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How can the probability of a fumble decrease linearly with more dice?


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9












$begingroup$


I'm working on a simplified RPG system that uses only D6s, and I want a mechanic for fumbles/critical fails.



Depending on how good the player character is, they have 1-5 dice to roll and they have to beat a difficulty set by the DM. I thought it would be fun to have players fail if they roll all 1s, but realized it makes it way too hard to fail if you have 5 dice, and a bit too easy if you have 1. Is there a more linear way of defining critical fails?



This is what I get if fumbles are on all dice showing 1s:



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{16.67%} \
text{2} & text{2.78%} \
text{3} & text{0.46%} \
text{4} & text{0.08%} \
text{5} & text{0.01%} \
hline
end{array}
$



What I would like (approximately, exact numbers are not that important):



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{18%} \
text{2} & text{15%} \
text{3} & text{12%} \
text{4} & text{9%} \
text{5} & text{6%} \
hline
end{array}
$










share|improve this question









New contributor




Himmators is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
    $endgroup$
    – Ifusaso
    9 hours ago










  • $begingroup$
    In your second table, when you say 'Probability of all 1's' you really mean 'Probability of failure', right? Given that you states you don't want to use the all 1 condition, some other failure condition that satisfies those general probabilities would work?
    $endgroup$
    – GreySage
    8 hours ago










  • $begingroup$
    @lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
    $endgroup$
    – Himmators
    8 hours ago










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    8 hours ago






  • 2




    $begingroup$
    How extensible do you need this table? Does it need to handle more than 5 d6 dice rolled at once, or is it capped at 5 dice for any possible roll?
    $endgroup$
    – Xirema
    8 hours ago
















9












$begingroup$


I'm working on a simplified RPG system that uses only D6s, and I want a mechanic for fumbles/critical fails.



Depending on how good the player character is, they have 1-5 dice to roll and they have to beat a difficulty set by the DM. I thought it would be fun to have players fail if they roll all 1s, but realized it makes it way too hard to fail if you have 5 dice, and a bit too easy if you have 1. Is there a more linear way of defining critical fails?



This is what I get if fumbles are on all dice showing 1s:



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{16.67%} \
text{2} & text{2.78%} \
text{3} & text{0.46%} \
text{4} & text{0.08%} \
text{5} & text{0.01%} \
hline
end{array}
$



What I would like (approximately, exact numbers are not that important):



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{18%} \
text{2} & text{15%} \
text{3} & text{12%} \
text{4} & text{9%} \
text{5} & text{6%} \
hline
end{array}
$










share|improve this question









New contributor




Himmators is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
    $endgroup$
    – Ifusaso
    9 hours ago










  • $begingroup$
    In your second table, when you say 'Probability of all 1's' you really mean 'Probability of failure', right? Given that you states you don't want to use the all 1 condition, some other failure condition that satisfies those general probabilities would work?
    $endgroup$
    – GreySage
    8 hours ago










  • $begingroup$
    @lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
    $endgroup$
    – Himmators
    8 hours ago










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    8 hours ago






  • 2




    $begingroup$
    How extensible do you need this table? Does it need to handle more than 5 d6 dice rolled at once, or is it capped at 5 dice for any possible roll?
    $endgroup$
    – Xirema
    8 hours ago














9












9








9





$begingroup$


I'm working on a simplified RPG system that uses only D6s, and I want a mechanic for fumbles/critical fails.



Depending on how good the player character is, they have 1-5 dice to roll and they have to beat a difficulty set by the DM. I thought it would be fun to have players fail if they roll all 1s, but realized it makes it way too hard to fail if you have 5 dice, and a bit too easy if you have 1. Is there a more linear way of defining critical fails?



This is what I get if fumbles are on all dice showing 1s:



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{16.67%} \
text{2} & text{2.78%} \
text{3} & text{0.46%} \
text{4} & text{0.08%} \
text{5} & text{0.01%} \
hline
end{array}
$



What I would like (approximately, exact numbers are not that important):



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{18%} \
text{2} & text{15%} \
text{3} & text{12%} \
text{4} & text{9%} \
text{5} & text{6%} \
hline
end{array}
$










share|improve this question









New contributor




Himmators is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I'm working on a simplified RPG system that uses only D6s, and I want a mechanic for fumbles/critical fails.



Depending on how good the player character is, they have 1-5 dice to roll and they have to beat a difficulty set by the DM. I thought it would be fun to have players fail if they roll all 1s, but realized it makes it way too hard to fail if you have 5 dice, and a bit too easy if you have 1. Is there a more linear way of defining critical fails?



This is what I get if fumbles are on all dice showing 1s:



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{16.67%} \
text{2} & text{2.78%} \
text{3} & text{0.46%} \
text{4} & text{0.08%} \
text{5} & text{0.01%} \
hline
end{array}
$



What I would like (approximately, exact numbers are not that important):



$begin{array}{|c|c|}
hline
textbf{Number of Dice} & textbf{Probability of Fumble} \
hline
text{1} & text{18%} \
text{2} & text{15%} \
text{3} & text{12%} \
text{4} & text{9%} \
text{5} & text{6%} \
hline
end{array}
$







dice game-design statistics critical-fail






share|improve this question









New contributor




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Check out our Code of Conduct.











share|improve this question









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share|improve this question




share|improve this question








edited 5 hours ago









Ruse

6,24311251




6,24311251






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asked 9 hours ago









HimmatorsHimmators

1485




1485




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New contributor





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Check out our Code of Conduct.












  • $begingroup$
    Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
    $endgroup$
    – Ifusaso
    9 hours ago










  • $begingroup$
    In your second table, when you say 'Probability of all 1's' you really mean 'Probability of failure', right? Given that you states you don't want to use the all 1 condition, some other failure condition that satisfies those general probabilities would work?
    $endgroup$
    – GreySage
    8 hours ago










  • $begingroup$
    @lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
    $endgroup$
    – Himmators
    8 hours ago










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    8 hours ago






  • 2




    $begingroup$
    How extensible do you need this table? Does it need to handle more than 5 d6 dice rolled at once, or is it capped at 5 dice for any possible roll?
    $endgroup$
    – Xirema
    8 hours ago


















  • $begingroup$
    Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
    $endgroup$
    – Ifusaso
    9 hours ago










  • $begingroup$
    In your second table, when you say 'Probability of all 1's' you really mean 'Probability of failure', right? Given that you states you don't want to use the all 1 condition, some other failure condition that satisfies those general probabilities would work?
    $endgroup$
    – GreySage
    8 hours ago










  • $begingroup$
    @lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
    $endgroup$
    – Himmators
    8 hours ago










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    8 hours ago






  • 2




    $begingroup$
    How extensible do you need this table? Does it need to handle more than 5 d6 dice rolled at once, or is it capped at 5 dice for any possible roll?
    $endgroup$
    – Xirema
    8 hours ago
















$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
9 hours ago




$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
9 hours ago












$begingroup$
In your second table, when you say 'Probability of all 1's' you really mean 'Probability of failure', right? Given that you states you don't want to use the all 1 condition, some other failure condition that satisfies those general probabilities would work?
$endgroup$
– GreySage
8 hours ago




$begingroup$
In your second table, when you say 'Probability of all 1's' you really mean 'Probability of failure', right? Given that you states you don't want to use the all 1 condition, some other failure condition that satisfies those general probabilities would work?
$endgroup$
– GreySage
8 hours ago












$begingroup$
@lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
$endgroup$
– Himmators
8 hours ago




$begingroup$
@lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
$endgroup$
– Himmators
8 hours ago












$begingroup$
@GreySage Thanks, sloppy copy :P
$endgroup$
– Himmators
8 hours ago




$begingroup$
@GreySage Thanks, sloppy copy :P
$endgroup$
– Himmators
8 hours ago




2




2




$begingroup$
How extensible do you need this table? Does it need to handle more than 5 d6 dice rolled at once, or is it capped at 5 dice for any possible roll?
$endgroup$
– Xirema
8 hours ago




$begingroup$
How extensible do you need this table? Does it need to handle more than 5 d6 dice rolled at once, or is it capped at 5 dice for any possible roll?
$endgroup$
– Xirema
8 hours ago










5 Answers
5






active

oldest

votes


















11












$begingroup$

A close approximation to the percentages you want would use something like this:



$begin{array}{|c|c|c|}
hline
textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
hline
text{1} & text{1} & text{1/6 (16.7%)} \
text{2} & text{2-4} & text{6/36 (16.7%)} \
text{3*} & text{3-7} & text{35/216 (16.2%)} \
& text{3-6} & text{20/216 (9.3%)} \
text{4} & text{4-9} & text{126/1296 (9%)} \
text{5} & text{5-11} & text{457/7776 (5.9%)} \
hline
end{array}
$



* (3 dice could go either way)



In terms of gameplay, simpler rules are frequently better than strictly matching the desired probability distribution. I might suggest something like $N$ dice fumble on a result $le 2times N$, with a special case that a single die only fumbles on a 1 (unless you want a 1/3 chance of a fumble in the 1d case). That would give you something like:



$begin{array}{|c|c|c|}
hline
textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
hline
text{1} & text{1} & text{1/6 (16.7%)} \
text{2} & text{2-4} & text{6/36 (16.7%)} \
text{3} & text{3-6} & text{20/216 (9.3%)} \
text{4} & text{4-8} & text{70/1296 (5.4%)} \
text{5} & text{5-10} & text{252/7776 (3.2%)} \
hline
end{array}
$






share|improve this answer










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Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$













  • $begingroup$
    +1 for mathemagics!
    $endgroup$
    – Harper
    5 hours ago



















12












$begingroup$

Fumble if the leftmost, unique die is a 1.



(Hear me out.)



N dice are rolled on the table. One of those dice is unique--say it's black with white pips and the rest are numbered dice. If the unique die is both showing a 1 and is farthest left (from the roller's POV), that's your fumble. In case of a leftmost-tie, let the closer (to the roller) die win.



It's not linear, but it's a lot closer than the original method (all 1s) while being simple and memorable.



begin{array}{rl}
N & P(text{fumble}) \
hline
1 & 16.67% \
2 & 8.33% \
3 & 5.55% \
4 & 4.16% \
5 & 3.34% \
end{array}






share|improve this answer











$endgroup$













  • $begingroup$
    @CraigMeier found it. Thanks for catching me earlier.
    $endgroup$
    – nitsua60
    2 hours ago










  • $begingroup$
    So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
    $endgroup$
    – Ryan Thompson
    2 hours ago












  • $begingroup$
    @RyanThompson Yeah, a dice cup.
    $endgroup$
    – KorvinStarmast
    2 hours ago










  • $begingroup$
    The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
    $endgroup$
    – VLAZ
    19 mins ago



















3












$begingroup$

It can be done but it's messy.



You need two special dice: a red die and a yellow die. If you roll 1d6, roll the red die. If you roll two or more, roll the red and yellow. Any additional dice are "green" and can't make you fumble.



Fumble conditions depend on the number of dice:




  • 1 die: Fumble on a 1.

  • 2 dice: Fumble on a red 1 and a yellow 1-5.

  • 3 dice: Fumble on a red 1 and a yellow 1-4.

  • 4 dice: Fumble on a red 1 and a yellow 1-3.

  • 5 dice: Fumble on a red 1 and a yellow 1-2.

  • 6 dice: Fumble on a red 1 and a yellow 1.

  • 7 or more: No chance of a fumble.


The fumble chance is (7-N)/36. Exactly which values count as a fumble is arbitrary, but I picked the outcomes that involve the lowest total values of the red and yellow dice to minimize the chance of rolling a success that's also a fumble.






share|improve this answer











$endgroup$





















    1












    $begingroup$

    Have one unique die (the red die) that players roll in addition to 0–4 other dice. If the red die is a 1, roll it a second time: if this second roll is less than the number of dice the player rolled (to start), then no fumble, but if the second roll is at least the number of dice the player rolled, then fumble.



    For example, if the player only got to roll one die (the red one), then a 1 is always a fumble. If the player got to roll five dice, then a 1 is a fumble if the reroll is 5 or 6 but not a fumble if the reroll is 1, 2, 3, or 4. This gives literally a linear sequence of probabilities:



    $begin{array}{|c|c|}
    hline
    textbf{Number of Dice} & textbf{Probability of Fumble} \
    hline
    text{1} & text{16.67%} \
    text{2} & text{13.89%} \
    text{3} & text{11.11%} \
    text{4} & text{8.33%} \
    text{5} & text{5.56%} \
    hline
    end{array}
    $






    share|improve this answer








    New contributor




    Greg Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$





















      -3












      $begingroup$

      Percentile dice are an example of linear probability with more than one die. The first d10 represents tens and the second d10 represents ones so you get a linear result, from 1-100. You could do something similar with other dice, but the end result would be non-trivial to understand. For example, you could use 2d4, where the first d4 is the tens and the second d4 is the ones, but since you don't have 5-10, you can only get results of



      11
      12
      13
      14
      21
      22
      23
      24
      31
      32
      33
      34
      41
      42
      43
      44


      But, within the possible outcomes, the probability is equal, or linear.



      A better, more useable version is to have one die, like say a d6, and on 1-3 equals 0, and on 4-6 equals the maximum of the next die, like say 12. Then you could produce 1-24 linearly with two dice.



      More intuitively, if the maximum number is greater than ten, make the first die a 10, then the second die determines whether you add to it. Like for 1-17, roll a d10 for ones, and then roll a d-whatever, where half or less is 0 and over half is +7.






      share|improve this answer











      $endgroup$













      • $begingroup$
        Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
        $endgroup$
        – Himmators
        8 hours ago










      • $begingroup$
        just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
        $endgroup$
        – Wyrmwood
        8 hours ago








      • 1




        $begingroup$
        This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
        $endgroup$
        – SevenSidedDie
        4 hours ago











      Your Answer





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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11












      $begingroup$

      A close approximation to the percentages you want would use something like this:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3*} & text{3-7} & text{35/216 (16.2%)} \
      & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-9} & text{126/1296 (9%)} \
      text{5} & text{5-11} & text{457/7776 (5.9%)} \
      hline
      end{array}
      $



      * (3 dice could go either way)



      In terms of gameplay, simpler rules are frequently better than strictly matching the desired probability distribution. I might suggest something like $N$ dice fumble on a result $le 2times N$, with a special case that a single die only fumbles on a 1 (unless you want a 1/3 chance of a fumble in the 1d case). That would give you something like:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3} & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-8} & text{70/1296 (5.4%)} \
      text{5} & text{5-10} & text{252/7776 (3.2%)} \
      hline
      end{array}
      $






      share|improve this answer










      New contributor




      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$













      • $begingroup$
        +1 for mathemagics!
        $endgroup$
        – Harper
        5 hours ago
















      11












      $begingroup$

      A close approximation to the percentages you want would use something like this:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3*} & text{3-7} & text{35/216 (16.2%)} \
      & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-9} & text{126/1296 (9%)} \
      text{5} & text{5-11} & text{457/7776 (5.9%)} \
      hline
      end{array}
      $



      * (3 dice could go either way)



      In terms of gameplay, simpler rules are frequently better than strictly matching the desired probability distribution. I might suggest something like $N$ dice fumble on a result $le 2times N$, with a special case that a single die only fumbles on a 1 (unless you want a 1/3 chance of a fumble in the 1d case). That would give you something like:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3} & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-8} & text{70/1296 (5.4%)} \
      text{5} & text{5-10} & text{252/7776 (3.2%)} \
      hline
      end{array}
      $






      share|improve this answer










      New contributor




      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$













      • $begingroup$
        +1 for mathemagics!
        $endgroup$
        – Harper
        5 hours ago














      11












      11








      11





      $begingroup$

      A close approximation to the percentages you want would use something like this:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3*} & text{3-7} & text{35/216 (16.2%)} \
      & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-9} & text{126/1296 (9%)} \
      text{5} & text{5-11} & text{457/7776 (5.9%)} \
      hline
      end{array}
      $



      * (3 dice could go either way)



      In terms of gameplay, simpler rules are frequently better than strictly matching the desired probability distribution. I might suggest something like $N$ dice fumble on a result $le 2times N$, with a special case that a single die only fumbles on a 1 (unless you want a 1/3 chance of a fumble in the 1d case). That would give you something like:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3} & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-8} & text{70/1296 (5.4%)} \
      text{5} & text{5-10} & text{252/7776 (3.2%)} \
      hline
      end{array}
      $






      share|improve this answer










      New contributor




      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$



      A close approximation to the percentages you want would use something like this:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3*} & text{3-7} & text{35/216 (16.2%)} \
      & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-9} & text{126/1296 (9%)} \
      text{5} & text{5-11} & text{457/7776 (5.9%)} \
      hline
      end{array}
      $



      * (3 dice could go either way)



      In terms of gameplay, simpler rules are frequently better than strictly matching the desired probability distribution. I might suggest something like $N$ dice fumble on a result $le 2times N$, with a special case that a single die only fumbles on a 1 (unless you want a 1/3 chance of a fumble in the 1d case). That would give you something like:



      $begin{array}{|c|c|c|}
      hline
      textbf{Dice} & textbf{Fumble Range} & textbf{Probability} \
      hline
      text{1} & text{1} & text{1/6 (16.7%)} \
      text{2} & text{2-4} & text{6/36 (16.7%)} \
      text{3} & text{3-6} & text{20/216 (9.3%)} \
      text{4} & text{4-8} & text{70/1296 (5.4%)} \
      text{5} & text{5-10} & text{252/7776 (3.2%)} \
      hline
      end{array}
      $







      share|improve this answer










      New contributor




      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this answer



      share|improve this answer








      edited 6 hours ago









      V2Blast

      23.3k375146




      23.3k375146






      New contributor




      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      answered 7 hours ago









      Craig MeierCraig Meier

      2564




      2564




      New contributor




      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Craig Meier is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      • $begingroup$
        +1 for mathemagics!
        $endgroup$
        – Harper
        5 hours ago


















      • $begingroup$
        +1 for mathemagics!
        $endgroup$
        – Harper
        5 hours ago
















      $begingroup$
      +1 for mathemagics!
      $endgroup$
      – Harper
      5 hours ago




      $begingroup$
      +1 for mathemagics!
      $endgroup$
      – Harper
      5 hours ago













      12












      $begingroup$

      Fumble if the leftmost, unique die is a 1.



      (Hear me out.)



      N dice are rolled on the table. One of those dice is unique--say it's black with white pips and the rest are numbered dice. If the unique die is both showing a 1 and is farthest left (from the roller's POV), that's your fumble. In case of a leftmost-tie, let the closer (to the roller) die win.



      It's not linear, but it's a lot closer than the original method (all 1s) while being simple and memorable.



      begin{array}{rl}
      N & P(text{fumble}) \
      hline
      1 & 16.67% \
      2 & 8.33% \
      3 & 5.55% \
      4 & 4.16% \
      5 & 3.34% \
      end{array}






      share|improve this answer











      $endgroup$













      • $begingroup$
        @CraigMeier found it. Thanks for catching me earlier.
        $endgroup$
        – nitsua60
        2 hours ago










      • $begingroup$
        So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
        $endgroup$
        – Ryan Thompson
        2 hours ago












      • $begingroup$
        @RyanThompson Yeah, a dice cup.
        $endgroup$
        – KorvinStarmast
        2 hours ago










      • $begingroup$
        The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
        $endgroup$
        – VLAZ
        19 mins ago
















      12












      $begingroup$

      Fumble if the leftmost, unique die is a 1.



      (Hear me out.)



      N dice are rolled on the table. One of those dice is unique--say it's black with white pips and the rest are numbered dice. If the unique die is both showing a 1 and is farthest left (from the roller's POV), that's your fumble. In case of a leftmost-tie, let the closer (to the roller) die win.



      It's not linear, but it's a lot closer than the original method (all 1s) while being simple and memorable.



      begin{array}{rl}
      N & P(text{fumble}) \
      hline
      1 & 16.67% \
      2 & 8.33% \
      3 & 5.55% \
      4 & 4.16% \
      5 & 3.34% \
      end{array}






      share|improve this answer











      $endgroup$













      • $begingroup$
        @CraigMeier found it. Thanks for catching me earlier.
        $endgroup$
        – nitsua60
        2 hours ago










      • $begingroup$
        So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
        $endgroup$
        – Ryan Thompson
        2 hours ago












      • $begingroup$
        @RyanThompson Yeah, a dice cup.
        $endgroup$
        – KorvinStarmast
        2 hours ago










      • $begingroup$
        The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
        $endgroup$
        – VLAZ
        19 mins ago














      12












      12








      12





      $begingroup$

      Fumble if the leftmost, unique die is a 1.



      (Hear me out.)



      N dice are rolled on the table. One of those dice is unique--say it's black with white pips and the rest are numbered dice. If the unique die is both showing a 1 and is farthest left (from the roller's POV), that's your fumble. In case of a leftmost-tie, let the closer (to the roller) die win.



      It's not linear, but it's a lot closer than the original method (all 1s) while being simple and memorable.



      begin{array}{rl}
      N & P(text{fumble}) \
      hline
      1 & 16.67% \
      2 & 8.33% \
      3 & 5.55% \
      4 & 4.16% \
      5 & 3.34% \
      end{array}






      share|improve this answer











      $endgroup$



      Fumble if the leftmost, unique die is a 1.



      (Hear me out.)



      N dice are rolled on the table. One of those dice is unique--say it's black with white pips and the rest are numbered dice. If the unique die is both showing a 1 and is farthest left (from the roller's POV), that's your fumble. In case of a leftmost-tie, let the closer (to the roller) die win.



      It's not linear, but it's a lot closer than the original method (all 1s) while being simple and memorable.



      begin{array}{rl}
      N & P(text{fumble}) \
      hline
      1 & 16.67% \
      2 & 8.33% \
      3 & 5.55% \
      4 & 4.16% \
      5 & 3.34% \
      end{array}







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 hours ago

























      answered 6 hours ago









      nitsua60nitsua60

      75.2k13309432




      75.2k13309432












      • $begingroup$
        @CraigMeier found it. Thanks for catching me earlier.
        $endgroup$
        – nitsua60
        2 hours ago










      • $begingroup$
        So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
        $endgroup$
        – Ryan Thompson
        2 hours ago












      • $begingroup$
        @RyanThompson Yeah, a dice cup.
        $endgroup$
        – KorvinStarmast
        2 hours ago










      • $begingroup$
        The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
        $endgroup$
        – VLAZ
        19 mins ago


















      • $begingroup$
        @CraigMeier found it. Thanks for catching me earlier.
        $endgroup$
        – nitsua60
        2 hours ago










      • $begingroup$
        So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
        $endgroup$
        – Ryan Thompson
        2 hours ago












      • $begingroup$
        @RyanThompson Yeah, a dice cup.
        $endgroup$
        – KorvinStarmast
        2 hours ago










      • $begingroup$
        The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
        $endgroup$
        – VLAZ
        19 mins ago
















      $begingroup$
      @CraigMeier found it. Thanks for catching me earlier.
      $endgroup$
      – nitsua60
      2 hours ago




      $begingroup$
      @CraigMeier found it. Thanks for catching me earlier.
      $endgroup$
      – nitsua60
      2 hours ago












      $begingroup$
      So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
      $endgroup$
      – Ryan Thompson
      2 hours ago






      $begingroup$
      So the probability of a 1 on the unique die is 1/6, and assuming the dice land randomly, the probability that the unique die is the leftmost is 1/N, which puts the probability of a fumble at 1/6 * 1/N = 1/(6N). That's definitely better than 1/(6^N). You would need to make sure that the dice land randomly though, which might be harder than it looks. Using a cup rather than one's hands would probably help a lot.
      $endgroup$
      – Ryan Thompson
      2 hours ago














      $begingroup$
      @RyanThompson Yeah, a dice cup.
      $endgroup$
      – KorvinStarmast
      2 hours ago




      $begingroup$
      @RyanThompson Yeah, a dice cup.
      $endgroup$
      – KorvinStarmast
      2 hours ago












      $begingroup$
      The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
      $endgroup$
      – VLAZ
      19 mins ago




      $begingroup$
      The only trouble I see with this is how much of a PIA would it be to track where the die lands. Does it need the location to be relevant? Can that factor be swapped with something else? Otherwise I do like the idea of distinct dies of some description.
      $endgroup$
      – VLAZ
      19 mins ago











      3












      $begingroup$

      It can be done but it's messy.



      You need two special dice: a red die and a yellow die. If you roll 1d6, roll the red die. If you roll two or more, roll the red and yellow. Any additional dice are "green" and can't make you fumble.



      Fumble conditions depend on the number of dice:




      • 1 die: Fumble on a 1.

      • 2 dice: Fumble on a red 1 and a yellow 1-5.

      • 3 dice: Fumble on a red 1 and a yellow 1-4.

      • 4 dice: Fumble on a red 1 and a yellow 1-3.

      • 5 dice: Fumble on a red 1 and a yellow 1-2.

      • 6 dice: Fumble on a red 1 and a yellow 1.

      • 7 or more: No chance of a fumble.


      The fumble chance is (7-N)/36. Exactly which values count as a fumble is arbitrary, but I picked the outcomes that involve the lowest total values of the red and yellow dice to minimize the chance of rolling a success that's also a fumble.






      share|improve this answer











      $endgroup$


















        3












        $begingroup$

        It can be done but it's messy.



        You need two special dice: a red die and a yellow die. If you roll 1d6, roll the red die. If you roll two or more, roll the red and yellow. Any additional dice are "green" and can't make you fumble.



        Fumble conditions depend on the number of dice:




        • 1 die: Fumble on a 1.

        • 2 dice: Fumble on a red 1 and a yellow 1-5.

        • 3 dice: Fumble on a red 1 and a yellow 1-4.

        • 4 dice: Fumble on a red 1 and a yellow 1-3.

        • 5 dice: Fumble on a red 1 and a yellow 1-2.

        • 6 dice: Fumble on a red 1 and a yellow 1.

        • 7 or more: No chance of a fumble.


        The fumble chance is (7-N)/36. Exactly which values count as a fumble is arbitrary, but I picked the outcomes that involve the lowest total values of the red and yellow dice to minimize the chance of rolling a success that's also a fumble.






        share|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          It can be done but it's messy.



          You need two special dice: a red die and a yellow die. If you roll 1d6, roll the red die. If you roll two or more, roll the red and yellow. Any additional dice are "green" and can't make you fumble.



          Fumble conditions depend on the number of dice:




          • 1 die: Fumble on a 1.

          • 2 dice: Fumble on a red 1 and a yellow 1-5.

          • 3 dice: Fumble on a red 1 and a yellow 1-4.

          • 4 dice: Fumble on a red 1 and a yellow 1-3.

          • 5 dice: Fumble on a red 1 and a yellow 1-2.

          • 6 dice: Fumble on a red 1 and a yellow 1.

          • 7 or more: No chance of a fumble.


          The fumble chance is (7-N)/36. Exactly which values count as a fumble is arbitrary, but I picked the outcomes that involve the lowest total values of the red and yellow dice to minimize the chance of rolling a success that's also a fumble.






          share|improve this answer











          $endgroup$



          It can be done but it's messy.



          You need two special dice: a red die and a yellow die. If you roll 1d6, roll the red die. If you roll two or more, roll the red and yellow. Any additional dice are "green" and can't make you fumble.



          Fumble conditions depend on the number of dice:




          • 1 die: Fumble on a 1.

          • 2 dice: Fumble on a red 1 and a yellow 1-5.

          • 3 dice: Fumble on a red 1 and a yellow 1-4.

          • 4 dice: Fumble on a red 1 and a yellow 1-3.

          • 5 dice: Fumble on a red 1 and a yellow 1-2.

          • 6 dice: Fumble on a red 1 and a yellow 1.

          • 7 or more: No chance of a fumble.


          The fumble chance is (7-N)/36. Exactly which values count as a fumble is arbitrary, but I picked the outcomes that involve the lowest total values of the red and yellow dice to minimize the chance of rolling a success that's also a fumble.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 hours ago

























          answered 3 hours ago









          Mark WellsMark Wells

          6,42111745




          6,42111745























              1












              $begingroup$

              Have one unique die (the red die) that players roll in addition to 0–4 other dice. If the red die is a 1, roll it a second time: if this second roll is less than the number of dice the player rolled (to start), then no fumble, but if the second roll is at least the number of dice the player rolled, then fumble.



              For example, if the player only got to roll one die (the red one), then a 1 is always a fumble. If the player got to roll five dice, then a 1 is a fumble if the reroll is 5 or 6 but not a fumble if the reroll is 1, 2, 3, or 4. This gives literally a linear sequence of probabilities:



              $begin{array}{|c|c|}
              hline
              textbf{Number of Dice} & textbf{Probability of Fumble} \
              hline
              text{1} & text{16.67%} \
              text{2} & text{13.89%} \
              text{3} & text{11.11%} \
              text{4} & text{8.33%} \
              text{5} & text{5.56%} \
              hline
              end{array}
              $






              share|improve this answer








              New contributor




              Greg Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$


















                1












                $begingroup$

                Have one unique die (the red die) that players roll in addition to 0–4 other dice. If the red die is a 1, roll it a second time: if this second roll is less than the number of dice the player rolled (to start), then no fumble, but if the second roll is at least the number of dice the player rolled, then fumble.



                For example, if the player only got to roll one die (the red one), then a 1 is always a fumble. If the player got to roll five dice, then a 1 is a fumble if the reroll is 5 or 6 but not a fumble if the reroll is 1, 2, 3, or 4. This gives literally a linear sequence of probabilities:



                $begin{array}{|c|c|}
                hline
                textbf{Number of Dice} & textbf{Probability of Fumble} \
                hline
                text{1} & text{16.67%} \
                text{2} & text{13.89%} \
                text{3} & text{11.11%} \
                text{4} & text{8.33%} \
                text{5} & text{5.56%} \
                hline
                end{array}
                $






                share|improve this answer








                New contributor




                Greg Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Have one unique die (the red die) that players roll in addition to 0–4 other dice. If the red die is a 1, roll it a second time: if this second roll is less than the number of dice the player rolled (to start), then no fumble, but if the second roll is at least the number of dice the player rolled, then fumble.



                  For example, if the player only got to roll one die (the red one), then a 1 is always a fumble. If the player got to roll five dice, then a 1 is a fumble if the reroll is 5 or 6 but not a fumble if the reroll is 1, 2, 3, or 4. This gives literally a linear sequence of probabilities:



                  $begin{array}{|c|c|}
                  hline
                  textbf{Number of Dice} & textbf{Probability of Fumble} \
                  hline
                  text{1} & text{16.67%} \
                  text{2} & text{13.89%} \
                  text{3} & text{11.11%} \
                  text{4} & text{8.33%} \
                  text{5} & text{5.56%} \
                  hline
                  end{array}
                  $






                  share|improve this answer








                  New contributor




                  Greg Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  Have one unique die (the red die) that players roll in addition to 0–4 other dice. If the red die is a 1, roll it a second time: if this second roll is less than the number of dice the player rolled (to start), then no fumble, but if the second roll is at least the number of dice the player rolled, then fumble.



                  For example, if the player only got to roll one die (the red one), then a 1 is always a fumble. If the player got to roll five dice, then a 1 is a fumble if the reroll is 5 or 6 but not a fumble if the reroll is 1, 2, 3, or 4. This gives literally a linear sequence of probabilities:



                  $begin{array}{|c|c|}
                  hline
                  textbf{Number of Dice} & textbf{Probability of Fumble} \
                  hline
                  text{1} & text{16.67%} \
                  text{2} & text{13.89%} \
                  text{3} & text{11.11%} \
                  text{4} & text{8.33%} \
                  text{5} & text{5.56%} \
                  hline
                  end{array}
                  $







                  share|improve this answer








                  New contributor




                  Greg Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  answered 51 mins ago









                  Greg MartinGreg Martin

                  1113




                  1113




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                  New contributor





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                      -3












                      $begingroup$

                      Percentile dice are an example of linear probability with more than one die. The first d10 represents tens and the second d10 represents ones so you get a linear result, from 1-100. You could do something similar with other dice, but the end result would be non-trivial to understand. For example, you could use 2d4, where the first d4 is the tens and the second d4 is the ones, but since you don't have 5-10, you can only get results of



                      11
                      12
                      13
                      14
                      21
                      22
                      23
                      24
                      31
                      32
                      33
                      34
                      41
                      42
                      43
                      44


                      But, within the possible outcomes, the probability is equal, or linear.



                      A better, more useable version is to have one die, like say a d6, and on 1-3 equals 0, and on 4-6 equals the maximum of the next die, like say 12. Then you could produce 1-24 linearly with two dice.



                      More intuitively, if the maximum number is greater than ten, make the first die a 10, then the second die determines whether you add to it. Like for 1-17, roll a d10 for ones, and then roll a d-whatever, where half or less is 0 and over half is +7.






                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
                        $endgroup$
                        – Himmators
                        8 hours ago










                      • $begingroup$
                        just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
                        $endgroup$
                        – Wyrmwood
                        8 hours ago








                      • 1




                        $begingroup$
                        This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
                        $endgroup$
                        – SevenSidedDie
                        4 hours ago
















                      -3












                      $begingroup$

                      Percentile dice are an example of linear probability with more than one die. The first d10 represents tens and the second d10 represents ones so you get a linear result, from 1-100. You could do something similar with other dice, but the end result would be non-trivial to understand. For example, you could use 2d4, where the first d4 is the tens and the second d4 is the ones, but since you don't have 5-10, you can only get results of



                      11
                      12
                      13
                      14
                      21
                      22
                      23
                      24
                      31
                      32
                      33
                      34
                      41
                      42
                      43
                      44


                      But, within the possible outcomes, the probability is equal, or linear.



                      A better, more useable version is to have one die, like say a d6, and on 1-3 equals 0, and on 4-6 equals the maximum of the next die, like say 12. Then you could produce 1-24 linearly with two dice.



                      More intuitively, if the maximum number is greater than ten, make the first die a 10, then the second die determines whether you add to it. Like for 1-17, roll a d10 for ones, and then roll a d-whatever, where half or less is 0 and over half is +7.






                      share|improve this answer











                      $endgroup$













                      • $begingroup$
                        Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
                        $endgroup$
                        – Himmators
                        8 hours ago










                      • $begingroup$
                        just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
                        $endgroup$
                        – Wyrmwood
                        8 hours ago








                      • 1




                        $begingroup$
                        This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
                        $endgroup$
                        – SevenSidedDie
                        4 hours ago














                      -3












                      -3








                      -3





                      $begingroup$

                      Percentile dice are an example of linear probability with more than one die. The first d10 represents tens and the second d10 represents ones so you get a linear result, from 1-100. You could do something similar with other dice, but the end result would be non-trivial to understand. For example, you could use 2d4, where the first d4 is the tens and the second d4 is the ones, but since you don't have 5-10, you can only get results of



                      11
                      12
                      13
                      14
                      21
                      22
                      23
                      24
                      31
                      32
                      33
                      34
                      41
                      42
                      43
                      44


                      But, within the possible outcomes, the probability is equal, or linear.



                      A better, more useable version is to have one die, like say a d6, and on 1-3 equals 0, and on 4-6 equals the maximum of the next die, like say 12. Then you could produce 1-24 linearly with two dice.



                      More intuitively, if the maximum number is greater than ten, make the first die a 10, then the second die determines whether you add to it. Like for 1-17, roll a d10 for ones, and then roll a d-whatever, where half or less is 0 and over half is +7.






                      share|improve this answer











                      $endgroup$



                      Percentile dice are an example of linear probability with more than one die. The first d10 represents tens and the second d10 represents ones so you get a linear result, from 1-100. You could do something similar with other dice, but the end result would be non-trivial to understand. For example, you could use 2d4, where the first d4 is the tens and the second d4 is the ones, but since you don't have 5-10, you can only get results of



                      11
                      12
                      13
                      14
                      21
                      22
                      23
                      24
                      31
                      32
                      33
                      34
                      41
                      42
                      43
                      44


                      But, within the possible outcomes, the probability is equal, or linear.



                      A better, more useable version is to have one die, like say a d6, and on 1-3 equals 0, and on 4-6 equals the maximum of the next die, like say 12. Then you could produce 1-24 linearly with two dice.



                      More intuitively, if the maximum number is greater than ten, make the first die a 10, then the second die determines whether you add to it. Like for 1-17, roll a d10 for ones, and then roll a d-whatever, where half or less is 0 and over half is +7.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 8 hours ago

























                      answered 8 hours ago









                      WyrmwoodWyrmwood

                      5,38511540




                      5,38511540












                      • $begingroup$
                        Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
                        $endgroup$
                        – Himmators
                        8 hours ago










                      • $begingroup$
                        just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
                        $endgroup$
                        – Wyrmwood
                        8 hours ago








                      • 1




                        $begingroup$
                        This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
                        $endgroup$
                        – SevenSidedDie
                        4 hours ago


















                      • $begingroup$
                        Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
                        $endgroup$
                        – Himmators
                        8 hours ago










                      • $begingroup$
                        just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
                        $endgroup$
                        – Wyrmwood
                        8 hours ago








                      • 1




                        $begingroup$
                        This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
                        $endgroup$
                        – SevenSidedDie
                        4 hours ago
















                      $begingroup$
                      Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
                      $endgroup$
                      – Himmators
                      8 hours ago




                      $begingroup$
                      Thx, though, I'm looking for a dynamic that works when the player is already playing. A bit of a bummer to roll a extra crit-fail-roll for every roll :P
                      $endgroup$
                      – Himmators
                      8 hours ago












                      $begingroup$
                      just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
                      $endgroup$
                      – Wyrmwood
                      8 hours ago






                      $begingroup$
                      just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
                      $endgroup$
                      – Wyrmwood
                      8 hours ago






                      1




                      1




                      $begingroup$
                      This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
                      $endgroup$
                      – SevenSidedDie
                      4 hours ago




                      $begingroup$
                      This describes a way of reading arbitrary sided dice to get linear results, but it doesn’t once mention fumbles, or how the 1-5 skill rating in the question can be factored in. It’s not obvious how to use this, let alone determine fumbles when using this system, so this seems a very incomplete post to adequately answer the question.
                      $endgroup$
                      – SevenSidedDie
                      4 hours ago










                      Himmators is a new contributor. Be nice, and check out our Code of Conduct.










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                      Himmators is a new contributor. Be nice, and check out our Code of Conduct.
















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