Just Simple Questions on Dice

[TickTock Man]

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

[Ron Edwards]

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

[James_Nostack]

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.

[Julian]

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.)

[Julian]

Oh yeah, that was for 10-siders.

[Ron Edwards]

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

[Julian]

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

[TickTock Man]

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

[Trevis Martin]

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

[Walt Freitag]

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

[Peter Nordstrand]

Hi Walt,

Here is where I get stuck:

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.

I don't think I know what "a limit value" is. The term means nothing to me, which makes understanding the rest of your text hard to understand for me. Would you mind explaining?

All the best,

/Peter

[Peter Nordstrand]

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?

[Marco]

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

[Darren Hill]

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!

[Walt Freitag]

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