Help Me with RPG Dice Combinatorics

The Game Room
1

My rudimentary understanding of dice analysis consists of the following:

[ul]
[li]The average roll for a given die is the top value plus one, divided by two. [/li][li]The probability that a combined event will occur is determined by multiplying the probability of each element not occurring, then subtracting that from 1.[/li][/ul]

Everything else I pretty much have to do by brute force generation of a histogram. But in games like Savage Worlds or HackMaster you get these exploding dice rules that I am powerless to analyze. Somewhere I got the data I needed to chart for Savage Worlds the results attainable vs. the die type used. But I’m now puzzling over HackMaster, for which I don’t see anybody having blazed the trail further than to have simulated die rolls. There, the results you get differ every time you attempt the analysis, and are subject to the vagaries that exist between random computer generation and the dice themselves.

With exploding dice systems, you need those kind of equations with N+1’s, brackets and ellipses. I don’t know how to do that. More importantly, I don’t even know how to brute force that with a spreadsheet. Well, okay, I can approximate the odds of results from one die. But I’m not sure how to even approximate the additive results of multiple results from one die type, much less multiple die types.

For HackMaster, there is a rule of Penetration. A roll of 1d4p means that you roll a 4-sided die, and if the result is the maximum 4, you roll again and add the next roll -1, and so on until you stop rolling the maximum value of the die (always only -1 per roll after the first).

From this, how is it possible to work out the likelihood of a given total result of multiple such die rolls added together?

2

Oh, and also, how to compute an expected average would be useful to understand.

3

We can get the expected value (mean) for such a roll fairly easily. First, for simplicity, let’s assume that our d4 is marked 0,1,2,3 instead of 1,2,3,4. This means that for our final answer we’ll have to add 1 for the first roll, but all our extra rolls, we won’t have to do the subtract 1 step.

Now, let’s define that M is the expected value of rolling this penetrating die, and see if we can write an equation. Well, we have a 1/4 chance each of rolling a 0,1, or 2, and a 1/4 chance of rolling a 3 and continuing to penetrate. So M = .25*(0) + .25*(1) + .25*(2) + .25*(3+M). Notice that M shows up in that last term, also, since if we roll a 3, we have to start the process over.

Well, we can rearrange this equation, to put all the Ms on the same side. If we do this, we get .75M = .25(0) + .25*(1) + .25*(2) + .25*(3), or simplifying, .75M = .256, or .75*M = 1.5, or M = 2. Adding the +1 back in to turn our first die roll from 0,1,2,3 to 1,2,3,4 , we find that the expected value of 1d4p is 3, a little more than the 2.5 you’d expect from 1d4 (no penetration).

Add enough such rolls together, and you’ll get something that can be well-approximated as a Gaussian normal distribution. To do this, you’d need to know not only the mean of a single roll, but the standard deviation. That’s a straightforward calculation, but not one that I feel like bothering with right now.

In any event, once you have the standard deviation of a single roll, you can get the variance of that single roll by squaring the standard deviation, then you add together the variances of all of your rolls to get the total variance, and then you take the square root of that total variance to get the standard deviation of the total.

4

Moved to the Game Room.

Colibri
General Questions Moderator

5

I think I can just about wrap my mind around how this manages to make the infinite recursion expressible in a straightforward equation, but that’s odd.

6

Hey, How ya doing?! I’m a long time player and GM of Hackmaster and long time member of the Kenzer&Co message boards(creators of HM)(NOTE: I’m not an employee of the company, just a big/long time fan of K&C0) and can answer your questions about it.

HM Dice Penetration odds have been figured out many times, new people the board over at K&Co often try to duplicate the previous work and always come out with the same results. Basically the penetration system of HM Dice essentially add “0.5” to the average roll of any die roll. I can’t remember the exact calculation but 0.5 is rounded off and very close for just about any set of die rolls.

Please note that there are two versions of the HackMaster RPG; “HackMaster 4.0” published in 2001 and the new version of HackMaster(sometimes called HackMaster 5 by fans). Both have the same basic penetration rules but the new version makes a slight change, a d20 penetrates with a d6 and a d100 penetrates with a d20, so the odds of a d100 or d20 penetration are slightly different from any other die roll penetration result.
I can answer any other questions about the game.

7

Well… if you want another suitably nerdy option for a single analysis: Poor Man’s Monte Carlo in Microsoft Excel.

-Model it in excel on a single row with a couple of columns, with IF statements and the RANDBETWEEN() function representing the dice outcomes and decision points
-Once you’ve got a row that randomly simulates your multiple dice throws, copy the row down about 10,000 times

Now that you’ve got 10,000 random scenarios, use Average(),Percentiles(), or other equations analyze the results.

I can never understand or remember all the statistical formulas, but it’s fairly easy to model randomness of different types in excel and observe the results, especially when it’s just dice and not funny stuff where you’d need actual Monte Carlo software.

Also, modeled your example in about 10 minutes as follows:

All on a single row
-1 set of 100 columns with RANDBETWEEN(1,4)
-1 set of 100 colmuns using an IF statement to determine the running Max
-1 set of 100 colmuns with IF statement converting the running Max set to 1’s and 0s, basically showing whether that roll should be counted or not
-1 column with a SUMPRODUCT of the 1’ and 0’s with the dice rolls to get the total penetration for that set of rolls

Copied down 5,000 times, I get:
Mean of 40
Median of 37
Stdev of 18
95/5 percentiles of 74/17
Maximum observed consecutive rolls was 52

Hope that helps, or at least is entertaining.:slight_smile: