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












9












$begingroup$


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











$endgroup$

















    9












    $begingroup$


    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











    $endgroup$















      9












      9








      9





      $begingroup$


      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 yesterday









      V2Blast

      24.6k383155




      24.6k383155










      asked yesterday









      David CoffronDavid Coffron

      37.6k3130265




      37.6k3130265






















          2 Answers
          2






          active

          oldest

          votes


















          21












          $begingroup$

          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.



          Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:



          output 3d[count {59..88} in 11d8 + 0]


          (The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])






          share|improve this answer











          $endgroup$









          • 2




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










          • $begingroup$
            @IlmariKaronen Alright, I appreciate the contribution
            $endgroup$
            – Sdjz
            yesterday





















          9












          $begingroup$

          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.



          I've posted my methodology here, which contains all of the individual calculations needed to come up with these trial numbers.






          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            may I ask what your independent method was?
            $endgroup$
            – Carmeister
            yesterday










          • $begingroup$
            @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
            $endgroup$
            – Sdjz
            13 hours ago










          • $begingroup$
            @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
            $endgroup$
            – Xirema
            11 hours ago










          • $begingroup$
            @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
            $endgroup$
            – Xirema
            10 hours ago











          Your Answer





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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

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          active

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          active

          oldest

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          21












          $begingroup$

          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.



          Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:



          output 3d[count {59..88} in 11d8 + 0]


          (The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])






          share|improve this answer











          $endgroup$









          • 2




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










          • $begingroup$
            @IlmariKaronen Alright, I appreciate the contribution
            $endgroup$
            – Sdjz
            yesterday


















          21












          $begingroup$

          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.



          Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:



          output 3d[count {59..88} in 11d8 + 0]


          (The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])






          share|improve this answer











          $endgroup$









          • 2




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










          • $begingroup$
            @IlmariKaronen Alright, I appreciate the contribution
            $endgroup$
            – Sdjz
            yesterday
















          21












          21








          21





          $begingroup$

          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.



          Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:



          output 3d[count {59..88} in 11d8 + 0]


          (The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])






          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.



          Alternatively, as proposed by Ilmari Karonen, you could also use, for some more versatility in the results, this function:



          output 3d[count {59..88} in 11d8 + 0]


          (The + 0 makes AnyDice sum the 11d8 roll before passing it to [count VALUES in SEQUENCE])







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited yesterday

























          answered yesterday









          SdjzSdjz

          13.3k464108




          13.3k464108








          • 2




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










          • $begingroup$
            @IlmariKaronen Alright, I appreciate the contribution
            $endgroup$
            – Sdjz
            yesterday
















          • 2




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










          • $begingroup$
            @IlmariKaronen Alright, I appreciate the contribution
            $endgroup$
            – Sdjz
            yesterday










          2




          2




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




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












          $begingroup$
          @IlmariKaronen Alright, I appreciate the contribution
          $endgroup$
          – Sdjz
          yesterday






          $begingroup$
          @IlmariKaronen Alright, I appreciate the contribution
          $endgroup$
          – Sdjz
          yesterday















          9












          $begingroup$

          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.



          I've posted my methodology here, which contains all of the individual calculations needed to come up with these trial numbers.






          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            may I ask what your independent method was?
            $endgroup$
            – Carmeister
            yesterday










          • $begingroup$
            @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
            $endgroup$
            – Sdjz
            13 hours ago










          • $begingroup$
            @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
            $endgroup$
            – Xirema
            11 hours ago










          • $begingroup$
            @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
            $endgroup$
            – Xirema
            10 hours ago
















          9












          $begingroup$

          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.



          I've posted my methodology here, which contains all of the individual calculations needed to come up with these trial numbers.






          share|improve this answer











          $endgroup$









          • 3




            $begingroup$
            may I ask what your independent method was?
            $endgroup$
            – Carmeister
            yesterday










          • $begingroup$
            @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
            $endgroup$
            – Sdjz
            13 hours ago










          • $begingroup$
            @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
            $endgroup$
            – Xirema
            11 hours ago










          • $begingroup$
            @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
            $endgroup$
            – Xirema
            10 hours ago














          9












          9








          9





          $begingroup$

          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.



          I've posted my methodology here, which contains all of the individual calculations needed to come up with these trial numbers.






          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.



          I've posted my methodology here, which contains all of the individual calculations needed to come up with these trial numbers.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 10 hours ago

























          answered yesterday









          XiremaXirema

          21.4k263126




          21.4k263126








          • 3




            $begingroup$
            may I ask what your independent method was?
            $endgroup$
            – Carmeister
            yesterday










          • $begingroup$
            @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
            $endgroup$
            – Sdjz
            13 hours ago










          • $begingroup$
            @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
            $endgroup$
            – Xirema
            11 hours ago










          • $begingroup$
            @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
            $endgroup$
            – Xirema
            10 hours ago














          • 3




            $begingroup$
            may I ask what your independent method was?
            $endgroup$
            – Carmeister
            yesterday










          • $begingroup$
            @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
            $endgroup$
            – Sdjz
            13 hours ago










          • $begingroup$
            @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
            $endgroup$
            – Xirema
            11 hours ago










          • $begingroup$
            @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
            $endgroup$
            – Xirema
            10 hours ago








          3




          3




          $begingroup$
          may I ask what your independent method was?
          $endgroup$
          – Carmeister
          yesterday




          $begingroup$
          may I ask what your independent method was?
          $endgroup$
          – Carmeister
          yesterday












          $begingroup$
          @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
          $endgroup$
          – Sdjz
          13 hours ago




          $begingroup$
          @Carmeister Based on the numbers, it seems to be a brute force method, simply counting every single possibility for the results of 3 rolls of 11d8 dice. (In other words, 8^33 trials total, consistent with the table).
          $endgroup$
          – Sdjz
          13 hours ago












          $begingroup$
          @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
          $endgroup$
          – Xirema
          11 hours ago




          $begingroup$
          @Carmeister I have a program that autogenerates a whole lot of tables that eventually are composed together to get the final results. I'm working on a function to spit out the decomposed tables to show the step-by-step process. Watch this space.
          $endgroup$
          – Xirema
          11 hours ago












          $begingroup$
          @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
          $endgroup$
          – Xirema
          10 hours ago




          $begingroup$
          @Carmeister Alright, I put a link in the answer to a chatroom I created solely to contain the methodology.
          $endgroup$
          – Xirema
          10 hours ago


















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