How can the probability of a fumble decrease linearly with more dice?Is it possible to produce a bowl-shaped...
Can you determine if focus is sharp without diopter adjustment if your sight is imperfect?
Microtypography protrusion with Polish quotation marks
Why is it that Bernie Sanders is always called a "socialist"?
Broad Strokes - missing letter riddle
Is there a file that always exists and a 'normal' user can't lstat it?
Boss asked me to sign a resignation paper without a date on it along with my new contract
Do authors have to be politically correct in article-writing?
Coworker asking me to not bring cakes due to self control issue. What should I do?
Not a Long-Winded Riddle
Am I correct in stating that the study of topology is purely theoretical?
What to do with threats of blacklisting?
What species should be used for storage of human minds?
What's this assembly doing?
The No-Straight Maze
Categorical Unification of Jordan Holder Theorems
Prevent Nautilus / Nemo from creating .Trash-1000 folder in mounted devices
If angels and devils are the same species, why would their mortal offspring appear physically different?
Is there a way to not have to poll the UART of an AVR?
Single-row INSERT...SELECT much slower than separate SELECT
How vim overwrites readonly mode?
Stuck on a Geometry Puzzle
Reading Mishnayos without understanding
Why is the "Domain users" group missing from this Powershell AD Query?
Find the smallest value of the function
How can the probability of a fumble decrease linearly with more dice?
Is it possible to produce a bowl-shaped probability curve with dice rolls?What is the probability for a thousand with 3 D10Calculating dice pool probability with limited rerollsHow do I calculate dice probability in the A Song of Ice and Fire system?Dice pool success probability with one exploding dieHow do I approach the probability of a d100, using 2 d100's?Improving “fumbles” when using pools dice in TROS, part 2How often should I be meeting monsters?How are a dice pool's probabilities affected by being allowed/forced to add dice then drop high/low dice?Help with probability for a complicated dice pool mechanic
$begingroup$
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
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
|
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 1 hour ago
Ruse
6,24311251
6,24311251
New contributor
asked 5 hours ago
HimmatorsHimmators
1285
1285
New contributor
New contributor
$begingroup$
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
4 Answers
4
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
2 hours ago
add a comment |
$begingroup$
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%.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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}
$endgroup$
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.
$endgroup$
$begingroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "122"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Himmators is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2frpg.stackexchange.com%2fquestions%2f141939%2fhow-can-the-probability-of-a-fumble-decrease-linearly-with-more-dice%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
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
2 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
2 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 3 hours ago
V2Blast
23.3k375146
23.3k375146
New contributor
answered 4 hours ago
Craig MeierCraig Meier
2263
2263
New contributor
New contributor
$begingroup$
+1 for mathemagics!
$endgroup$
– Harper
2 hours ago
add a comment |
$begingroup$
+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
add a comment |
$begingroup$
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%.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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%.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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%.
$endgroup$
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%.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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}
$endgroup$
add a comment |
$begingroup$
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}
$endgroup$
add a comment |
$begingroup$
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}
$endgroup$
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}
answered 2 hours ago
nitsua60♦nitsua60
75.2k13309431
75.2k13309431
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
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
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
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
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 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
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
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
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.
Thanks for contributing an answer to Role-playing Games Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2frpg.stackexchange.com%2fquestions%2f141939%2fhow-can-the-probability-of-a-fumble-decrease-linearly-with-more-dice%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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