# Earthdawn dice roll probabilities

After ad-hoc running Misguided Ambitions with my regular DnD 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 ;-) ) and that characters were pretty capable if you looked at the averages and didn't only focus at rolling 1s or 2s with your single d8 with the pre-made characters from 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 characters.

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

//Edit
The mechanics are pretty straight forward. Task resolution is handled by rolling a number of dice against a set target number (the difficult of the skill check, the monster's physical defense, etc.). The number and sides of the dice rolled depend on your "step". The step rolled for any task is influenced by the character's attributes, his skills/talents, magic items, buffing/debuffing spells, and so on.

``````Step (Dice)
1 = d6-3
2 = d6-2
3 = d6-1
4 = d6
5 = d8
6 = d10
7 = d12
8 = 2d6
9 = d8 + d6
10 = 2d8
11 = d10 + d8
12 = 2d10
``````

Step 13 to 19 is the same as step 6 to 12 but +d12, step 20 to 26 adds +2d12, and so on. For example, step 36 is 4d12 + 2d6.

Dice are rolled normally, but each time a die comes up with the max number it "explodes". When a die explodes it is rerolled and the results are added together. For example, if you roll step 22 (2d12 + 2d6) and roll 4, 12 on the d12s and 1, 6 on the d6s you reroll the 12 and the 6 and add the results. In the example, if the rerolled d12 came up with a 3 and the rerolled d6 showed a 5 the total result for the check would be 4 + 12 + 3 + 1 + 6 + 5 = 31. Dice can explode repeatedly, and it can happen to have a d6 or d8 explode 3 or 4 times in a row.

Damage is also handled by rolling an appropriate number of step dice, with an optional rule to cap the maximum possible result at Damage Step * 3. For example, if an attack would deal damage step 11 (d10 + d8) the attack could inflict at most 33 points of damage, no matter how many dice exploded.

Average step values for starting characters:

• Attributes: 4 - 7
• Skills/Talents: 5 - 9 (same as the associated Attribute plus 1 or two steps from ranks in the Skill or Talent)
• Damage: 6 - 11 (usually the Strength step plus the bonus from the used weapon)

Average health for starting characters ranges from around 30 to as high as 40 for really tough characters. Basic armor the characters can afford provides from 4 to up to 8 points of armor, which reduces dealt damage on a 1-to-1 basis. For example, a character with 6 points of armor would only take 3 damage from an attack that dealt 9 points of damage.

//Edit#2
The questions I intended to ask (but messed up when writing the original post) were:

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

//Edit#3
Wow, lots of very good answers. I wished I could upvote more than once. The link provided by Lokathor turned out to be the most useful answer for me. Anyway, kudos to all other answers that explained the math behind it. :)

-

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 that RedBrick wrote on their website. 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.

-
Very nice article. Thanks for the link, haven't seen that one before. :) Regarding DnD, the DSCC (and apparently MM3, too) did significantly increase the monster's damage. In the DSCC there's a level 1 brute that can dish out 4d6+4 at +6vsAC three times per encounter. I think one of the main problems was that they were fighting the troll "endboss" of the adventure while his darkness aura was running and everyone was attack at a -5 step penalty. But I kinda like the higher randomness of the ED system compared to 4E because it keeps the players on their toes and prevents many stupid actions. –  user660 Oct 7 '10 at 19:23
Oh yeah, I tried to get my group to run away several round in a row against that guy; the obsidiman swordmaster backed off, but the t'skrang swordmaster refused to back down even after being put exactly 1 away from unconscious. The next attack missed, but the one after that killed him instantly. Next session, the same player's new character got knocked unconscious instantly by a lightning lizard, and then later after healing an insane obsidiman punched him dead in 1 strike (41 damage armor defeating). Some people just have bad luck >_> –  Lokathor Oct 7 '10 at 22:40
@SevenSidedDie: if you're interested, the adventure I was running can be downloaded for free from DriveThruRPG or RPGNow. Here's the relevant page on Redbrick's homepage. –  user660 Oct 10 '10 at 20:32
@Lokathor Point well taken. :) Still, players need to know how to pick their battles! That's one major thing that I lament about D&D 4e: it teaches players new to roleplaying that an encounter can always be defeated if they just keep fighting. No other game system works like that, and it trips up players who move from D&D 4e to anything else. –  SevenSidedDie Oct 11 '10 at 5:09
@Baelnorn Thanks for the link. I've always been interested in Earthdawn, but never played. This should be good reading. –  SevenSidedDie Oct 11 '10 at 5:10

## 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)))
``````

-
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

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

-
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

I started trying to play with the math here, and got bogged down. I've got a start of a google doc at https://spreadsheets.google.com/ccc?key=0AvNIsC5oiKpndGR2UndRZVlJaHdKLTJicWp1cTlfVXc&hl=en if anyone would like to contribute

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Very nice work with the spreadsheet! :D –  user660 Oct 7 '10 at 19:23

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.

-

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.

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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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
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
@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
@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
@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