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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













3












$begingroup$


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











$endgroup$











  • $begingroup$
    @Sdjz added. Good thought
    $endgroup$
    – David Coffron
    2 hours ago















3












$begingroup$


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











$endgroup$











  • $begingroup$
    @Sdjz added. Good thought
    $endgroup$
    – David Coffron
    2 hours ago













3












3








3





$begingroup$


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 37 mins ago









V2Blast

24.6k383155




24.6k383155










asked 3 hours ago









David CoffronDavid Coffron

37.6k3129263




37.6k3129263











  • $begingroup$
    @Sdjz added. Good thought
    $endgroup$
    – David Coffron
    2 hours ago
















  • $begingroup$
    @Sdjz added. Good thought
    $endgroup$
    – David Coffron
    2 hours ago















$begingroup$
@Sdjz added. Good thought
$endgroup$
– David Coffron
2 hours ago




$begingroup$
@Sdjz added. Good thought
$endgroup$
– David Coffron
2 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

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











$endgroup$




















    4












    $begingroup$

    This function appears to be correct



    The output of the function in Anydice is



    beginarrayl
    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











    $endgroup$












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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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











      $endgroup$

















        3












        $begingroup$

        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











        $endgroup$















          3












          3








          3





          $begingroup$

          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 1 hour ago

























          answered 2 hours ago









          SdjzSdjz

          13k462106




          13k462106























              4












              $begingroup$

              This function appears to be correct



              The output of the function in Anydice is



              beginarrayl
              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











              $endgroup$

















                4












                $begingroup$

                This function appears to be correct



                The output of the function in Anydice is



                beginarrayl
                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











                $endgroup$















                  4












                  4








                  4





                  $begingroup$

                  This function appears to be correct



                  The output of the function in Anydice is



                  beginarrayl
                  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



                  beginarrayl
                  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 2 hours ago

























                  answered 2 hours ago









                  XiremaXirema

                  21.3k263126




                  21.3k263126



























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