          • 2




            $begingroup$
            A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
            $endgroup$
            – Ilmari Karonen
            2 hours ago











          • $begingroup$
            Reason #473 I should really learn how AnyDice actually works..... Good answer.
            $endgroup$
            – Xirema
            3 mins ago


















          7












          $begingroup$

          This function appears to be correct



          The output of the function in Anydice is



          beginarray
          hline
          textNumber & textProbability \ hline
          0 & 68.19% \ hline
          1 & 27.85% \ hline
          2 & 3.79% \ hline
          3 & 0.17% \ hline
          endarray



          Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:



          beginarrayl
          hline
          textNumber & textTrials & textProbability \ hline
          0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
          1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
          2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
          3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
          endarray



          So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.






          share|improve this answer











          $endgroup$












            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            11












            $begingroup$

            The function is correct but can be made simpler



            The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.



            However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.



            A simpler and faster way to do the same function would be this anydice function:



            output 3d(11d8 > 58)


            Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.



            Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.






            share|improve this answer











            $endgroup$








            • 2




              $begingroup$
              A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
              $endgroup$
              – Ilmari Karonen
              2 hours ago











            • $begingroup$
              Reason #473 I should really learn how AnyDice actually works..... Good answer.
              $endgroup$
              – Xirema
              3 mins ago















            11












            $begingroup$

            The function is correct but can be made simpler



            The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.



            However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.



            A simpler and faster way to do the same function would be this anydice function:



            output 3d(11d8 > 58)


            Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.



            Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.






            share|improve this answer











            $endgroup$








            • 2




              $begingroup$
              A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
              $endgroup$
              – Ilmari Karonen
              2 hours ago











            • $begingroup$
              Reason #473 I should really learn how AnyDice actually works..... Good answer.
              $endgroup$
              – Xirema
              3 mins ago













            11












            11








            11





            $begingroup$

            The function is correct but can be made simpler



            The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.



            However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.



            A simpler and faster way to do the same function would be this anydice function:



            output 3d(11d8 > 58)


            Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.



            Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.






            share|improve this answer











            $endgroup$



            The function is correct but can be made simpler



            The function makes sense from a theoretical point of view (you are counting the number of times one of the listed numbers appears when doing 11d8 three times) and from practical corroboration from Xirema's answer.



            However, it seems to be a particularly taxing function for anydice. Indeed, when changing it to 4 times instead of 3, anydice refused to provide results because it exceeded the maximum execution time.



            A simpler and faster way to do the same function would be this anydice function:



            output 3d(11d8 > 58)


            Which is equivalent, also helps show that your initial thoughts were correct but is much faster and you can also use it for bigger values without anydice complaining.



            Naturally this is not as versatile as your specific listing of each possible result but since you only care about results bigger than the creature's HP in this case it may work better.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 5 hours ago

























            answered 5 hours ago









            SdjzSdjz

            13.1k462107




            13.1k462107







            • 2




              $begingroup$
              A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
              $endgroup$
              – Ilmari Karonen
              2 hours ago











            • $begingroup$
              Reason #473 I should really learn how AnyDice actually works..... Good answer.
              $endgroup$
              – Xirema
              3 mins ago












            • 2




              $begingroup$
              A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
              $endgroup$
              – Ilmari Karonen
              2 hours ago











            • $begingroup$
              Reason #473 I should really learn how AnyDice actually works..... Good answer.
              $endgroup$
              – Xirema
              3 mins ago







            2




            2




            $begingroup$
            A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
            $endgroup$
            – Ilmari Karonen
            2 hours ago





            $begingroup$
            A slightly more versatile (but still fast) version would be 3d[count 59..88 in 11d8 + 0]. The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE]. In any case, the major speedup in both solutions comes from moving the counting / comparison inside the 3d(...).
            $endgroup$
            – Ilmari Karonen
            2 hours ago













            $begingroup$
            Reason #473 I should really learn how AnyDice actually works..... Good answer.
            $endgroup$
            – Xirema
            3 mins ago




            $begingroup$
            Reason #473 I should really learn how AnyDice actually works..... Good answer.
            $endgroup$
            – Xirema
            3 mins ago













            7












            $begingroup$

            This function appears to be correct



            The output of the function in Anydice is



            beginarray
            hline
            textNumber & textProbability \ hline
            0 & 68.19% \ hline
            1 & 27.85% \ hline
            2 & 3.79% \ hline
            3 & 0.17% \ hline
            endarray



            Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:



            beginarrayl
            hline
            textNumber & textTrials & textProbability \ hline
            0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
            1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
            2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
            3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
            endarray



            So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.






            share|improve this answer











            $endgroup$

















              7












              $begingroup$

              This function appears to be correct



              The output of the function in Anydice is



              beginarray
              hline
              textNumber & textProbability \ hline
              0 & 68.19% \ hline
              1 & 27.85% \ hline
              2 & 3.79% \ hline
              3 & 0.17% \ hline
              endarray



              Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:



              beginarrayl
              hline
              textNumber & textTrials & textProbability \ hline
              0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
              1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
              2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
              3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
              endarray



              So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.






              share|improve this answer











              $endgroup$















                7












                7








                7





                $begingroup$

                This function appears to be correct



                The output of the function in Anydice is



                beginarray
                hline
                textNumber & textProbability \ hline
                0 & 68.19% \ hline
                1 & 27.85% \ hline
                2 & 3.79% \ hline
                3 & 0.17% \ hline
                endarray



                Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:



                beginarrayl
                hline
                textNumber & textTrials & textProbability \ hline
                0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
                1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
                2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
                3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
                endarray



                So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.






                share|improve this answer











                $endgroup$



                This function appears to be correct



                The output of the function in Anydice is



                beginarray
                hline
                textNumber & textProbability \ hline
                0 & 68.19% \ hline
                1 & 27.85% \ hline
                2 & 3.79% \ hline
                3 & 0.17% \ hline
                endarray



                Using an independent method, I was able to calculate [within an acceptable Margin of Error] identical odds:



                beginarrayl
                hline
                textNumber & textTrials & textProbability \ hline
                0 & 432224441369339628206761607168 (4.322 * 10^29) & 68.1930% \ hline
                1 & 176489190284293025714047057920 (1.765 * 10^29) & 27.8415% \ hline
                2 & 24021805421679657469725081600 (2.402 * 10^28) & 3.7900% \ hline
                3 & 1089863038802389357817856000 (1.090 * 10^27) & 0.1720% \ hline
                endarray



                So because these two methods appear to have identical outcomes, I'm led to conclude that the function you've provided is correct.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 5 hours ago

























                answered 6 hours ago









                XiremaXirema

                21.3k263126




                21.3k263126



























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