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After an Earthdawn one-off running Misguided Ambitions with my regular D&D 4E group — because one guy was MIA — I had a discussion with one of the players about Earthdawn's dice mechanics. The player argued that the mechanics in ED were too unpredictable because of exploding dice and that the characters were far too weak because of the (supposedly) high chance of just rolling low.

My counterargument was that this was equally true for the characters as well as the monsters; that the randomness was part of the game world (after all, the characters are Adepts and not Superman — although some get pretty close at high Circles!) and that characters were pretty capable if you looked at the averages and didn't only focus on rolling 1s or 2s with your single d8 from the pre-made characters in the adventure.

Well, it ended with the player insisting on his point of view and stating that he'll never again play Earthdawn because he couldn't pre-calculate a character's chances and abilities (unlike 4E) and that the game system was too unfairly weighted against the player characters.

Now I'm wondering if there are any actual statistics and analyses of the Earthdawn dice mechanics. I tried Google but didn't really find anything useful.

So:

  • Are the ED dice mechanics as unfair and weighted as my player claims them to be?
  • What are the mathematical pitfalls of the Step System and how do they affect combat/skill tests?
  • If the mechanics are unbalanced, what are suitable changes to make the system fair?
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5 Answers 5

up vote 25 down vote accepted

As a fellow GM of Earthdawn, and former GM/Player of DnD 4e I have some good news and some bad news:

  1. Your player is being somewhat silly if he's actually hardcore about statistics: It's easy enough to perform a numeric analysis on Earthdawn mechanics if you really want to. There's even an article, "The Bare Bones", that RedBrick commissioned for their website; after RedBrick shut down their old site, the author republished it under the title "Step by Step". The guy uses basic excel and such to get his results, it's not hard if you have the time. Note the part in the article where it explains that the easiest way to understand why the various Result Levels of each target number have such odd progressions sometimes has to do with maintaining an X% chance of Y result level on step Z, which can be slightly complicated.

  2. Your player is also correct about the difficulty of statics on the fly: performing a statistics review of your chances to do X action within Earthdawn while the game is running is pretty much impossible unless you've setup all the formulas and such on a calculator before hand. The best you can really do is to know that for every +1 the target number is over your step number you're 1 category "less likely" than before to get what you want (and the reverse for TNs under your step number). Similar to Shadowrun, the exact odds for anything other than an average roll is largely a black box of sorts. Unlike in Battletech or DnD, there isn't a single dice expression to compare all your target numbers against, so having an index card saying "X% of getting Y or more on 1d20/2d6/3d6/whatever" isn't a thing you can do. You could do it by having a separate card for each step number against all target numbers, but bleh to that man.

  3. Your player is largely correct about the dying thing (from a DnD 4e perspective): It's far far far easier for a player to randomly die in Earthdawn than it is for them to randomly die in 4e. On the other hand, Earthdawn has the Last Chance Salve to try to bring people back, and it's available for a simple 60 silver each if someone in the party has Alchemy (otherwise they're 600 silver each, ouch).

Earthdawn's step system makes it so that occasionally you'll roll much higher than expected, and occasionally you'll roll much lower than expected. At the low step numbers (and low Circles) you'll roll much higher than expected more often than much lower than expected simply because there's a minimum bound on the number rolled (1, or 2 if you're at step 8 to whatever, then it's 3, and so on), and your expected outputs aren't too high. At the same time, when everyone is rolling low then mostly things just don't happen much, but when people are all rolling high then people die horribly.

To put it another way, there's infinitely more NPCs than PCs, and so when the PCs are faced off against a small percent chance of horrible death with every attack they're targeted by, eventually they're suffer a horrible death with no prior warning. The same thing might happen to an NPC opponent, but they weren't expected to survive anyways so it's not a huge deal. When a PC suffers an unexpected death with no warning at all it seems extremely unfair because the players expect their characters to mostly live from one adventure to the next.

In addition to the chance of horrible death from plain old high damage rolls, there's also the issue of armor defeating hits. As play progresses into the higher circles, characters have more and more Physical and Mystic Armor, but their Physical and Spell defense doesn't really go up at the same rate. It becomes a contest of who can armor defeat who first and get a ton of unsoaked damage. Wild West battles might be your thing, but they're not everyone's thing. DnD 4e is a game where bringing down an opponent, PC or NPC, generally requires several successful hits in a row, and the opponent can generally tell that they're going to die soon and turn on any preventative measures. So, it's a very different dynamic to get used to.

One fix for just the issue of "all these weird dice steps" and the minor but odd probability break-points they cause, is to replace the die steps with Xd3−X, where X is the Step Number. This always gives results 0–2X, with X being the most likely result in the centre of a nice bell curve. It makes the odds much more predictable on-the-fly without altering the system much at all, at the expense of having to roll unusual dice (or reading d6s differently) and an extra mathematical operation every roll. Some Earthdawn players cite simplicity and just liking to get to roll all the polyhedrals as a reason to stick with the original Steps, but if your group prefers the statistical ideal enough to deal with more complicated rolling procedures, this might be a good tweak.

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I'm not sure what the real question is. I'm thinking at least part of it is "Help me understand the mechanics of exploding dice" so I'll start there.

Where N is the size of the die.

Average # of rolls = N/(N - 1)

Average result of an exploding die = N*(N+1) / (2 * (N-1))

Die    Average        Average
        Rolls          Value
d4    1.3333334      3.3333333
d6    1.2            4.2
d8    1.1428572      5.142857
d10   1.1111112      6.111111
d12   1.0909091      7.090909
d20   1.0526316     11.052631

Once you know the averages, or the formulas, you can better eyeball your odds of any particular result at the table.

Gory math details are here

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Yeah, I wouldn't do too much more work than this - basically the answer is "your average result is around the usual average of the die - a little more, but less than one point more." More than that is not useful for understanding/enjoying a game. –  mxyzplk Oct 7 '10 at 4:00

If you have an exploding N-sided die that adds another die when it's at max, the average A is:

A = (1+2+...N+A)/N

This is the same as:

A = (N(N+1) + A)/2N

This, in turn, can be simplified to:

2N*A = N(N+1) + A

And that in turn, can be simplified to what Pat Ludwig has so kindly tabulated.

A similar method can be used to calculate averages when the highest roll of a die is substituted by rolling 2 (or more) of the same dice-type.

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To address your player's concerns, I've prepared a paper which I've posted on Scribd. UPDATE: For the benefit of anyone who prefers not to use scribd, I've posted both the PDF and the LaTeX source files of this paper on MediaFire.

The paper analyzes Earthdawn's dice roll probabilities. Its two main conclusions are

  1. Earthdawn's dice mechanics seem complicated, but it is still possible to pre-calculate a character's chances. Knowing the probability mass function of an exploding n-sided die, it is quite easy to calculate the mean and variance at any step based on the linearity of the expectation operator and the Bienayme formula. We provide a table of means and variances for steps 1 through 12 on page 2. Furthermore, it is possible to use the operations of horizontal shifting and convolution to pre-calculate the probability of hitting a given target number at each step. We provide such a table on page 7 that precalculates probabilities for target numbers 1 through 50 for steps 1 through 12.
  2. While it's technically possible to succeed with a low step number and yet fail on a high step number, the probability of succeeding on a high step is still greater than on a low step.

If you have any corrections or questions about this paper, please leave comments here on stackexchange. Hope this analysis helps convince your player!

UPDATE: At the suggestion of Simon Withers, I've posted a revised version of the document (now 8 pages) on both Scribd and MediaFire, which shows in detail how to compute the odds of hitting a given target number for each step. Page 7 now has a table showing the odds for hitting target numbers 1 to 50 at steps 1 to 12. That table is a bit too big to paste here. But here I will give the table from page 2 of means and variances for steps 1 to 12:

Step  Dice Throw  Mean      Variance
1     d6-3         1.2      10.64
2     d6-2         2.2      10.64
3     d6-1         3.2      10.64
4     d6           4.2      10.64
5     d8           5.14286  14.449
6     d10          6.11111  19.0123
7     d12          7.09091  24.281
8     2d6          8.4      21.28
9     d8 + d6      9.34286  25.089
10    2d8         10.2857   28.898
11    d10 + d8    11.254    33.4613
12    2d10        12.2222   38.0247

N.B. The mean at step number n is just n plus a tiny fraction (ranging from 0.09091 to 0.4). So that's a real easy way to memorize it.

The variances are the same from steps 1 to 4; they increase at each step from 4 to 12 except going from step 7 to step 8 (because step 7 is a roughly uniform distribution [wide variance] and step 8 is a roughly triangular distribution [narrower variance]).

Explanation of how to calculate higher steps: Means are additive. Note that the mean of step 9 (d8+d6) is just the mean of step 4 (d6) plus the mean of step 5 (d8). Ditto for variances.

If you need the spread (standard deviation), then just take the square root of the variance. But note that unlike variances, standard deviations are not additive.

In the paper, I only used the kind of math you'd get in a Precalculus class (functions, shifting graphs of functions and infinite series) and an introductory class in probability theory (definitions of random variables, probability mas functions, expectations, means variances). But I understand why a lot of people might not be interested in these details. The paper is written for anyone who wants to deepen their understanding of the underlying math. Hopefully, that's many of you reading and responding to this type of question on stackexchange.

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3  
Actually, the most complicated stuff in that paper isn't calculus-level, and in any case is only the "showing your work" for the statements that come later. They don't have to be understood unless you want to verify the conclusion's correctness. You can just skip to the table of probabilities on page 2. –  SevenSidedDie Oct 10 '10 at 16:00
1  
I suspect that I'm close to being able to use the math in your document... but I have one main question - I think that the variance can be used to compute the odds of achieving a given target number, but I'm not sure. For me at least, computing the odds of success is the main goal here. –  Simon Withers Oct 10 '10 at 20:02
1  
@Simon Withers: To calculate the success at a step against a target number T, you have two main choices: 1- You can approximate your chances using the mean and variance. To do this you can use a normal approximation. However, this method is not so good esp. when the 1 die is thrown but improves as more dice are rolled together. You'd find the area under a standard normal curve from (T-mean)/stddeviation to infinity. (If you use a standard normal table, be sure to take the complement of the tabled entry!) Continued... –  A. N. Other Oct 10 '10 at 21:06
1  
@Simon Withers: ...Or 2- you can calculate the probability exactly, by working out the pmf, using the techniques in the paper. The probability is the the discrete sum of the pmf from T to infinity. Practically, you only need to work out the partial sum with say 100 terms because it converges so quickly. This is the method I recommend. Its more trouble, but its exact. Good luck! –  A. N. Other Oct 10 '10 at 21:06
1  
@Simon Withers: A new revision of the paper on scribd now has a table with the odds of achieving target numbers 1 to 50 for steps 1 to 12. In addition, it details how these calculations were done. –  A. N. Other Oct 11 '10 at 8:37

Preliminary

Dice Table

     6              8              10     12
1    0.166666667    0.125          0.1    0.083333333
2    0.166666667    0.125          0.1    0.083333333
3    0.166666667    0.125          0.1    0.083333333
4    0.166666667    0.125          0.1    0.083333333
5    0.166666667    0.125          0.1    0.083333333
6    0              0.125          0.1    0.083333333
7    0.027777778    0.125          0.1    0.083333333
8    0.027777778    0              0.1    0.083333333
9    0.027777778    0.015625       0.1    0.083333333
10   0.027777778    0.015625       0      0.083333333
11   0.027777778    0.015625       0.01   0.083333333
12   0              0.015625       0.01   0
13   0.00462963     0.015625       0.01   0.006944444
14   0.00462963     0.015625       0.01   0.006944444
15   0.00462963     0.015625       0.01   0.006944444
16   0.00462963     0              0.01   0.006944444
17   0.00462963     0.001953125    0.01   0.006944444
18   0              0.001953125    0.01   0.006944444
19   0.000771605    0.001953125    0.01   0.006944444
20   0.000771605    0.001953125    0      0.006944444
21   0.000771605    0.001953125    0.001  0.006944444
22   0.000771605    0.001953125    0.001  0.006944444
23   0.000771605    0.001953125    0.001  0.006944444
24   0              0              0.001  0
25   0.000128601    0.000244141    0.001  0.000578704
26   0.000128601    0.000244141    0.001  0.000578704
27   0.000128601    0.000244141    0.001  0.000578704
28   0.000128601    0.000244141    0.001  0.000578704
29   0.000128601    0.000244141    0.001  0.000578704
30   0              0.000244141    0      0.000578704

Calculated using Excel formula

=IF(0=MOD(TOTAL,DIE_SIDES), 0, 1/POWER(DIE_SIDES, 1+FLOOR(TOTAL/DIE_SIDES,1)))

Feel free to use this in your answer

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Before you vote me down, I know this is hardly an answer to the overall question. It'll take several hours of work to do that, and I wanted to get something usable up before that. –  C. Ross Oct 7 '10 at 0:09
    
Maybe make this answer a community wiki then, since it's a resource for the other answers. –  SevenSidedDie Oct 7 '10 at 0:14
    
@SevenSidedDie good point. –  C. Ross Oct 7 '10 at 0:16

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