Greetings!

Do you use a standard die type in your games?  What type of die do you use and why?  Is there a Sorcerer "standard" size?

Thanks!

Angelo

Hello,

I typically use d10s. However, d6s have worked fine for me in the past. Reports from various actual play have confirmed that d20s are fine; I don't know if anyone's bothered using d4s or d12s.

The text is accurate and, I think, pretty clear about the differences in using differently-sized dice.

One person has pointed out that the basic claim - that for chances of win-lose, die size doesn't matter - is not valid when odd-sided dice are used (e.g. d3, d7). So stick with any one of d4, d6, d8, d10, d12, or d20, which are what I had in mind when designing the system anyway. d7 (which wouldn't work) and d30 (which would) had not been invented when I wrote it. The whole point of the system was to be quick and usable, so the step of processing d3 (which wouldn't work) is counter-indicated anyway.

Best,
Ron

Following up on this: Ron, do you (or any of the other Sorcerer fans out there) have an idea of the probability curves involved?  Like, what odds are generally necessary to snag X number of successes? 

In my limited experience the number of successes hovers mostly around 1-2 for dice pools of roughly comparable size, with the occasional 3-4 successes occuring maybe 10% of the time.  (When things get really mismatched--one guy rolling 1 die, and the other guy rolling 6+, the reverse is probably true.)

I used to take a bunch of math courses, and I probably should be able to figure it out if I grind at it hard enough, but I wondered if someone had already done the math.  It's just idle curiosity, I suppose.

The bigger the dice, the better the underdog's chances.

Our experience is that d10's are already pretty high variance. Twenties would make life very random. I'd stick to the 6-10 range, depending on how random you want things to be.

As for the actual probabilities, the math is hard, so I just wrote a simulator.

(Across is your number of dice. Down is the number of dice you're
opposing. Top left corner is 1 vs 1. The numbers on the diagonal would
be zero if I were doing the math, rather than cheating.)

Wu-Ming-Yi:~/perl jl8e$ ./sorcdice.pl
-0.00  0.62  1.15  1.66  2.11  2.59  3.08  3.53  4.00  4.47
  ----  -0.00  0.55  1.02  1.44  1.86  2.27  2.69  3.07  3.47
  ----  ----  -0.00  0.46  0.86  1.25  1.61  1.93  2.27  2.62
  ----  ----  ----  -0.02  0.38  0.75  1.08  1.39  1.70  2.00
  ----  ----  ----  ----  0.00  0.35  0.66  0.97  1.26  1.53
  ----  ----  ----  ----  ----  0.00  0.30  0.60  0.88  1.14
  ----  ----  ----  ----  ----  ----  -0.01  0.27  0.56  0.81
  ----  ----  ----  ----  ----  ----  ----  0.01  0.27  0.51
  ----  ----  ----  ----  ----  ----  ----  ----  0.01  0.25
  ----  ----  ----  ----  ----  ----  ----  ----  ----  0.00

This also assumes all the assumptions I made about weird edge cases are correct. (Most likely one to be wrong - what happens when all your dice tie all the other guy's dice, but he's got some extra dice.)

Oh yeah, that was for 10-siders.

Hiya,

To be clear, when you say "the better the underdog's chances," you're not talking about the base chance of win/lose, but rather, the degree of success if the underdog wins.

Best,
Ron

No, the base chance of winning.

From a quick program, rolling a two vs one test 100,000 times for each die size:

Wu-Ming-Yi:~/perl jl8e$ ./sorctest.pl
10: 28496
12: 29386
20: 30652
30: 31702
4: 21972
6: 25275
8: 27444

Thanks for the responses.  I have been using d10's and I wondered if I was in the norm, or if maybe someone had switched to d8's or d6's and had a significant difference in the feel of the game.  Sorcerer is definitely clear about the implications of the change, but I thought I would see if there was any anecdotal evidence one way or the other in how the game has been affected (positively, negatively or just differently).

Thanks again,

Angelo

I have spent one entire game (about 8 to 10 sessions per game) using d10's and d12's  I can't say I noticed any significant difference other than I like d12's better  for their shape.  Perhaps the step was too small to matter.

Trevis

Hi Ron and Julian,

Having heard conflicting claims about the behavior of the Sorcerer dice mechanism now and then over the years, I decided to look into it this time. Here's what I've determined.

First, a review of the system we're analyzing here. Those who know the basics can skip past this. I'm only going to discuss two-party rolls (one player against one other player OR the GM).

Each side rolls a number of dice, at least one. All dice are the same size. The side rolling the highest single die wins. Each of the winner's dice that rolled higher than the loser's highest die indicates one success achieved by the winner.

Example: using d10s, John rolls 8 6 5 3 and Mary rolls 10 9 5. Mary wins, with two successes. Her 10 and 9 are both higher than John's highest, 8.

Example: using d20s, John rolls 11 7 6 3 1 and Mary rolls 6 4. John wins, with two successes. His 11 and 7 are higher than Mary's 6.

Example: using d8s, John rolls 6 5 5 2 and Mary rolls 8 8 8 4. Mary wins with three successes. Mary's 8 beats John's highest, 6, and her other two 8's do also.

If the two sides are tied for highest die, the tied dice are both disregarded and the remainder of the roll is evaluated according to the above rules. If the next highest dice are also tied, they're disregarded too, and the remainder of the roll is evaluated according to the above rules. And so forth.

Example: using d8s, John rolls 7 4 3 and Mary rolls 7 5 5 3. Mary wins, with two successes. The 7s are disregarded and Mary's two 5's beat John's highest, 4.

Example: using d6's, John rolls 5 4 4 2 1 and Mary rolls 5 4 4 3. Mary wins with one success. The 5's tie and are disregarded, the 4's tie and are disregarded, and Mary's 3 beats John's next highest, 2.

Nuance #1: When dice are tied, they are only disregarded on a one-for-one basis. Other dice in the same roll that also show the same number as the tied dice are not disregarded, unless they are also tied with other dice in the opponent's roll.

Example: using d10's, John rolls 10 8 and Mary rolls 10 10 3. Mary wins with one success. One of Mary's 10's ties with John's 10 and is disregarded along with John's 10. Mary's other 10 beats John's next highest, 8.

Example: using d6's, John rolls 5 4 4 4 1 and Mary rolls 5 4 4 3. John wins with one success. The 5 and two 4's from each roll are disregarded, and John's remaining 4 beats Mary's next highest, 3.

Nuance #2: If after disregarding tied dice, one side has no dice left, all of the other side's dice count as successes (because regardless of what they rolled, they're higher than "nothing").

Example: using d20s, John rolls 18 2 and Mary rolls 18. The 18's are disregarded and John's remaining die, 2, automatically beats Mary's nothing-at-all.

Example: using d10s, John rolls 8 6 6 and Mary rolls 8 6 6 6 3 1. Mary wins with 3 successes. The 8 and two 6's from each side are disregarded, and John has nothing left to match Mary's remaining 6, 3, and 1.

An additional rule specifies that if all the dice on both sides tie, which due to Nuance #2 can only happen if both sides are rolling the same number of dice, then the result is a complete tie and must either be rerolled, or the outcome determined by a different way.

-- end of system review

What I've found, looking at the system as defined by those rules, is that Julian is correct in saying that the underdog's chance of winning is better when larger dice are rolled. However, Julian is wrong in concluding from this that d20's "would make life very random" as compared with d10s. Actually what happens is that for any given roll of X dice vs. Y dice, the underdog's chance gets closer and closer to a limit value as the size of the dice increase. The larger the dice being rolled, the closer the underdog's chance is to the limit. The more dice the underdog is rolling (that is, the larger min(X, Y) is), the closer the actual probability is to the limit for a given die size.

The limit of the underdog's chance of winning is very easy to calculate. Let's look only at rolls in which there is no tie for highest single die. Suppose, for instance, that John is rolling 3 dice and Mary is rolling 2 dice. If there is no tie for highest single die, then one of the 5 dice on the table is the highest one. Each of the 5 dice on the table has an equal chance of being the highest one. What's the chance that it's one of Mary's 2 dice -- that is to say, that Mary is the winner? Obviously, it's 2 out of 5, or .4. John's chance is 3 out of 5, or .6.

In general, by the same logic, each player's chance of winning if there is no tie for high die roll is

(number of dice the player rolls) / (total number of dice rolled by both players).

Note that this calculation of the limit probabilities is entirely independent of the size of the dice. If there is no tie for highest die, the underdog's chance of winning in a 2 dice vs. 3 dice contest is 2/5, no matter if the dice have six sides or six million.

If the dice were extemely large (so large that the chance of tied high dice becomes negligible), the players' actual chances to win in the system would be very close to this "limit" probability. Also, if the rule were that rolls with ties for high die get rerolled, the players' actual chances to win in the system would be exactly the limit probability. (Though many rerolls would likely be required, when small dice and large pools were involved.)

In the actual system, the possibility of ties for highest roll reduce the underdog's chances below the limit value (and increase the favorite's chances above it). But the underdog's chances are never higher than the limit value, no matter how big the dice get. If there were million-sided dice, a 2 dice vs. 3 dice underdog would still have (infinitesimally less than) a .4 chance to win.

Why does the possibility of ties for highest roll reduce the underdog's chance to win? Because once the high dice are disregarded from each side, the ratio of dice remaining in the pools tilts farther against the player who had fewer to begin with. (If the pools start out equal, then of course there is no underdog in the first place and the possiblity of ties doesn't shift either player's odds away from even.) For instance, if there's a tie in a 4 vs. 2 roll, then the remaining dice are 3 vs. 1, and if the second highest dice also tie, it's now 2 vs. 0 and the underdog cannot win at all.

The larger and closer to even the dice pools start out with, the less pronounced this effect is, even though rolling more dice increases the chance of a tie, because disregarding a die from each side shifts the odds less and the chance of ties nullifying all of the underdog's dice decreases.

The larger the die size, the lower the chance of ties and the closer to the limit value the actual odds get. If you were rolling million-sided dice, for instance, there would be only a few chances in a million of a tie for high roll.

So the question remains, how significant are these effects at die sizes and die pool sizes relevant to Sorcerer? To start with, you must understand that "X against 1" (where X >= 2) cases are the worst case for this effect. In the 2 against 1 case, Julian's Monte Carlo results in his post above are pretty close to the truth.

2 vs. 1 underdog's chance to with rolling:

d4 -- .219
d6 -- .255
d10 -- .285
d20 -- .309
d50 -- .323
d100 -- .328
limit value -- .333

However, even in this case, the difference between d10's and d20's isn't very significant, especially in the context of Sorcerer play. Compared to d10s or d20s, the difference using d4s or d6s would be noticeable (with d6's, for instance, one roll in every 6 will result in a tie for high roll, and the underdog loses all of these) but certainly not extreme.

With larger, more balanced rolls the die-size effect diminishes. For instance, if we go from 2 vs. 1 to 3 vs. 2, the odds work out this way:

3 vs. 2 Underdog's chance of winning using:

d4 -- .303
d6 -- .341
d10 -- .368
d20 -- .386
d50 -- .395
limit value -- .400

Compared to the 2 vs. 1 roll, the 3 vs. 2 starts out (at smaller die sizes) closer to the limit success probabilities and converges on them faster (as the die size is increased). There's not much difference between d10s and d20s, and d6s are right in the ballpark.

I should also mention, I haven't been able to figure out any reason why dice with an odd number of sides would make any difference in the behavior.

Sorry for the verbosity. I hope it pays off in generating some clarity.

- Walt

Hi Walt,

Here is where I get stuck:

Walt wrote: Actually what happens is that for any given roll of X dice vs. Y dice, the underdog's chance gets closer and closer to a limit value as the size of the dice increase.

Oh, wait a minute. If there is no tie for highest die, the underdog's chance in a 2 dice vs. 3 dice situation is 2/5. The larger the dice, the closer his actual chance gets to that number. So, in this specific case, the limit value is 2/5, or 0.4.  Right?

We're using D20's. I think there's a nostaglia kick of going "I rolled a 20! Beat that!" They roll well and there's something about a giant basin full of D20's that says RPG to me :)

(not that I'm not a *huge* fan of the d6 though).

-Marco

Peter: yes, that's right. You've got it.
Walt: great post. I've fiddled with sorcerer probabilities a lot, and now feel stupid for missing that (my dice) / (my dice + your dice) ratio!

Just to confirm what Darren said -- Peter, you have it right.

Generally, but loosely speaking, a limit value is a finite value that a property of a mathematical system gets very close to, but never reaches exactly, as some parameter of the system is increased without limit (or, in some cases, decreased toward zero).

For instance, consider a series of regular polygons, each with a distance S between their centers and their vertices (corner points), starting with a triangle and progressing through increasing numbers of sides through square, pentagon, hexagon, and so forth. The triangle's area is different from the square's area which is different from the pentagon's area and so forth. However, there is a limit value to the areas of the polygons as their number of sides N increases without limit. The limit area is pi * S^2 (pi * S squared), which is the exact area of a circle of radius S. That's because the more sides the polygon has, the more it resembles a circle of the same size. There is no number of sides N, no matter how large, for which the area of the N-sided polygon has area of exactly pi * S^2, but the larger the N, the closer the actual area will be to that value. We say that the area of the polygon "approaches pi * S^2 as N increases."

A bit more technically, what establishes pi * S^2 as the limit value is that you can specify any number "epsilon" -- which can be extremely small -- and there will always be some number of sides N (which may be extremely large if epsilon is extremely small) for which the difference between the area of the polygon and the area given by the formula pi * S^2 is less than epsilon. In other words, it's not just that the area gets "closer and closer" to the limit value as N increases, it's that it gets closer than any distance you can name if N is large enough. Say that you need a polygon whose area is less than .0000000000000000000001% away from pi * S^2, there's guaranteed to be some number X for which that will be true for all polygons whose number of sides N >= X.

In this case, you've surmised correctly that the limit value is the player's chance of success -- comparable to the area of the polygon, in the example; the formula for that limit is (player's number of dice) / (total number of dice for both players) -- comparable to the formula pi * S^2 in the example; and the chance of success approaches that limit as the number of sides on the dice increase -- comparable to the number N of sides on the polygon in the example.

Darren, your experience (leaving aside any need for you to feel stupid about anything) brings up a very deep issue, regarding game design and game criticism, of what it means to fiddle around with probabilities. I believe it also hints at why some game designers (I'll name no names, and let any who want to fess up do so) tend to be disdainful of some methods of fiddling, such as Julian's approach of testing 100,000 rolls and counting up the results in a computer program. I think that's beyond the scope of this thread, but it's something I now plan to talk about in the series of "what game designers really need to know about probability" articles I've been working on.

- Walt

Walt, where will these articles appear? This is a subject I find intensely fascinating.

Walt wrote,

... brings up a very deep issue, regarding game design and game criticism, of what it means to fiddle around with probabilities. I believe it also hints at why some game designers (I'll name no names, and let any who want to fess up do so) tend to be disdainful of some methods of fiddling, such as Julian's approach of testing 100,000 rolls and counting up the results in a computer program

Ron wrote:
One person has pointed out that the basic claim - that for chances of win-lose, die size doesn't matter - is not valid when odd-sided dice are used (e.g. d3, d7).

Eh.  Hadn't read to the end and seen the rather more in detail math than mine (and the fact that Ron had "closed" the thread) -- sorry about that.

Hi,

The thread isn't closed. Didn't say it, didn't imply it, don't know what you're talking about.

I have our emails too, but my reading of them is a little different. Rather than debate that, which doesn't strike me as productive, I prefer to let Walt's analysis stand as the reference for all future questions.

Best,
Ron

Walt wrote:
I believe it also hints at why some game designers (I'll name no names, and let any who want to fess up do so) tend to be disdainful of some methods of fiddling, such as Julian's approach of testing 100,000 rolls and counting up the results in a computer program. I think that's beyond the scope of this thread, but it's something I now plan to talk about in the series of "what game designers really need to know about probability" articles I've been working on.

Hi guys,

I can't tell you when the article will appear. It has been, and will continue to be, a challenging project. The idea is to juxtapose exposition of mathematical probability theory, starting from the most elementary level, with a treatise on "how to think about" probability and design issues related to probability that's much harder to explain. Kind of the reverse of the usual approach, which is to preset the theory (the easiest part to explain) while leaving it mostly up to the reader to figure out what it means and how to apply it.

Here's part of what I mean, conceptually. Take a typical probability "gotcha!" question and consider the different ways someone might answer it:

Q: You're rolling 2d10. If either of the dice comes up 10, what's the chance that the other die also comes up 10?

Possible answer 1: "One in ten. Duh!"
Possible answer 2: "I took out some dice and rolled a bunch of times, it looks like it's about once out of every twenty times."
Possible answer 3: "I [pick one: tabulated all the possible rolls / ran a Monte Carlo program with 100,000 tries / looked up a formula and plugged in the numbers / posted the question on a bulletin board] and the answer looks like it's one in nineteen. Is that really right?"
Possible answer 4: "One in nineteen. Duh!"

and let's not neglect the all-important

Possible answer 0: "I have no idea."

One key point is that the fact that answer 1 is dead wrong and answer 4 is right shouldn't blind us to the fact that they're very similar answers, in terms of how the answerer comes up with them. In both cases the "duh!" conceals the act of creating and analyzing a mental model of what's going on. For answer 1, it's "the 'other die' has ten sides, each side is equally likely to come up, so..." For answer 4, it's "twenty -- no, wait, nineteen -- rolls in every hundred have at least one 10, but only one roll in 100 comes up double-10, so..." Without such a mental model it's hard to trust, apply, or extend any results we obtain by other means (as I worded answers 2 and 3 to try to illustrate). But of course, without some kind of verification as per answer 2 or 3, it's easy to mistakenly trust a flawed model. Also, the types of testing listed in answers 2 and 3 -- some much more than others -- can be used to help derive a good model in the first place, especially when starting from answer 0.

There's a synergy, of which the principles and formulas of probability theory are a vital part but far from the only part. If you calculate or test but don't try to achieve an intuitive understanding, you can (correctly) see that the underdog's chance in Sorcerer increases with the die size but miss that the increases converge on a limit, and so mistakenly conclude that the game would get wildly random with larger dice than the ones you've actually tested. If you think you have an inuitive understanding, but you don't calculate or test, you'll never notice when you fall into any of numerous pitfalls relating to unwarranted assumptions, as the now-classic Monty Hall probem demonstrates. (See Wikipedia for a thorough treatment of that problem, if you're not familiar with it).

A case in point: in my long post above I made some statements that further calculation and testing have revealed to be mistaken. The calculated probabilities that I listed, and my main conclusions and description of the behavior of the system, still stand. But along the way I tossed off some side comments that turn out to be wrong. Specifically, I said that if the dice don't roll a high tie, then the players' chances of winning are equal to the limit values (player's # of dice / total # of dice rolled). And that therefore, if rolls with ties for high die were rerolled, then likewise the players' chances of winning would be equal to the limit values. Turns out those assertions are wrong. The error is that once you exclude (or reroll) rolls with ties for high die, it's no longer true that each die on the table has an equal chance of being the high die. On the contrary, the higher the underdog rolls, the more likely the roll will be a tie, so once rolls with ties are excluded or rerolled, the underdog's dice are slightly less likely, die for die, to be the highest.

So, that's what I'm up against. Darren, Ron, and Ben, thanks for your interest; I'll keep at it and keep you posted. And any comments or suggestions are welcome.

- Walt

A random aside.  I suspect that the odd sided dice = meltdown idea comes from treating the Sorcerer system as tossing high/low dice and comparing number of highs to number of highs.  Since Sorcerer actually doesn't reduce to that, except in the degenerate case where you flip coins instead of rolling dice, anything you get from that assumption is wrong.

One question one can ask with this stuff that is interesting is what methods one can come up with to produce similar results to a system via a different means.

FREX, one of our players, Beth, dislikes the Sorceror system asthetically -- she finds that the multiple comparisons ("Wha'ts your highest die? 8. "Ok, second highest?" "6" "I have 2 successes") interfere with her fun.  I don't, and nor do some of the other players, but the question remains whether one could find an elegant die-rolling system that has similar properties re probability, but typically involves fewer comparison (my gut instinct says that this may be true of a "higher than 5/lower than 6"  successes comparison system with high die as a tie-breaker will produce similar-enough results with (typically) 1 comparison, but that's a case of 6: "my first wild guess is that it's 1/20ish (or 1/20ish)" (on the above set of examples)).

Ron, I'm a little nervous about walking into the "this thread is closed" type landmine, given the site's practice of "closed by convention" thread closings, so I erred on the side of caution when encountering "I 'fess. No real need to discuss it further, pending Walt's articles."  Sorry about that.

We're all good! This thread is huggy-bunnies, so far.

Best,
Ron

Huggy-bunnies?

Words I never thought to hear Ron utter.

I have recovered from my mortification at missing the old (my dice)/(total dice) ratio, but that observation triggers a new question.
One thing I really like about the Sorcerer dice system is the way that successes gained on a roll don't scale linearly with number of dice rolled. In Vampire and all the other dice pool systems I can think of off-hand, if you roll twice as many dice, your expected average number of successes doubles. Roll twice as many dice, get three times as many successes. (On average, with all that implies.) Most of these games have pretty clunky ways of dealing with the problems this causes (along with the observation that getting a number of 'successes' isn't always 'successful').

Sorcerer doesn't work like this, which is fortunate for its currency system. I haven't found an easy way to calculate the chance of getting, say, exactly 1 success, or 3 or more successes, on a roll.
What's the simplest formulaic way to do this kind of calculation?

Great analysis Walt. The "limit value" description that I find easier to think about is to start with any number and divide it by 2. Now divide that number by 2, and repeat. It will approach a limit value of 0 but never reach it.

Either there's no simple formula, or there's a simple formula but there's no simple way (that is, simple enough for me to have accomplished it) to derive it. I suspect there's no simple formula -- that is, the exact formula will be very complex, especially if the formula includes the number of sides on the dice as a variable. But I can't be sure either way.

The best I can do formulaically, so far, is to derive another limit ratio akin to the (my dice)/(total dice) ratio, which will converge on the real probabilities only if the number of sides on the dice are very large. That is, the derivation ignores the possibility of ties so it's only accurate when the number of sides on the dice are large enough to make the probability of ties small.

In that simplified model, if I get "two or more" successes, it simply means that the two highest dice are on my side of the table. The chances of that happening are the number of different ways to distribute the two highest dice that put them on my side, over the total number of different ways to distribute the two highest dice.

m = the number of dice I roll ("my" dice)
t = total number of dice
s = the number of successes "or more" I want to calculate the probability for

probability of s or more successes on my side = binomial (m, s) / binomial (t, s) = C(m, s) / C(t,s)

(Wikipedia "binomial coefficient" if you're not familiar with the binomial coeffiicent function C.)

Note that when s  = 1, the formula devolves to C(m, 1) / C(t, 1) = m / t as we had before.

If I get exactly two successes, it means that the two highest dice are on my side of the table, AND the next highest die is NOT on my side of the table. Overall, the chance of the (s + 1)th highest die being on my opponent's side is (t - m) / t, but I can't just multiply that by C(m, s) / C(t,s) because the two conditions are not independent. What I need is the chance of the (s + 1)th highest die being on the opponents side IF it's already established that the s highest dice are on my side. That's (t-m) / (t - s) -- that is, the number of dice on the opponent's side over the total number of dice not counting the s higher dice we're already assuming are on my side.

So, my probability of exactly s successes = C(m, s) / C(t, s) * (t-m) / (t - s)

Note that when s = m, (t-m) / (t-s) = 1, which means that the chance of getting m succcesses (all the dice I roll being successes) is the same as the chance of getting that many successes "or more," which makes sense because it's not possible to get more than m successes.

Note also that 1 - (t-m) / (t-1) is the probability of getting more than one success assuming you win (get at least 1 success) in the first place. Clearly the more of the dice are yours (the smaller t-m is), the higher that probability is.

Don't forget, though, that all this is only precisely true if we're rolling infinite-sided dice. Ties make it all much more complicated. For instance, if the (s+1)th highest die happens to be a tie with the (s)th highest die, then the probability of getting exactly s successes suddenly becomes zero (assuming also that no higher dice in the roll are disregarded due to ties, and that the (s+2)th die isn't also a tie), because it's either on my side and must represent an additional success, or it's on the other side and nullifies (at least) my (s)th success. Come up with exact formulas that cover these cases correctly? Sure, right after I finish off this little unifying-relativity-with-QM problem I've been working on.

However, even without exact formulas there's another way to calculate exact probabilities if you want to. All you need is a program that iterates through all the possible die rolls for a given m and t (and number of die sides n), evaluates each one, and counts up the wins and the numbers of successes. Dice probabilities being what they are, this is guaranteed to yield exact correct answers (without even rounding errors, if you express the results as fractions over n ^ t, the total number of possible distinct die rolls). Most programming languages on most computers should be able to handle rolls of 8 d10s (100 million rolls to score) in reasonable time. To get beyond that you might need to use C or another speed-optimized compiled language and/or do overnight or multi-day runs and/or iterate through the die rolls in a more sophisticated way. As an example of the latter, you can iterate through tuples where for instance (1, 3, 0, 0, 4, 0, 0, 1, 1, 2) represents a roll of 12d10 in which there's one 1, three 2s, four 5's, an 8, a 9, and two 10s. When the number of dice in the roll is close to or greater than the number of sides on the dice, there are far fewer such tuples than there are possible distinct die rolls. However, for each tuple you have to calculate how many different rolls generate that tuple, and weight the results counts accordingly. For the Sorcerer system you have to generate a separate tuple for each side's dice. Some general advice for these kinds of computations: Never use a built-in binomial coefficient function; instead, create a lookup table for the range of binomial coefficients you actually need (which is usually only for arguments up to the total number of dice in the roll). If possible use high-range integers rather than reals for all the results counts, dividing the various totals by the n^t denominator only at the end. Often, reducing those fractions will give you a better feel for the results than just displaying them as rounded real numbers.

- Walt