What is the probability of the Maze Engine from Out of the Abyss activating this effect?











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In Out of the Abyss, there is something called the Maze Engine which, once activated, requires the DM to roll on a d100 table for random possible effects (pp. 186-188).



One of these entries is:




81—00: The engine emits a flash of violet-white light. All extraplanar creatures within 100 miles of the engine instantly return to their native planes of existence.




The Maze Engine is slowly sliding into magma below and will take 12 rounds to sink into the magma and be destroyed. One of these random effects happens per turn. So if I'm rolling on the d100 table with the hope of landing on the 81-00 option, I have 12 rolls before the Maze Engine is destroyed.



I suck as statistics, so can someone who understand statistics better than I tell me: what is the probability of rolling a 81-00 at least once on the d100 table over 12 attempts?



The reason I want this number is so that a modron (tridrone) NPC can tell the PCs exactly how likely it is to send them home with a precise-sounding decimal number (this is how I'm running the modron's motivation for helping us find the "Orderer" a.k.a. Maze Engine; it's so we can help them get back to Mechanus). And yes, as DM I can just fudge the dice roll so that it lands on this entry; this is more about wanting the exact probability for the purposes of the modron's in-game dialog.










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  • Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
    – David Coffron
    1 hour ago










  • @DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
    – NathanS
    58 mins ago

















up vote
7
down vote

favorite












In Out of the Abyss, there is something called the Maze Engine which, once activated, requires the DM to roll on a d100 table for random possible effects (pp. 186-188).



One of these entries is:




81—00: The engine emits a flash of violet-white light. All extraplanar creatures within 100 miles of the engine instantly return to their native planes of existence.




The Maze Engine is slowly sliding into magma below and will take 12 rounds to sink into the magma and be destroyed. One of these random effects happens per turn. So if I'm rolling on the d100 table with the hope of landing on the 81-00 option, I have 12 rolls before the Maze Engine is destroyed.



I suck as statistics, so can someone who understand statistics better than I tell me: what is the probability of rolling a 81-00 at least once on the d100 table over 12 attempts?



The reason I want this number is so that a modron (tridrone) NPC can tell the PCs exactly how likely it is to send them home with a precise-sounding decimal number (this is how I'm running the modron's motivation for helping us find the "Orderer" a.k.a. Maze Engine; it's so we can help them get back to Mechanus). And yes, as DM I can just fudge the dice roll so that it lands on this entry; this is more about wanting the exact probability for the purposes of the modron's in-game dialog.










share|improve this question
























  • Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
    – David Coffron
    1 hour ago










  • @DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
    – NathanS
    58 mins ago















up vote
7
down vote

favorite









up vote
7
down vote

favorite











In Out of the Abyss, there is something called the Maze Engine which, once activated, requires the DM to roll on a d100 table for random possible effects (pp. 186-188).



One of these entries is:




81—00: The engine emits a flash of violet-white light. All extraplanar creatures within 100 miles of the engine instantly return to their native planes of existence.




The Maze Engine is slowly sliding into magma below and will take 12 rounds to sink into the magma and be destroyed. One of these random effects happens per turn. So if I'm rolling on the d100 table with the hope of landing on the 81-00 option, I have 12 rolls before the Maze Engine is destroyed.



I suck as statistics, so can someone who understand statistics better than I tell me: what is the probability of rolling a 81-00 at least once on the d100 table over 12 attempts?



The reason I want this number is so that a modron (tridrone) NPC can tell the PCs exactly how likely it is to send them home with a precise-sounding decimal number (this is how I'm running the modron's motivation for helping us find the "Orderer" a.k.a. Maze Engine; it's so we can help them get back to Mechanus). And yes, as DM I can just fudge the dice roll so that it lands on this entry; this is more about wanting the exact probability for the purposes of the modron's in-game dialog.










share|improve this question















In Out of the Abyss, there is something called the Maze Engine which, once activated, requires the DM to roll on a d100 table for random possible effects (pp. 186-188).



One of these entries is:




81—00: The engine emits a flash of violet-white light. All extraplanar creatures within 100 miles of the engine instantly return to their native planes of existence.




The Maze Engine is slowly sliding into magma below and will take 12 rounds to sink into the magma and be destroyed. One of these random effects happens per turn. So if I'm rolling on the d100 table with the hope of landing on the 81-00 option, I have 12 rolls before the Maze Engine is destroyed.



I suck as statistics, so can someone who understand statistics better than I tell me: what is the probability of rolling a 81-00 at least once on the d100 table over 12 attempts?



The reason I want this number is so that a modron (tridrone) NPC can tell the PCs exactly how likely it is to send them home with a precise-sounding decimal number (this is how I'm running the modron's motivation for helping us find the "Orderer" a.k.a. Maze Engine; it's so we can help them get back to Mechanus). And yes, as DM I can just fudge the dice roll so that it lands on this entry; this is more about wanting the exact probability for the purposes of the modron's in-game dialog.







dnd-5e published-adventures statistics out-of-the-abyss






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edited 19 mins ago









Sdjz

10.4k34993




10.4k34993










asked 3 hours ago









NathanS

22.1k6102238




22.1k6102238












  • Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
    – David Coffron
    1 hour ago










  • @DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
    – NathanS
    58 mins ago




















  • Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
    – David Coffron
    1 hour ago










  • @DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
    – NathanS
    58 mins ago


















Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago




Is the tridrone planning to account for the chance that the other results of the maze engine could affect the chance of them lasting ling enough to get sent home?
– David Coffron
1 hour ago












@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago






@DavidCoffron No, the tridrone is assuming that they can survive the other various effects like the one that tries to disintegrate them or throws a green slaad at them, etc.
– NathanS
58 mins ago












2 Answers
2






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up vote
13
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The probability of not rolling 81–100 on a single attempt is 0.8 (80%), on 12 attempts the chance of never rolling it is 0.8^12 = 0.068719476736. So the chance of rolling 81–100 at least once is 1 − 0.068719476736 = 0.931280523264, or 93.1280523264% – as precise as it can get here.






share|improve this answer






























    up vote
    10
    down vote













    The probability of an event is 1 minus the probability of the inverse of the event. Therefore, the probability that you do not roll an 81-100 is 1 minus the probability that you do roll an 81-100 in those turns.



    Since the dice rolls are independent this is



    p(X < 81)^12 = (0.80)^12
    ~= 0.069


    And therefore the probability that the dice roll was an 81 or higher on any of the rolls is 1 - 0.069 which is approximately 93%, or exactly 93.1280523264% to 10 d.p. (derived from 227363409/244140625).






    share|improve this answer























    • I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
      – mklingen
      1 hour ago










    • It surprises me that 1-(4/5)^12 results in an even denominator
      – Yakk
      1 hour ago












    • My math was wrong. Fixed the rational.
      – mklingen
      1 hour ago











    Your Answer





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    2 Answers
    2






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    2 Answers
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    up vote
    13
    down vote













    The probability of not rolling 81–100 on a single attempt is 0.8 (80%), on 12 attempts the chance of never rolling it is 0.8^12 = 0.068719476736. So the chance of rolling 81–100 at least once is 1 − 0.068719476736 = 0.931280523264, or 93.1280523264% – as precise as it can get here.






    share|improve this answer



























      up vote
      13
      down vote













      The probability of not rolling 81–100 on a single attempt is 0.8 (80%), on 12 attempts the chance of never rolling it is 0.8^12 = 0.068719476736. So the chance of rolling 81–100 at least once is 1 − 0.068719476736 = 0.931280523264, or 93.1280523264% – as precise as it can get here.






      share|improve this answer

























        up vote
        13
        down vote










        up vote
        13
        down vote









        The probability of not rolling 81–100 on a single attempt is 0.8 (80%), on 12 attempts the chance of never rolling it is 0.8^12 = 0.068719476736. So the chance of rolling 81–100 at least once is 1 − 0.068719476736 = 0.931280523264, or 93.1280523264% – as precise as it can get here.






        share|improve this answer














        The probability of not rolling 81–100 on a single attempt is 0.8 (80%), on 12 attempts the chance of never rolling it is 0.8^12 = 0.068719476736. So the chance of rolling 81–100 at least once is 1 − 0.068719476736 = 0.931280523264, or 93.1280523264% – as precise as it can get here.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 3 hours ago

























        answered 3 hours ago









        black_fm

        1,90211321




        1,90211321
























            up vote
            10
            down vote













            The probability of an event is 1 minus the probability of the inverse of the event. Therefore, the probability that you do not roll an 81-100 is 1 minus the probability that you do roll an 81-100 in those turns.



            Since the dice rolls are independent this is



            p(X < 81)^12 = (0.80)^12
            ~= 0.069


            And therefore the probability that the dice roll was an 81 or higher on any of the rolls is 1 - 0.069 which is approximately 93%, or exactly 93.1280523264% to 10 d.p. (derived from 227363409/244140625).






            share|improve this answer























            • I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
              – mklingen
              1 hour ago










            • It surprises me that 1-(4/5)^12 results in an even denominator
              – Yakk
              1 hour ago












            • My math was wrong. Fixed the rational.
              – mklingen
              1 hour ago















            up vote
            10
            down vote













            The probability of an event is 1 minus the probability of the inverse of the event. Therefore, the probability that you do not roll an 81-100 is 1 minus the probability that you do roll an 81-100 in those turns.



            Since the dice rolls are independent this is



            p(X < 81)^12 = (0.80)^12
            ~= 0.069


            And therefore the probability that the dice roll was an 81 or higher on any of the rolls is 1 - 0.069 which is approximately 93%, or exactly 93.1280523264% to 10 d.p. (derived from 227363409/244140625).






            share|improve this answer























            • I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
              – mklingen
              1 hour ago










            • It surprises me that 1-(4/5)^12 results in an even denominator
              – Yakk
              1 hour ago












            • My math was wrong. Fixed the rational.
              – mklingen
              1 hour ago













            up vote
            10
            down vote










            up vote
            10
            down vote









            The probability of an event is 1 minus the probability of the inverse of the event. Therefore, the probability that you do not roll an 81-100 is 1 minus the probability that you do roll an 81-100 in those turns.



            Since the dice rolls are independent this is



            p(X < 81)^12 = (0.80)^12
            ~= 0.069


            And therefore the probability that the dice roll was an 81 or higher on any of the rolls is 1 - 0.069 which is approximately 93%, or exactly 93.1280523264% to 10 d.p. (derived from 227363409/244140625).






            share|improve this answer














            The probability of an event is 1 minus the probability of the inverse of the event. Therefore, the probability that you do not roll an 81-100 is 1 minus the probability that you do roll an 81-100 in those turns.



            Since the dice rolls are independent this is



            p(X < 81)^12 = (0.80)^12
            ~= 0.069


            And therefore the probability that the dice roll was an 81 or higher on any of the rolls is 1 - 0.069 which is approximately 93%, or exactly 93.1280523264% to 10 d.p. (derived from 227363409/244140625).







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 1 hour ago

























            answered 3 hours ago









            mklingen

            2184




            2184












            • I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
              – mklingen
              1 hour ago










            • It surprises me that 1-(4/5)^12 results in an even denominator
              – Yakk
              1 hour ago












            • My math was wrong. Fixed the rational.
              – mklingen
              1 hour ago


















            • I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
              – mklingen
              1 hour ago










            • It surprises me that 1-(4/5)^12 results in an even denominator
              – Yakk
              1 hour ago












            • My math was wrong. Fixed the rational.
              – mklingen
              1 hour ago
















            I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
            – mklingen
            1 hour ago




            I didn't know that was possible. If it is possible then the distribution is going to be way more complicated.
            – mklingen
            1 hour ago












            It surprises me that 1-(4/5)^12 results in an even denominator
            – Yakk
            1 hour ago






            It surprises me that 1-(4/5)^12 results in an even denominator
            – Yakk
            1 hour ago














            My math was wrong. Fixed the rational.
            – mklingen
            1 hour ago




            My math was wrong. Fixed the rational.
            – mklingen
            1 hour ago


















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