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












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?






dnd-5e anydice







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?






    dnd-5e anydice







    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?






      dnd-5e anydice







      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




      dnd-5e anydice






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 4 hours ago









      V2Blast

      24.6k383155




      24.6k383155










      asked 7 hours ago









      David CoffronDavid Coffron

      37.6k3129264




      37.6k3129264






















          2 Answers
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          active

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          12












          $begingroup$

          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
            51 mins ago










          • $begingroup$
            @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
            $endgroup$
            – Sdjz
            13 mins ago



















          7












          $begingroup$

          This function appears to be correct



          The output of the function in Anydice is



          begin{array}{|l|l|}
          hline
          text{Number} & text{Probability} \ hline
          0 & 68.19% \ hline
          1 & 27.85% \ hline
          2 & 3.79% \ hline
          3 & 0.17% \ hline
          end{array}



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



          begin{array}{|l|l|l|}
          hline
          text{Number} & text{Trials} & text{Probability} \ 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
          end{array}



          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











          $endgroup$













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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            12












            $begingroup$

            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
              51 mins ago










            • $begingroup$
              @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
              $endgroup$
              – Sdjz
              13 mins ago
















            12












            $begingroup$

            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
              51 mins ago










            • $begingroup$
              @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
              $endgroup$
              – Sdjz
              13 mins ago














            12












            12








            12





            $begingroup$

            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 6 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
              51 mins ago










            • $begingroup$
              @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
              $endgroup$
              – Sdjz
              13 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
              51 mins ago










            • $begingroup$
              @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
              $endgroup$
              – Sdjz
              13 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
            51 mins ago




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












            $begingroup$
            @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
            $endgroup$
            – Sdjz
            13 mins ago




            $begingroup$
            @IlmariKaronen I think that is an improvement worth an answer if you want to have that. I could also include this method in mine if you prefer.
            $endgroup$
            – Sdjz
            13 mins ago













            7












            $begingroup$

            This function appears to be correct



            The output of the function in Anydice is



            begin{array}{|l|l|}
            hline
            text{Number} & text{Probability} \ hline
            0 & 68.19% \ hline
            1 & 27.85% \ hline
            2 & 3.79% \ hline
            3 & 0.17% \ hline
            end{array}



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



            begin{array}{|l|l|l|}
            hline
            text{Number} & text{Trials} & text{Probability} \ 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
            end{array}



            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



              begin{array}{|l|l|}
              hline
              text{Number} & text{Probability} \ hline
              0 & 68.19% \ hline
              1 & 27.85% \ hline
              2 & 3.79% \ hline
              3 & 0.17% \ hline
              end{array}



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



              begin{array}{|l|l|l|}
              hline
              text{Number} & text{Trials} & text{Probability} \ 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
              end{array}



              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



                begin{array}{|l|l|}
                hline
                text{Number} & text{Probability} \ hline
                0 & 68.19% \ hline
                1 & 27.85% \ hline
                2 & 3.79% \ hline
                3 & 0.17% \ hline
                end{array}



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



                begin{array}{|l|l|l|}
                hline
                text{Number} & text{Trials} & text{Probability} \ 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
                end{array}



                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



                begin{array}{|l|l|}
                hline
                text{Number} & text{Probability} \ hline
                0 & 68.19% \ hline
                1 & 27.85% \ hline
                2 & 3.79% \ hline
                3 & 0.17% \ hline
                end{array}



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



                begin{array}{|l|l|l|}
                hline
                text{Number} & text{Trials} & text{Probability} \ 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
                end{array}



                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 6 hours ago

























                answered 6 hours ago









                XiremaXirema

                21.3k263126




                21.3k263126






























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