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How can the probability of a fumble decrease linearly with more dice?
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$begingroup$
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}
$
dice game-design statistics critical-fail
New contributor
$endgroup$
|
show 3 more comments
$begingroup$
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}
$
dice game-design statistics critical-fail
New contributor
$endgroup$
$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
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
$begingroup$
@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
|
show 3 more comments
$begingroup$
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}
$
dice game-design statistics critical-fail
New contributor
$endgroup$
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
dice game-design statistics critical-fail
New contributor
New contributor
edited 5 hours ago
Ruse
6,24311251
6,24311251
New contributor
asked 8 hours ago
HimmatorsHimmators
1485
1485
New contributor
New contributor
$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
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
$begingroup$
@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
|
show 3 more comments
$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
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
$begingroup$
@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
8 hours ago
$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
$endgroup$
– Ifusaso
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
$begingroup$
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
|
show 3 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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}
$
New contributor
$endgroup$
$begingroup$
+1 for mathemagics!
$endgroup$
– Harper
5 hours ago
add a comment |
$begingroup$
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}
$endgroup$
$begingroup$
@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.
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– Ryan Thompson
2 hours ago
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@RyanThompson Yeah, a dice cup.
$endgroup$
– KorvinStarmast
1 hour ago
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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
7 mins ago
add a comment |
$begingroup$
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.
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add a comment |
$begingroup$
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}
$
New contributor
$endgroup$
add a comment |
$begingroup$
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.
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$begingroup$
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
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– 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/…
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– 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
add a comment |
Your Answer
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5 Answers
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5 Answers
5
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$begingroup$
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}
$
New contributor
$endgroup$
$begingroup$
+1 for mathemagics!
$endgroup$
– Harper
5 hours ago
add a comment |
$begingroup$
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}
$
New contributor
$endgroup$
$begingroup$
+1 for mathemagics!
$endgroup$
– Harper
5 hours ago
add a comment |
$begingroup$
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}
$
New contributor
$endgroup$
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}
$
New contributor
edited 6 hours ago
V2Blast
23.3k375146
23.3k375146
New contributor
answered 7 hours ago
Craig MeierCraig Meier
2564
2564
New contributor
New contributor
$begingroup$
+1 for mathemagics!
$endgroup$
– Harper
5 hours ago
add a comment |
$begingroup$
+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
add a comment |
$begingroup$
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}
$endgroup$
$begingroup$
@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
1 hour 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
7 mins ago
add a comment |
$begingroup$
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}
$endgroup$
$begingroup$
@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
1 hour 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
7 mins ago
add a comment |
$begingroup$
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}
$endgroup$
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}
edited 2 hours ago
answered 6 hours ago
nitsua60♦nitsua60
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
1 hour 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
7 mins ago
add a comment |
$begingroup$
@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
1 hour 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
7 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
1 hour ago
$begingroup$
@RyanThompson Yeah, a dice cup.
$endgroup$
– KorvinStarmast
1 hour 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
7 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
7 mins ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 2 hours ago
answered 3 hours ago
Mark WellsMark Wells
6,42111745
6,42111745
add a comment |
add a comment |
$begingroup$
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}
$
New contributor
$endgroup$
add a comment |
$begingroup$
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}
$
New contributor
$endgroup$
add a comment |
$begingroup$
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}
$
New contributor
$endgroup$
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}
$
New contributor
New contributor
answered 39 mins ago
Greg MartinGreg Martin
1113
1113
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
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– Himmators
8 hours ago
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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
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– Himmators
8 hours ago
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just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
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– Wyrmwood
8 hours ago
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just choose the percentage, so less than X that would produce that percent is a fail, like this d20srd.org/srd/variant/adventuring/…
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– Wyrmwood
8 hours ago
1
1
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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.
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– SevenSidedDie♦
4 hours ago
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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.
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– SevenSidedDie♦
4 hours ago
add a comment |
Himmators is a new contributor. Be nice, and check out our Code of Conduct.
Himmators is a new contributor. Be nice, and check out our Code of Conduct.
Himmators is a new contributor. Be nice, and check out our Code of Conduct.
Himmators is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6?
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– Ifusaso
8 hours ago
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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?
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– GreySage
8 hours ago
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@lfusaso, nope, I only need something that reduces with a fixed number of percent per added die.
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– Himmators
8 hours ago
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@GreySage Thanks, sloppy copy :P
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– Himmators
8 hours ago
2
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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?
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– Xirema
8 hours ago