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


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5












$begingroup$


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




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$endgroup$












  • $begingroup$
    Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
    $endgroup$
    – Ifusaso
    5 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
    5 hours ago










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










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    5 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
    5 hours ago
















5












$begingroup$


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
    5 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
    5 hours ago










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










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    5 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
    5 hours ago














5












5








5





$begingroup$


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 1 hour ago









Ruse

6,24311251




6,24311251






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asked 5 hours ago









HimmatorsHimmators

1285




1285




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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.






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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
    5 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
    5 hours ago










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










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    5 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
    5 hours ago


















  • $begingroup$
    Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
    $endgroup$
    – Ifusaso
    5 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
    5 hours ago










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










  • $begingroup$
    @GreySage Thanks, sloppy copy :P
    $endgroup$
    – Himmators
    5 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
    5 hours ago
















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




$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
5 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
5 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
5 hours ago












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




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












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




$begingroup$
@GreySage Thanks, sloppy copy :P
$endgroup$
– Himmators
5 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
5 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
5 hours ago










4 Answers
4






active

oldest

votes


















8












$begingroup$

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
    2 hours ago



















15












$begingroup$

I couldn't match your targets exactly, but here's a different approach



Linearity is difficult to exactly model with only d6s as you desire, but there is a very simple method to describe that closely resembles the approximations you posit:




A character fumbles if more 1s are showing than 6s.




The resulting probabilities are:



begin{array}{rl}
text{Number of Dice} & text{Probability} \
hline
1 & 16.7% \
2 & 13.9% \
3 & 10.2% \
4 & 6.6% \
5 & 4.6% \
end{array}



For what it's worth, this method is basically linear anyway. The r-squared value (how much of the data can be approximated by a straight line) is 99%.






share|improve this answer











$endgroup$













  • $begingroup$
    Seems something wrong with your table? thx!
    $endgroup$
    – Himmators
    2 hours ago










  • $begingroup$
    @himmators fixed
    $endgroup$
    – David Coffron
    2 hours ago










  • $begingroup$
    Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
    $endgroup$
    – Mark Wells
    6 mins ago










  • $begingroup$
    @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
    $endgroup$
    – David Coffron
    3 mins ago





















12












$begingroup$

Fumble if exactly one die shows a 1.



N dice are rolled on the table. If exactly one shows a 1, then it's a fumble.



begin{array}{rl}
N & P(text{fumble}) \
hline
1 & 16.67% \
2 & 13.89% \
3 & 11.57% \
4 & 9.65% \
5 & 8.04% \
end{array}






share|improve this answer









$endgroup$





















    -2












    $begingroup$

    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
      5 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
      5 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
      1 hour ago











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    8












    $begingroup$

    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
      2 hours ago
















    8












    $begingroup$

    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
      2 hours ago














    8












    8








    8





    $begingroup$

    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 3 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 4 hours ago









    Craig MeierCraig Meier

    2263




    2263




    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
      2 hours ago


















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
















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




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













    15












    $begingroup$

    I couldn't match your targets exactly, but here's a different approach



    Linearity is difficult to exactly model with only d6s as you desire, but there is a very simple method to describe that closely resembles the approximations you posit:




    A character fumbles if more 1s are showing than 6s.




    The resulting probabilities are:



    begin{array}{rl}
    text{Number of Dice} & text{Probability} \
    hline
    1 & 16.7% \
    2 & 13.9% \
    3 & 10.2% \
    4 & 6.6% \
    5 & 4.6% \
    end{array}



    For what it's worth, this method is basically linear anyway. The r-squared value (how much of the data can be approximated by a straight line) is 99%.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Seems something wrong with your table? thx!
      $endgroup$
      – Himmators
      2 hours ago










    • $begingroup$
      @himmators fixed
      $endgroup$
      – David Coffron
      2 hours ago










    • $begingroup$
      Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
      $endgroup$
      – Mark Wells
      6 mins ago










    • $begingroup$
      @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
      $endgroup$
      – David Coffron
      3 mins ago


















    15












    $begingroup$

    I couldn't match your targets exactly, but here's a different approach



    Linearity is difficult to exactly model with only d6s as you desire, but there is a very simple method to describe that closely resembles the approximations you posit:




    A character fumbles if more 1s are showing than 6s.




    The resulting probabilities are:



    begin{array}{rl}
    text{Number of Dice} & text{Probability} \
    hline
    1 & 16.7% \
    2 & 13.9% \
    3 & 10.2% \
    4 & 6.6% \
    5 & 4.6% \
    end{array}



    For what it's worth, this method is basically linear anyway. The r-squared value (how much of the data can be approximated by a straight line) is 99%.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Seems something wrong with your table? thx!
      $endgroup$
      – Himmators
      2 hours ago










    • $begingroup$
      @himmators fixed
      $endgroup$
      – David Coffron
      2 hours ago










    • $begingroup$
      Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
      $endgroup$
      – Mark Wells
      6 mins ago










    • $begingroup$
      @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
      $endgroup$
      – David Coffron
      3 mins ago
















    15












    15








    15





    $begingroup$

    I couldn't match your targets exactly, but here's a different approach



    Linearity is difficult to exactly model with only d6s as you desire, but there is a very simple method to describe that closely resembles the approximations you posit:




    A character fumbles if more 1s are showing than 6s.




    The resulting probabilities are:



    begin{array}{rl}
    text{Number of Dice} & text{Probability} \
    hline
    1 & 16.7% \
    2 & 13.9% \
    3 & 10.2% \
    4 & 6.6% \
    5 & 4.6% \
    end{array}



    For what it's worth, this method is basically linear anyway. The r-squared value (how much of the data can be approximated by a straight line) is 99%.






    share|improve this answer











    $endgroup$



    I couldn't match your targets exactly, but here's a different approach



    Linearity is difficult to exactly model with only d6s as you desire, but there is a very simple method to describe that closely resembles the approximations you posit:




    A character fumbles if more 1s are showing than 6s.




    The resulting probabilities are:



    begin{array}{rl}
    text{Number of Dice} & text{Probability} \
    hline
    1 & 16.7% \
    2 & 13.9% \
    3 & 10.2% \
    4 & 6.6% \
    5 & 4.6% \
    end{array}



    For what it's worth, this method is basically linear anyway. The r-squared value (how much of the data can be approximated by a straight line) is 99%.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 1 hour ago

























    answered 3 hours ago









    David CoffronDavid Coffron

    36.2k3123251




    36.2k3123251












    • $begingroup$
      Seems something wrong with your table? thx!
      $endgroup$
      – Himmators
      2 hours ago










    • $begingroup$
      @himmators fixed
      $endgroup$
      – David Coffron
      2 hours ago










    • $begingroup$
      Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
      $endgroup$
      – Mark Wells
      6 mins ago










    • $begingroup$
      @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
      $endgroup$
      – David Coffron
      3 mins ago




















    • $begingroup$
      Seems something wrong with your table? thx!
      $endgroup$
      – Himmators
      2 hours ago










    • $begingroup$
      @himmators fixed
      $endgroup$
      – David Coffron
      2 hours ago










    • $begingroup$
      Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
      $endgroup$
      – Mark Wells
      6 mins ago










    • $begingroup$
      @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
      $endgroup$
      – David Coffron
      3 mins ago


















    $begingroup$
    Seems something wrong with your table? thx!
    $endgroup$
    – Himmators
    2 hours ago




    $begingroup$
    Seems something wrong with your table? thx!
    $endgroup$
    – Himmators
    2 hours ago












    $begingroup$
    @himmators fixed
    $endgroup$
    – David Coffron
    2 hours ago




    $begingroup$
    @himmators fixed
    $endgroup$
    – David Coffron
    2 hours ago












    $begingroup$
    Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
    $endgroup$
    – Mark Wells
    6 mins ago




    $begingroup$
    Am I understanding this right? For N=2, this rule gives a fumble if one die is a 1 and the other is not a 6. That's 9 outcomes = 0.25.
    $endgroup$
    – Mark Wells
    6 mins ago












    $begingroup$
    @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
    $endgroup$
    – David Coffron
    3 mins ago






    $begingroup$
    @MarkWells not quite. there is also the chance that there are 2 1's. 11,12,13,14,15 is 5 outcomes out of 36 = 13.9% (mind you I solved them literally, so... (1/6)*(4/6) = chance of a 1 times chance of a non-6, non-1 ;plus the 1/36 chance of both 1s)
    $endgroup$
    – David Coffron
    3 mins ago













    12












    $begingroup$

    Fumble if exactly one die shows a 1.



    N dice are rolled on the table. If exactly one shows a 1, then it's a fumble.



    begin{array}{rl}
    N & P(text{fumble}) \
    hline
    1 & 16.67% \
    2 & 13.89% \
    3 & 11.57% \
    4 & 9.65% \
    5 & 8.04% \
    end{array}






    share|improve this answer









    $endgroup$


















      12












      $begingroup$

      Fumble if exactly one die shows a 1.



      N dice are rolled on the table. If exactly one shows a 1, then it's a fumble.



      begin{array}{rl}
      N & P(text{fumble}) \
      hline
      1 & 16.67% \
      2 & 13.89% \
      3 & 11.57% \
      4 & 9.65% \
      5 & 8.04% \
      end{array}






      share|improve this answer









      $endgroup$
















        12












        12








        12





        $begingroup$

        Fumble if exactly one die shows a 1.



        N dice are rolled on the table. If exactly one shows a 1, then it's a fumble.



        begin{array}{rl}
        N & P(text{fumble}) \
        hline
        1 & 16.67% \
        2 & 13.89% \
        3 & 11.57% \
        4 & 9.65% \
        5 & 8.04% \
        end{array}






        share|improve this answer









        $endgroup$



        Fumble if exactly one die shows a 1.



        N dice are rolled on the table. If exactly one shows a 1, then it's a fumble.



        begin{array}{rl}
        N & P(text{fumble}) \
        hline
        1 & 16.67% \
        2 & 13.89% \
        3 & 11.57% \
        4 & 9.65% \
        5 & 8.04% \
        end{array}







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        nitsua60nitsua60

        75.2k13309431




        75.2k13309431























            -2












            $begingroup$

            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
              5 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
              5 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
              1 hour ago
















            -2












            $begingroup$

            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
              5 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
              5 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
              1 hour ago














            -2












            -2








            -2





            $begingroup$

            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 4 hours ago

























            answered 5 hours ago









            WyrmwoodWyrmwood

            5,38711540




            5,38711540












            • $begingroup$
              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
              5 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
              5 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
              1 hour 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
              5 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
              5 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
              1 hour 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
            5 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
            5 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
            5 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
            5 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
            1 hour 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
            1 hour ago










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