There are a number of spells and abilities that involve large amounts of dice rolling. For example, a high level rogue can have up to 10d6 in Sneak Attack, and a high level Wizard can have a staggering 40d6 for Disintegrate. This is DnD, but I'm fairly sure there are other games with similar issues.

I don't really think that rolling and adding up a d6 for several minutes is fun to watch.

Now for the 40d6, I figured out: the least you can roll is 40, the most is 240, so maybe you can roll percentages twice, add them up and add 40, but that's probably not as accurate and would have a different distribution.

So, how do you deal with rolling this many dice? More specifically: what methods exist that can reduce the amount of dice you need to roll with minimal loss of the randomness properties of the original roll?


20 Answers 20


At the moment I can only see two solutions to this particular problem. The unfortunate thing is that both solutions offer their own challenges and difficulties as well. I agree with you that rolling a bunch of dice and adding it all together is tedious- let's see what we can do about that!

Let the Robots take care of it

For large sums of dice getting rolled together, a great way to speed up the process is to use a Dice Roller program that can be found through a quick Google search (I'm at my office right now and can't link you, sorry.) You type in how many dice you want to roll, select the value, and the computer instantaneously gives you the sum.

You can either use truly random dice generators like Random.org's, which are based off atmospheric data. Other generators will be pseudorandom number generators which run off a mathematical algorithm. The difference between either doesn't really matter unless you're doing cryptography or quantum mechanics, so either is just as good as the real dice for people playing RPG at a table.


You could also, when there is a ridiculously large amount of dice to be rolled, use average value instead. So 40d6 becomes 140 damage (3.5 x 40). No rolling, just flat damage.

The Problem

People like rolling dice. It is the part of the game that is most "tactile" and therefore is very satisfying to many people who enjoy the hobby. Clicking a button on the computer can sort of emulate the feeling, but just doing average value cheapens the experience. Not to mention that some players may balk at the idea of not getting the chance to do more than average damage with their awesome spell/ability.

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    \$\begingroup\$ For all intents and purposes (of the layman) the random numbers output by any computer are "truely random". So unless you do cryptographic calculations, quantum mechanics or the like you don't need to worry about the pseudo randomness of your computers results. \$\endgroup\$
    – fgysin
    Commented Mar 24, 2015 at 12:15

The more dice you roll, the more the outcome focusses on the average, as the high and low rolls start evening out. While there's a reasonable chance of getting a 2 on 2d6 (1 in 36) there's no chance in hell you're getting a 240 out of 40d6 (1/6^40, vastly less than one in a trillion, I don't even know how to describe this number)

The copout

The easiest fix is to simple average them (ie; 40d6 simple deals 140 damage, the average outcome). However, this removes the dice entirely and is also considered to be "not fun" by many (because rolling a fistful of dice is awesome)

So here's two other ways to keep some randomness without requiring lots of dice rolling.

Average out part of the dice

Any dice over a preset amount are averaged, you only roll up to X dice. So if you say "any dice over 5 are discarded" then rolling a 40d6 disintigrate means you average 35d6 (for a total of 122 damage) and then roll and add an extra 5d6. This puts your damage output between 127 and 152; less severe than if you roll but realistically speaking the odds of going outside these bounds are really small anyway.

Divide and multiply

This is more swingy, which some people say is a good thing and others say is not. You divide the number of dice by an amount to make it more managable, roll those, add them up, then multiply back.

So for example, instead of rolling 40d6, you instead roll 4d6 and then multiply the total by 10. This still gives you the full range of 40-240, but makes extreme (high or low) outcomes vastly more likely.

For an example of the swinginess: Consider that with 4d6, you'll have an 0.08% chance of rolling 240 (and the same odds for 40). An 0.08% chance is over a trillion times more likely than with 40d6, so using divide and multiply you'll probably actually see such results. The closer you get to 1d6, the more uniform the results get, and the more likely the extremes get.

  • \$\begingroup\$ How do you divide and multiply dice pools that are not multiples of 5? This method looks a little odd. \$\endgroup\$ Commented Mar 24, 2015 at 22:27
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    \$\begingroup\$ Roll a couple spare dice. 53d6, roll 5d6*10+3d6. MATH!!! \$\endgroup\$
    – mxyzplk
    Commented Mar 25, 2015 at 0:38
  • 1
    \$\begingroup\$ You can pick any divisor you like. In the case of a prime number, either average out the rest or roll them seperately. (ie; 23d6 = 4d6x5 + 3d6 or 4d6x5 + 10) \$\endgroup\$
    – Erik
    Commented Mar 26, 2015 at 14:13
  • \$\begingroup\$ @pureferret obviously the more you cluster, the more "wiggle" you get in the graph, despite averages staying the same. anydice.com/program/587c \$\endgroup\$
    – mxyzplk
    Commented Mar 26, 2015 at 14:44
  • \$\begingroup\$ @mxyzplk: You can actually get even the amount of "wiggle" (i.e. variance) to match, by combining the "average out" and "multiply" methods in the correct proprotion; see Glen_b's and my answer below for the math. \$\endgroup\$ Commented Sep 10, 2015 at 12:51

I'll add one more to the list:

Fast Counting Bulk Rolls

As has been pointed out the biggest problem with this number of dice is the addition. You can make this easier by grouping the dice into sets of 10 points after the roll. Even with a low number of dice I have found this to speed up counting.

1) Roll your huge pile-o-damage.

2) Sort your dice into groups of 10 points.

3) Count the groups and then add the remaining dice.

Then Counting 11 sets plus 8 points for 118 is way faster than trying to add the dice up one at a time and keep the running total. After trying it a couple of times it becomes second nature to just grab all the 6's and 4's to set up groups and match all the 5's together. Give it a try, you might be surprised at how it works out.


There's really 3 solutions:

  • Skip rolling entirely, and use the expected result (similar to DnD's "Take 10" mechanic). In the long run, this should not produce any statistical advantage or disadvantage for your players. However, in the short term your players will act differently since it is more deterministic.
  • Approximate the same distribution with a less laborious process.
  • Use a computer to roll your dice for you.

I'm assuming you don't have access to a computer, so I'll skip the last option.

"Take 10"

The average result of a d6 roll is 3.5, so the average of 40d6 will be 40*3.5=140 . If you don't mind taking luck out of the equation, you can just offer your player the option of taking 140, and your player will have neither gained or lost anything by taking that.

It so happens that most of the time, 40d6 will give a result very close to 140 anyway, because adding together many dice rolls reduces variance.


Rolling multiple dice and adding up their results approximates a normal (aka Gaussian) distribution. All Gaussian distributions are characterized by two variables: The mean (expected value) and standard deviation (spread). You can transform any Gaussian distribution into another one by multiplying (changes SD) and adding numbers (changes mean).

If you have access to a random number generator that outputs numbers following a Gaussian distribution with mean M and standard deviation S, you can use that to easily generate numbers of any Gaussian distribution. If you want a distribution with mean m and SD s, you simply take the output from your M/S distribution, multiply by s/S, and add m-M. You can easily measure M and S (since you only have to do it once for each RNG device), but knowing m and s (which change for each desired distribution) is a problem you would need to solve.

There are many natural processes that generate Gaussian numbers. As said above, summed die rolls is one kind of these: You could, for instance, use 4d6 as your Gaussian RNG. There are better ones, for example throwing a needle on a grid of parallel lines or applying the Box-Muller transform to a uniform RNG like d100 (you would realistically create a pre-computed table for this).

One quick and dirty way would be as follows:

  1. Generate a table of 100 Gaussians
  2. Number them from 1 to 100
  3. Print them out
  4. Roll d100 to choose one at random
  5. Transform by multiplying/adding

You would need to pick around 100 numbers this way before it starts behaving differently from an actual Gaussian distribution (and you could easily print a new table before every session).

With 4d6, you would start by comparing the two distributions (I used WolframAlpha):

enter image description here

So based on this, you can roll 4d6 instead 40d6, multiply by 3 (actually 10.8/3.416=3.16 but rounding is easier) to get the same spread/standard deviation, and add 140-3*14=98 to the result to get the same mean/expected value. So (3*4d6)+98 would approximate 40d6 quite well:

enter image description here

What you lose is resolution: If you zoom into the simulated bell curve, you will see coarser steps than with 40d6. The human mind is not capable of perceiving this difference, so your players won't be able to tell (unless you play with prodigious statisticians, but if that was the case, you would not be asking this question).


There is a way to make a single roll and keep the probability distribution, more or less. It will require some preparations from you, though.

For example, let's convert Rogue's 10d6 Sneak Attack into a single d20 roll. Go to anydice.com, put "output 10d6" and it will give you statistics for the roll. Select At Most and Table (or Export). You will get two columns: the value and the chance that the roll will be not more than this value. We can use this statistics to convert a result of a d20 roll (which is uniformly distributed from 1 to 20) into a reasonable approximation of a 10d6 roll.

One step of d20 is a 5% chance. Go through the column and pick the values that more or less correspond to chances that are multiples of 5%: e.g. you have 3.9% for 25 and 5.8% for 26, pick 26. The numbers are too dense in the middle, thus 33 will count for both 35% and 40%. In the end you will get a list of 20 numbers;

\begin{array}{rl| rl| rl| rl} \textbf{d20} & \textbf{result} &\textbf{d20} & \textbf{result} &\textbf{d20} & \textbf{result} &\textbf{d20} & \textbf{result} \\\hline 1& 26& 6& 32& 11& 35& 16& 39\\ 2& 28& 7& 33& 12& 36& 17& 40\\ 3& 29& 8& 33& 13& 37& 18& 42\\ 4& 30& 9& 34& 14& 37& 19& 43\\ 5& 31& 10& 35& 15& 38& 20& 60\\ \end{array}

You can see that this table is not symmetric, we can reach the maximum damage, but not the minimum. That's because we should've picked 2.5%, 7.5%… instead of 5%, 10%… But for rolls from 2 to 19 it won't make a change bigger that a single point of damage. The only big difference is for rolls of 1 and 20. There are two options:

  1. Keep things realistic and pick values corresponding to 2.5% and 97.5%: 24 and 45.

  2. Make things more exciting and allow for a chance of getting extreme values of damage, assigning 10 and 60 or something in between (15/55, 20/40…)

Since there are not that many rolls like this, you can prepare a table for each repeating the steps above. When you need to roll for damage, roll d20 and take the number. For 40d6 you can use the same table and roll 4 times.

  • 6
    \$\begingroup\$ It would probably be slightly more accurate to make the d20 rolls align with 2.5%, 7.5%, 12.5%... 87.5%, 92.5%, 97.5%. With your current distribution, it's possible to get the max, but not the minimum. This distribution associates each d20 side with the middle of each 5% subrange rather than the top of each subrange. \$\endgroup\$ Commented Mar 24, 2015 at 22:08

Building on Erik's excellent answer, here's a way to very precisely approximate large die pools using only about 25% of the dice:

  1. If the number of dice is not a multiple of 4, set aside 1 to 3 dice to make it so.

  2. Roll 1/4 of the dice, add them up, and double the result.

  3. Add half of the expected average of the original roll to the result.

  4. Roll the extra dice set aside in step 1 (if any) and add them to the result.

For example, for 40d6, you can roll 10d6, double them, and add 70.

More generally, for (4×N)d6, roll Nd6, double them, and add 7 × N.

Even more generally, for (4×N)dX, roll NdX, double them, and add (X+1) × N.

This method gives exactly the same mean and variance as rolling the full die pool. For N ≥ 3, it also approximates the higher moments of the probability distribution fairly well. The low and high tails of the probability distribution will be somewhat truncated, but for large numbers of dice, such very low or high rolls are astronomically unlikely anyway.

The most important way in which this approximation differs from rolling all the dice is that the result of step 2 will always be even. Thus, if you don't have any remainder dice set aside in step 1, the parity of the outcome will always be fixed: a roll of 2 × 40d6 + 70 may yield 138 or 140 or 142, but never 139 or 141!

This only happens if the number of dice to be rolled is an exact multiple of four, so that no dice are set aside in step 1; even a single undoubled roll is enough to nearly perfectly smooth out the distribution. If the number of dice is a multiple of four, and you feel that the fixed parity could be an issue, you can always choose to set aside four dice instead of none in step 1 to avoid it.

Ps. Here's a similar method that only needs about 1/9 of the dice compared to a normal roll:

  1. If the number of dice is not a multiple of 9, set aside enough dice to make it so. (If it's a multiple of 10, you can just set aside 1/10 of the dice.)

  2. Roll 1/9 of the dice, add them up, and triple the result.

  3. Add 2/3 of the expected average of the original roll to the result.

  4. Roll the extra dice set aside in step 1 (if any) and add them to the result.

Thus, for 99d6, you can roll 11d6, triple them, and add 231 = 7 × 3 × 11. (To get 100d6, roll one more d6 and add it to the result.) More generally, for (9×N)dX, roll NdX, triple it, and add (X+1) × 3 × N.

If you want, you can extend this to even larger multipliers. The only requirement is that the total number of dice in the pool must be divisible by a square (2² = 4, 3² = 9, 4² = 16, 5² = 25, etc.). Specifically, if the number of dice N is a multiple of n², then you can roll N / n² dice, multiply the result by n, and add in the average of the remaining N × (n − 1) / n dice.

The reason this trick works is that the mean of a roll of N dice is proportional to N, while the standard deviation is proportional to the square root of N. Meanwhile, multiplying the outcome of a roll by a constant C multiplies both the mean and the standard deviation by C, while adding a constant obviously increases the mean but does not change the standard deviation. Thus, dividing the number of dice in the pool by n², and then multiplying the outcome by n, yields the correct standard deviation, but the mean is only 1/n of what we want. Adding in a constant amount equal to (n−1) / n of the expected mean then corrects this, giving both the correct mean and the correct standard deviation.

Pps. Here's an AnyDice program showing different approximations of 40d6. You can verify that the mean and the standard deviation both match; the full distributions are also very similar, except of course that the "70 + double 10d6" method never gives an odd result (and thus makes even results approximately twice as likely).

  • 1
    \$\begingroup\$ Why not recommend setting aside four-not-zero remainder dice on the first pass, guaranteeing better results? Later in the post you could explain why setting aside zero is not preferred. (And I agree with your comment on Glen_B's answer: this presentation is clearer and the answer, I think, the better for it.) \$\endgroup\$
    – nitsua60
    Commented Feb 16, 2016 at 3:03

Count 4,5,6

Counting dice with a specific result is quite fast, even more so when you group multiple results together. So instead of adding up all results, quickly count the number of dice greater or equal than 4.

Statistically, rolls of 4, 5 and 6 average out to 5, so count every such result as 5 (which is a very nice number for doing math with, especially multiplication).

The rest of the rolls average out to 2, but counting them as such actually makes the distribution a bit weird, since you used the random variable "count of {4,5,6}" already and "count of {1,2,3}" is just the "conjugate". Therefore, we just say that on average, half of the dice are 1, 2, or 3, and count them all as 2, effectively adding the number of dice rolled to the previous result.

For a theoretical 40d6 roll, you might end up adding together 25 × 5, and 15 × 2 (15 being the remaining number of dice), to get 155.

Taking it all together, we end up with

 5*[count {4,5,6} in Xd6] + X
 `----------.-----------´   |
  high rolls count as 5    avg. of rest, on avg.

You can check the distribution with AnyDice. The average result matches quite well, although the spread of the distribution is around 1.5 that of the real one, making this method a little more swingy (which might not actually be a problem, depending on your groups preferences).

Note that this technique only works nicely for d6. For example, with a d8, the average of the four high rolls is 6.5, half the average of the low ones is 1.25, which makes the math really awkward and requires flooring/rounding in the end.

  • \$\begingroup\$ For a d8 just count the low rolls as 2 and the high rolls as 7. It's both closer on spread than using the average of the high numbers or the average of the low numbers --- and easier because the numbers are round. The correct standard deviation is just below 2.3, using 2 and 7 gives 2.5 instead, and using 2.5 and 6.5 gives a standard deviation of 2. \$\endgroup\$
    – Glen_b
    Commented Apr 20, 2023 at 8:02
  • \$\begingroup\$ In general you want to pick a value a bit beyond the middle of each half, because the squaring in a standard deviation weights it more toward the outside values. \$\endgroup\$
    – Glen_b
    Commented Apr 20, 2023 at 8:34

Groups of tens

In Starwars d6 we sometimes deal with dice rolls of 10-13 d6 (extreme cases) - adding these up is fairly easy I think.

  • What I do personally is quickly grouping them in sums of 10. E.g. I'd group a 6 and a 4, two 5s, two 3s and a 4, ... and eventually add up the rest. For me this method is quick and reliable.

Huge numbers of dice > go digital

The only experience I have with rolling huge amounts of dice is from the Warhammer table top games (Fantasy and 40k). There the problem is solved by basically filtering the throws over multiple levels, e.g. you roll 24 attacks, keep all above 3, then roll for example 11 resistance rolls, keep all dice above 4, etc. ... Thus dice get fewer after several rolls. But you still basically need a lot of dice to make this feasible, I remember having around 20 or 30 small d6.

  • So for 20-30+ d6 that you simply need to add up I'd suggest getting a mobile app that does that for you (I'm sure there are many) or use an online dice tool (there are even more).

This way you'll also ensure that the distribution/spread and the variance or your result is exactly the one specified by the rules.


Let 'em roll. Part of the fun, excitement, and suspense was always watching and counting the results. Rarely did a player have enough d6 to make a roll such as that, so we'd group them into a reasonable number, say 10d6, four times.

If the total damage by roll 3 was enough to kill the target, then you're done. As DM, I'd declare that all that was left was a pile of cinder and move on. No need to do the fourth lot.

But, if batches two and three were particularly poor, and there was a chance that the target would survive, well that piqued everyone's interest.

I know the statistics, and inference by distribution and probability never appealed to those I played with. The tactile feel and sound of rolling dice was also a part of the game; the true, in your face, participation of "the Fates" in the outcome.

I played with one DM who has a special d20 that he reserved for suspenseful situations, where the outcome could mean life or death, or major changes in the outcome. That d20 was large (about a 1.5 inch 'diameter'), and so worn that there weren't any sharp edges left. With a gentle roll, that danged thing would wander around on the table for a good 10-15 seconds before finally settling to a stop. The outcome was whichever facet was closest to "face up". We all watched it closely, but there was never any rhyme or reason to the result, it was equally beat up on all the edges, and still sufficiently random. But the suspense it generated was well worth the 10-15 seconds to get the answer.

  • 6
    \$\begingroup\$ In every group I've ever played with, we still roll that last set of d6s, just to see how fine the ash was. ... we ... may have problems ... \$\endgroup\$
    – minnmass
    Commented Mar 25, 2015 at 5:58

Share it out

If you have 40d6 to roll; have yourself and three others each roll 10d6 and add them up, then add the totals.

Saves some time I've found, only really used this for D&D fireballs however; otherwise I use a phone dice roller.


If you are willing to enter the dark domain of digital dice rollers, and have an iOS device, I cannot recommend RPG Roller enough. It is completely free with no in-app purchases or advertisements of any kind, and it is very high quality software. I've used it many times and the author seems to be very serious about making a good dice roller. It randomizes with accelerometer data, and provides rich dice formula generators, configurable pages of "hot buttons" to represent different characters, items or scenarios, roll history, formula history and even randomness analysis. All for free.

And, no, I didn't write it, nor am I involved with the author in any way.

RPG Roller screens


There are a lot of great solutions already, but I have one more to add:

Roll Them Beforehand

The problem is counting dice while playing, right? So roll and add up the dice 20 times between sessions, writing down the results. No fudging! When it comes time to roll damage, roll a d20 and use that result to pick from the list. That way, the player still gets to roll their dice for damage, but you all save the time of having to count up large numbers of dice.

Just make sure to roll new numbers each session. You can (and probably should) keep the results list hidden from the players.

And if they want the feeling of rolling forty dice all at once, let them roll the dice - just don't count them, and use the d20 table result instead.


The "Big dice-small dice" approach

This gives you really close to the right distribution but saves a lot of rolling; I'll only describe how to deal with d6's here, and only by using d6's.

You can do way more than what I cover here, but what I cover here is easy to remember.

Method 1 -- Tripling for 10 dice:

Take some large d6s (the big dice) and the same number of small d6s (e.g. 4 of each),

Triple the big dice, add the small dice and add 21 (7 x 3) for every big die.

This only works well for at least 3 big dice (and its better for 4 big dice).

The result has almost the same distribution as rolling as many lots of 10d6 as the big dice you have - same mean, same variance, almost the same distribution shape. You use 1/5 th as many dice.

So for 30d6, roll 3 big dice, 3 small dice and add 63

For 40d6, roll 4 big dice, 4 small dice and add 84

For 50 dice, roll 5 big dice, 5 small dice and add 105

For say 42 dice, you'd do the trick for 40 dice but roll 2 more small dice. Easy. You could do 39d6 by rolling one fewer of the small dice than for 40d6 (i.e. 3 small rather than 4), but you should try to always leave at least a couple of small dice in there. So rather than doing 37d6 as 4 big and 1 small do it as 3 big and 3 small + 63 (for 30 dice), plus 7 more small. Thirteen dice is still a lot better than 37.

[Of course they don't have to be physically different in size, just distinguishable (e.g. red vs white works as well as big vs small), but if you do have two different sizes that helps remind you how you're supposed to treat them.]

Here's how that looks for 40 dice:

enter image description here

As you can see it comes really close. (The noise makes that look a little worse than it is -- the actual underlying distributions are closer together than that.)

Method 2: Quintupling for 25 dice

Quintuple the big dice, add 70 (7 x 10) for every big die. Here you only add as many small dice as you need beyond the multiple of 25.

This is most useful for dealing with hundreds of dice, and shouldn't be used with less than three big dice.

So for example if you have more than 100 dice, replace 100 of the dice by 4d6 x 5 + 280.

Again, you should try to have at least a couple of small dice in the mix or the results will be "clumpy" (but still right on average).

Here's an example with 160 dice.

Six 25's make 150, so take six big dice, quintupled, + 6 x 70 to replace 150d6, and add the remaining 10d6 as 10 small dice. It's a lot easier to do that than roll 160 dice.

enter image description here

[Tricks like grouping into sets of 10 when finding totals can still be done with these approaches.]

(Edited in response to comments from nitsua60 about it not being clear how this works]

How this works

Sums of multiple dice (3 or more) have a distribution* which is approximately normal. The critical thing in trying to make the distribution of some other rolling scheme than rolling all the dice the same is then to get something that's also approximately normal and has the same variance (and so the same standard deviation - which is how spread-out it is). You can then make the means the same by shifting the result.

*(I mean the cdf, specifically)

From basic properties of variance, when you multiply the result on a die by a number, you multiply its variance by the square of that number, and when you add a number to the result, you don't change the variance.

A single die with s sides has average (s+1)/2 and variance (s+1)(s-1)/12, so for example a d6 has average 3.5 and variance 35/12.

When you sum the results on multiple dice you add both the averages and the variances for the individual dice.

So if you triple a big die the variance is multiplied by 9, and when you then add a small die, you make that combination have the variance of 10 small dice (since we want to replace a group of small dice that's a round number). But the average of 10 small dice is 10 times the average of 1 die, while the average for one tripled die and one normal die is only four times the average of 1 die -- so you have to add on the average of six dice for each "one tripled die and one normal die" combination in your total.

For example, 30d6 has an average of 30 x 3.5 = 105 and a variance of 30x35/12 = 350/4. Each "tripled big die + small die" has the same variance as 10d6, but the average is 14 rather than 35 -- you need to add 21 for each "tripled big die + small die" combination. That is the mean and variance of 10d6 is the same as one "tripled big die + small die + 21".

Now to get close to that normal shape, you need at least 3 big dice (preferably more, though if there are more small dice you can sometimes get away with fewer).

The quintupling works the same but the variance is 25 times as big which is already a round number, so we don't need to add any small dice to make it round -- as long as there are plenty of small dice in the mix, and at least 3 big dice it should work too.


Based on experience with Warhammer, which uses buckets of dice, sorting by result speeds things up.

You roll 40 dice, then quickly sort them by result: all dice showing 1 together, all dice showing 2 together, etc. Then you can quickly sum them by multiplying number of dice by the face result of that set, and adding the totals of the sets together.

For example, you might roll 40 dice and sort them into eight 1s (8), ten 2s (20), six 3s (18), seven 4s (28), four 5s (20), five 6s (30), for a total of 124.


If you are really strapped for cash, this isn't a good option, but you could simply buy more dice. You still need to add up all the numbers, but at least you can roll 5, 10, or more dice at once. And if your whole group pitches in, the cost wouldn't be that significant. For example, you can currently get a pack of 100 d6's for only $10 on amazon. You can also get a pack of 100 random polyhedral dice for $20.

Personally, I really enjoy the action of rolling dice, and it feels really great when my character levels up, and I get to roll 5 die instead of 4 at once, for example.


I seldom have to roll more than 8D6 on a regular basis, but to do that, I take 8 D6, shake them for a few seconds, roll and add. There's multiple ways of adding fast ("grouping into tens", "grouping into sevens", "count each individual result, multiply and add" or simply just doing a running tally).

If I needed to roll 40D6, I would probably do it as 4 consecutive rolls of 10D6, taking probably around 2 minutes, in total (10D6 fits in my hands, 40D6 would need a shaker cup, dice tower or similar, which I normally don't have; plus I don't actually have 40D6 with me on a regular basis). If speed was of the essence, I'd probably roll 10D6 once, then multiply by 4. It's not an identical distribution, but the overall spread and likelihood are close enough that I'd accept that as a decent speed-up.

  • \$\begingroup\$ As to what my "daily dice carry" is, I have a box of 12 D6, slightly transparent; Chessex, I believe. I have 3D6 that were custom-designed by one of the players in my GURPS group and another 5-or-so D6, for a total of about 20 D6 in a pocket in my backpack. \$\endgroup\$
    – Vatine
    Commented Sep 10, 2015 at 10:36
  • \$\begingroup\$ I timed this yesterday, rolling and summing 10D6 took me about 25 seconds (varied between 21 and 27 seconds, over six trials). Based on that, 40D6 should be rollable-and-summable in "under two minutes". \$\endgroup\$
    – Vatine
    Commented Sep 14, 2015 at 9:56

Depending on the way your style of play, you may be able to fudge dice rolls.

However you should only ever do this if it makes the game more fun for the players.

So if counting large pools of dice is taking a lot of time and taking fun away from the game, then fudging it is a valid solution.

For example, if they roll 40 dice and they appear at a glance to be decent rolls - or even poor rolls if what they are trying to do is easy, you can just say the target disintegrated without actually counting them.

If they seem low, and they are trying to do something difficult, you can describe how the target was wounded, and in your notes substitute what seems to you to be a generous number for the damage.

However it's worth reiterating that you shouldn't generally do this - except for the purpose of fun.

Chris Perkins the DM of acquisitions incorporated has done this on the show when large dice rolls were called for. Though there he has additional concerns to the regular DM, I still think he sets a good example.

  • \$\begingroup\$ Whether fudging is acceptable or even possible depends on the way you play. It is not always okay. The first sentence is quite simplified and does not reflect reality well. \$\endgroup\$
    – Tommi
    Commented Feb 16, 2016 at 17:56
  • 1
    \$\begingroup\$ People don't only play for fun (especially immediate fun). There are some games that provide very powerful experiences, which are often unpleasant. (Most of these are Jeepform games or structured freeform games, but also some that use dice do exist, I think.) \$\endgroup\$
    – Tommi
    Commented Feb 16, 2016 at 18:05

When I have to roll extremely large amounts of dice, I just use https://www.wizards.com/dnd/dice/dice.htm It lets you roll an arbitrarily large amount of any dice as far as I can tell.

  • 1
    \$\begingroup\$ Yes, but several answers mentioned computer-aided rolling long ago and in rather more detail. \$\endgroup\$
    – user17995
    Commented Feb 17, 2016 at 1:00

If you've got a head for statistics (or programming), this is where a set of FUDGE dice can come in handy. They're specially marked six-sided dice: two sides have a + sign, two sides have a - sign, and two sides are blank. If you don't have them, you can easily simulate a FUDGE die with a standard d6: 1 and 2 are -, 3 and 4 are blank, and 5 and 6 are +. For this method, you'll need four FUDGE dice (4dF, to use their terms).

Set up a table of the mean (average) and standard deviation for rolls that you expect to use a lot. Do not count any kind of bonuses or penalties in this table: this is only for raw dice results. This requires a bit of statistics work up-front, but since you know your characters, you should have enough data to prepare a table well in advance. It should not need to change very often

To simulate the huge roll, break out your table:

  1. Set X to the average for this type of roll, and then roll 4dF.
  2. (optional) Discard blank dice. They have no effect.
  3. (optional) If you can make a pair out of a + and - result, discard them both; they cancel each other out. Keep doing this until you cannot make any more pairs.
  4. For every + result, add one standard deviation to X. For every - result, subtract one standard deviation from X.
  5. X is now the raw die result. Add bonuses or penalties, and you're done.

You don't have to do Steps 2 and 3, but it makes the math in Step 4 a lot simpler if you do. Either you go into Step 4 with no dice left (because you discarded them all), or all the dice left will say the same thing (because you discarded cancelling pairs).


As others have suggested, you will get easiest results if you accept a slightly different distribution of results or a slight approximation.

For instance, the two following are pretty close

4d6 min 4 ave 14 max 24


min 3 ave 14 max 25

Or for 10 d6

min 10 ave 35 max 60

6 d10 + 2 min 8 ave 35 max 62

Using bigger dice results in slightly more variability. Although you can counter that to some degree with higher flat offsets

40 d6 min 40 ave 140 max 240

22d10+19 min 41 ave 140 max 239

I also find d10s nice to add up (easy to do fgysin's method of grouping in 10s). Although availability of dice may be an issue depending on how many you own.

Edit: Note, even if the min, max and average are the same, the exact distribution will not be.


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