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

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} \$$

• Is there a reason your desired outcome doesn't begin with 16.67% failure rate on 1d6? – Ifusaso Feb 25 at 20:50
• 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? – Xirema Feb 25 at 21:12
• @Himmators That doesn't really answer AlexP's question. They roll 1 to 5 dice. What do they do afterwards? Do they sum all of the values from the dice rolled and compare to a target number set by the DM? Do they average the values from the dice rolled and compare to a target number set by the DM? Do they take the single highest value and compare to a target number set by the DM? Etc. – Anthony Grist Feb 26 at 14:35
• What happened to the answer that said "If the number of 1s exceed the number of 6s"? That was genius. Now I see all these "red/special die" answers that are weak and not half the answer that one or Craig Meyer's are... – Harper Feb 26 at 18:07
• @DavidCoffron deleted his own answer because he mis-applied some statistics, so the numbers were off. – Ifusaso Feb 26 at 18:14

## Fumble if ($$\2 \times\$$number of dice) $$\\gt\$$ (sum of dice except for any ones)

Equivalently: (number of ones)$$\+\$$ ($$\2 \times\$$number of dice) $$\\gt\$$ (sum of dice)

$$\\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{10.19%} \\ \text{4} & \text{7.48%} \\ \text{5} & \text{5.67%} \\ \hline \end{array} \$$

This is the most "linear" rule I've found without sacrificing too much practicality. When searching for a strategy, I wanted to avoid looking at any properties involving specially-marked dice or "ordered" dice. I limited myself to looking at linear sums of either the total number of dice, the counts of different values, or the sum of all the dice.

Assuming that the players are going to be calculating the sum of the dice anyways, they can check for a fumble by subtracting the number of 1s that were rolled, then subtracting twice the number of dice. If the result is negative then that is a fumble.

Edit: If you haven't finalized a decision on how exactly the dice rolls will determine success / failure, then I would recommend using "the sum of all dice excluding any 1s" as the value to be compared against the DM's target number. This way there is good integration of the above fumble mechanic into the rest of the dice-rolling mechanics.

## Fumble when (count of ones) $$\\gt 2 \times\$$ (count of fours, fives, and sixes)

$$\\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{8.80%} \\ \text{4} & \text{5.94%} \\ \text{5} & \text{4.45%} \\ \hline \end{array} \$$

The above rule is another practical rule I've found. A player can look at the dice results, count up how many 4/5/6s were rolled, multiply by 2, then subtract how many 1s were rolled. If they end up with a negative number, that's a fumble.

• I will try this one, Fumble when (count of ones) >2× (count of fours, fives, and sixes). This will be considered a risk of fumble, you have to then roll another skill-roll. If that also fails, you fumble, if you fumble again, you double fumble etc. – Himmators Feb 27 at 22:46
• Also, any idea about this but the inverse? That is, probability of critical success that increases with the number of die? – Himmators Feb 27 at 22:52
• @Himmators When the number of dice is 1, would you rather have a 16.7% chance of critical success or a 0% chance of critical success? You could try having critical success when count of sixes is strictly greater than count of ones, which has probabilities of ${16.7%, 25%, 29.6%, 32.5%, 34.4%}$. – PhiNotPi Feb 28 at 1:10
• I think 0% of crit on one die, I have described 1-die skill-level as "basically retarded", well a less offensive version that it but not sure what the english word would be... so more sixes than ones is nice. The probability of a crit gets a bit out of hand though at 34%. Any way to get it down a bit? – Himmators Mar 1 at 10:53

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 2\times 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} \$$

• This is equivalent to the average of the dice being two or less, so 2-1-3 is a fumble, 2-1-1-4 and 2-2-1-1-4 are fumbles. I think this would be pretty easy to see at a glance; you need more "points" above two than you have ones to avoid a fumble. – JollyJoker Feb 26 at 8:55
• +1 for simple, logical mechanics. Simpler rules are always better IMO. Saying that you fumble on any roll less than twice the number of die rolled seems pretty straightforward to me. – D.Spetz Feb 26 at 15:26

# 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.

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

* - at my table I often use the positions of dice, the ordering of dice, and even the orientation of dice as they fall to inform various effects. (I do't like to throw away good information.) I haven't yet had a player complain that it's hard to tell which die is on the left--they generally count/read left-to-right anyway.

• 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. – Ryan Thompson Feb 26 at 3:28
• @RyanThompson Yeah, a dice cup. – KorvinStarmast Feb 26 at 3:51
• @VLAZ it's gone because it served its purpose: it prompted me to re-think and re-explain something. I don't know why you didn't get an inbox notification, though. – nitsua60 Feb 28 at 15:16

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} \$$

• If if feels weird for high rolls to produce fumbles on the second roll, you could also say that a fumble is avoided if the total of the reroll and the number of dice rolled is greater than 7. This yields the same probabilities, but with fumbles always caused by lower rolls. The trade-off is that it makes the math a bit more complicated, of course. – Ryan Thompson Feb 26 at 19:10
• A simple variation of this, is that the roll is a fumble if the red die rolls a 1 and the overall roll is a failure. This removes the need for re-rolls, but changes the dynamics somewhat away from OP's desired fixed percentages. However it is a reasonable mechanic IMO, because player who rolled a red 1 and 3 normal 6's and would have succeeded well on the test, would feel somewhat robbed if that was a fumble. – Neil Slater Feb 27 at 8:35

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.

• You understate the elegance of this approach. Most of the other answers use techniques that are at least as complicated, yet yield results that are blatantly exponential. Yours just requires that they roll at least 8-N on the yellow die, and gives perfectly linear results. – Ray Feb 27 at 3:50

## More Elegance?

I read nitsua60's answer (which was the accepted answer at the time) and found that it's not very elegant to use the location of the dice. I'd prefer something that's simple, doesn't require rolling and doesn't require the extra bookkeeping of things like where the dice end up on the table. Therefor, I came up with something that meets those goals at the cost of being less precise. This solution does not adhere to your example percentages, nor is it actually linear, but on the scale from 1 to 5 dice, I think it'll feel close enough to linear.

## A red die

So, one of the dice is different from the others. The easiest is if it's a different color. We'll call this the red die. You always roll the red die. That means that if you roll one die, it'll be the red die and if you roll more than one die, it'll be the red die and a number of other dice.

Now, if the red die comes up a 1 and none of the dice comes up a 6, it's a fumble. The probability of fumbling will be as follows:

$$\\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{11.57%} \\ \text{4} & \text{9.65%} \\ \text{5} & \text{8.04%} \\ \hline \end{array} \$$

Note that adding a second die decreases the chance by approximately 2.8 percentage points, while adding the fifth die will decrease the chance by approximately 1.6 percentage points. This is not quite linear, but I do feel that it's close enough to feel more or less linear.

## Side effects

Originally, this answer had a solution that was "a red 1 and no other ones", but based on a suggestion in the comments by both @Nick543211 and @NathanHinchey I changed it to "a red 1 and no sixes". This has the same probability distribution, but seems to have fewer side effects. For example, the original system had rolls that feel like they should have been fumbles (like three ones) but weren't.

Another side effect is that this changes the probability distribution of success and failure rates somewhat. There may be some rolls that would have been a success without the proposed fumble system, but now end up being a fumble. For example - if we assume that the dice are simply added up and then compared to the target number - you could roll two fives and a red one when you're up against a target of 10. The change has made this problem occur far less and I'd say it's not that of a problem anymore. You could still try to balance it with a "critical success" system if you do feel it's a problem.

• This is a good answer, but I believe their side effects can be removed by looking at a 6 on the other dice instead of 1's. So if the Red Die is a 1 and no dice are a 6 than it is a critical fail. Similarly, if the Red Die is a 6 and no other die is a 1 than it is a critical success. This method would still keep the same probabilities. – RuggedMcNugget Feb 26 at 16:56
• A slight improvement to the feel would be to change if the red die comes up a 1 and none of the other dice does, it's a fumble to if the red die comes up a 1 and you don't also roll a 6, it's a fumble Probabilities remain the same, but 1s don't mean both success and failure. [EDIT: just noticed an answer that should be a comment that says this exact thing.] – Nathan Hinchey Feb 26 at 21:54
• @Nick543211 Weird side effect of this is that odds of critical success decrease with more dice. – Nathan Hinchey Feb 26 at 21:59
• @Nick543211 I’ve converted your post into the comment above. Cheers! – SevenSidedDie Feb 26 at 22:48
• Gotta upvote this: it's the correct implementation of what I was shooting for in my other answer (10K+ visible only). Though I do wish "elegance" were defined.... – nitsua60 Feb 27 at 4:49

Add an extra die of a different color, the fumble die.
You fumble if the fumble die is a 1 and all other dice are different. If there is no fumble the fumble die is removed and does not participate in the success/fail of the skill.

So, in example, if the player has skill 2 then 3 dice would be rolled. A fumble die and 2 skill dice. If the fumble die is a 1 and the other two are different the player has fumbled. Thus a 1 in fumble die and 1,2 in the skill dice would be a fumble. And so a 1 and 3,6. A 2 in fumble die would be no fumble regardless of the skill dice. A 1 in fumble die and 3,3 would also not be a fumble.

This method is very simple are requires no calculations.

Don't be bothered by all ones not being a fumble. Nobody is gonna complain about it when they roll that.

$$\\begin{array}{|c|c|c|} \hline \textbf{Skill} & \textbf{Wanted Prob.} & \textbf{Probability} & \textbf{Difference}\\ \hline \text{1} & \text{18%} & \text{16.7%} & \text{1.3%}\\ \text{2} & \text{15%} & \text{13.9%} & \text{1.1%} \\ \text{3} & \text{12%} & \text{9.3%} & \text{2.7%} \\ \text{4} & \text{9%} & \text{4.7%} & \text{4.3%} \\ \text{5} & \text{6%} & \text{1.5%} & \text{4.5%} \\ \text{6} & \text{3%} & \text{0.3%} & \text{2.7%} \\ \text{7} & \text{0%} & \text{0%} & \text{0%} \\ \hline \end{array} \$$

This method does not require having any special colored dice, keeping track of dice position, or performing extra rolls. It also doesn't require much math and the probabilities are somewhat close to the desired ones you listed.

# Rules:

• Whenever you roll any number of dice each die will fall into one of the following categories:
• {1}: fumble
• {2, 3}: neutral
• {4, 5}: save
• {6}: critical save
• If there is at least one [critical save] die then regardless of any other dice, a fumble is averted
• Otherwise, if the number of [fumble] dice is greater than the number of [save] dice then the roll is a fumble

• 1
• 1, 3
• 1, 1, 3
• 1, 1, 5
• 1, 2, 3

• 2
• 1, 4
• 2, 3
• 1, 1, 6
• 1, 2, 4

# Probabilities:

These have been determined experimentally with 1,000,000,000 rolls for each number of dice.

$$\\begin{array}{|c|c|c|} \hline \textbf{Dice} & \textbf{Probability of Fumble} \\ \hline \text{1} & \text{16.67%} \\ \text{2} & \text{13.89%} \\ \text{3} & \text{11.58%} \\ \text{4} & \text{9.34%} \\ \text{5} & \text{7.47%} \\ \hline \end{array} \$$

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Another way to do this is to just rule fumble as having anything below all 2s. That means for 2 die you would need at least 4 to avoid a fumble. This is memorable and can be used without a unique die.

$$\\begin{array}{|c|c|c|} \hline \textbf{Dice} & \textbf{Fumble Range} & \textbf{Probability} \\ \hline \text{1} & \text{1} & \text{16.7%} \\ \text{2} & \text{2-3} & \text{8.33%} \\ \text{3} & \text{3-5} & \text{4.63%} \\ \text{4} & \text{4-7} & \text{2.70%} \\ \text{5} & \text{5-9} & \text{1.62%} \\ \hline \end{array} \$$

For higher chances of fumble, just rule out all 2s too, except for one die situation (since that would make 33% chance of fumble).

$$\\begin{array}{|c|c|c|} \hline \textbf{Dice} & \textbf{Fumble Range} & \textbf{Probability} \\ \hline \text{1} & \text{1} & \text{16.7%} \\ \text{2} & \text{2-4} & \text{16.7%} \\ \text{3} & \text{3-5} & \text{9.26%} \\ \text{4} & \text{4-7} & \text{5.40%} \\ \text{5} & \text{5-9} & \text{3.24%} \\ \hline \end{array} \$$

• Note that the left-most die solution does not require unique dice. In fact, if the left-most die is ignored and you look at the second one from the left if the first number has been rolled more than once... (so you'd either need a set of five unique dice, or use, as suggested in that answer, look at the direction from left to right) – Jasper Feb 26 at 15:17
• @Jasper ah right I missed that part. I'll leave this here anyway for future reference :) – John Hamilton Feb 27 at 5:48
• Well, it's only a commentary of the first line of the answer, not of the proposed solution. As such, I'd recommend just editing the first line to say something like "another way to do this is.." as then the answer is simply a good one :) – Jasper Feb 27 at 10:23