Fuzzy on the statistics...

[darkling]

So the next campaign is going to use a home-brew variant drawing on Metamorphosis Alpha. To this end I am trying to make combat focus a lot more on what weapons you have and how you use them than just on level (introducing the weapon class system, etc.).

So one of my players had a suggestion and it is just too late at night for me to work through the statistics on it. Hopefully some of y'all kind folk can help. Basically the ideas was this: rather than tinkering with + and -, fighting with two weapons allows you to roll the attack roll twice and take the highest, fighting with a shield forces your opponent to roll the attack twice and take the lowest (with two weapon vs. shield effectively canceling), and fighting with a two handed weapon allows you to roll damage twice and take the highest.

Now, I get the general notion that this means that two weapons is much better when you have a lower chance to hit and that a two handed weapon is much better when you have a high chance to hit. But I am not sure at exactly what point it crosses over and whether or not there is a similar range of utility for each fighting style or if two weapon fighting will consistently eclipse two handed. Any advice?

P.S. if it matters, the weapon class vs. armor class table is set up so that the target numbers range from 20-2 so there is no 'negative' AC or guaranteed hits.

[Finarvyn]

I don't think it's an easy "statistics" problem to solve since there are so many variables, but it's a cool idea and that's what "old school" gaming is all about.

As you've noted, having two attempts to roll an attack (two weapons) certainly improves your odds of hiding a bad roll. Most RPGs are designed where a typical person has roughly 50% chance to hit a typical foe, so we can approximate it with flipping coins. Heads hit, tails miss. With one coin you hit half the time (H, T -- miss on T) but with two coins you hit more like 3/4 of the time (HH, HT, TH, TT -- miss on TT). With the shield forcing a character to take the low roll, it's more like having to hit twice so you only hit 1/4 of the time (HH, HT, TH, TT -- hit on HH). This is an oversimlification, but you get the idea. In reality, with variable AC types the hit-miss thing isn't as cut-and-dried as 50%. My guess is that you'd get similar effects, however.

Advantage: it's a simple rule.
Disadvantage: it might overbalance the effect of shields and/or two weapons.

I'd say give it a playtest and see how it works in a game. Then report back!

[Aplus]

Sounds like a fun way to do it. Like Marv said, try it out. If it doesn't work out, you can always fall back on the old +1 hit/dmg/AC (or whatever else floats your boat).

[kesher]

I like it---though I do have to ask what happens when a two-weaponed fighter is in combat with a shielded opponent...

You might want to consider making a DEX requirement for two-weapon fighting, and perhaps a STR requirement for shield use, otherwise you may have every single character fighting with two weapons...

[Harbinger]

To compare these two techniques you can calculate damage per round:

dpr = % chance to hit x average damage

Let's assume you're an OD&D 4th level fighter needing a 10 to hit leather (AC 7). That's 55% chance. Now if you roll two dice and take the best, that's the same as saying 'What is my chance of hitting at least once on two dice?'.

This is easier to calculate if you reverse the question: 'What is the chance that I not hit with two dice?'. This is %miss x %miss = 45% x 45% = 20.25%

So your chance to hit is 1 - 0.2025 = 0.7975

And average damage on a d6 is 3.5 so:

DPR for roll atk twice = 0.7975 x 3.5 = 2.79125 dmg per round

Now for the other case, average value for roll 2d6 and take highest is trickier to calculate. I basically used a spreadsheet to get the value 4.47.

DPR for roll dmg twice = 0.55 x 4.47 = 2.4585

Hopefully that gives you enough info to build out a spreadsheet to analyze things the way you want.

[starcraft]

I'm slow on the uptake, so just to clarify,

The extra attack roll increases the probability of hitting (and thusly scoring damage) to an extent that it is superior to a roll 2 take the best damage 'advantage'?

FWIW, while I have not tried to get as detailed into weapons and combat emulation as the OP, one of the things I have done to help my Holmes based game is to use a tweaked weapon vs Armor table.

I find it helps a great deal. I also give shields a +2 AC adjustment vs all missile attacks (obviously not from the flank/rear etc) This makes bow fire/hurled weapons much less useful against shield equipped enemies. I do not allow bucklers to be used in such a fashion, however.

Anyway, just an idea shield wise.

Another is that I allow for the shield-push style maneuver as well as a clubbing strike with a shield inspired by 300. The TV show Deadliest Warrior (which is occasionally downright laughable, granted) did measure the force of a shield-edge strike to the head and let me tell you, it is nothing to sneeze at. If I were in the unfortunate position of being cornered by a few orcs, I think I;d rather have the versatility of a shield over a offhand weapon TBH.

[Harbinger]

... and I just kept going... It looks like the crossover point is a to-hit value of 8 which means two-weapon fighting is better at lower levels and against heavier armored foes.

Edit - the horizontal axis is your 'to hit' number, and vertical is damage per round.

Attachments:

[Finarvyn]

Nice analysis, Harbinger. Much better than mine. An EXALT is in order.

[Harbinger]

Feb 14, 2012 14:18:35 GMT -6 starcraft said:

The extra attack roll increases the probability of hitting (and thusly scoring damage) to an extent that it is superior to a roll 2 take the best damage 'advantage'?

In general, rolling multiple dice where only one has to beat a target number greatly increases your odds. As you saw in my example, it jumped from 55% to 79% with just one die.

The other thing is that your chance to hit always goes up, and as it gets higher, the chance of hitting grows exponentially when you roll two dice. Whereas your rolling of two dice for damage doesn't increase in damage as you level up.

[Harbinger]

Feb 14, 2012 14:30:53 GMT -6 Harbinger said:

... and I just kept going... It looks like the crossover point is a to-hit value of 8 which means two-weapon fighting is better at lower levels and against heavier armored foes.

To keep some perspective on this, we're talking about less than 1/2 point of damage per round difference between the two. So really it won't make much difference in an OD&D combat.

'Feel' wise, I think it captures the 'light fighter' vs. 'heavy fighter' really well.

As it stands if I were faced with the three weapon options I would choose two-weapons fighter not because it does more damage, but because it partially negates a shield opponent's advantage. I would suggest you make the shield advantage apply regardless of opponent type.

Now it would be interesting to see how a shielded opponent would do...

[kesher]

The other thing is that your chance to hit always goes up, and as it gets higher, the chance of hitting grows exponentially when you roll two dice. Whereas your rolling of two dice for damage doesn't increase in damage as you level up.

That is a variable I'd never considered before with this sort of combat adjustment. Fascinating!

[darkling]

Thank you all, and especially harbinger for the lovely graph! I will tinker with it and be sure to let you know how it turns out.

Feb 14, 2012 14:41:39 GMT -6 Harbinger said:

The other thing is that your chance to hit always goes up, and as it gets higher, the chance of hitting grows exponentially when you roll two dice. Whereas your rolling of two dice for damage doesn't increase in damage as you level up.

To be clear, one of the reasons that I invoked Metamorphosis Alpha in my first post is because in my system your to-hit value is not based on level, but rather based on the make and condition of the weapons you are using. HD goes up with level, to-hit goes up when you find or forge a shiny new toy.

[aher]

Feb 14, 2012 14:07:14 GMT -6 Harbinger said:

average value for roll 2d6 and take highest is trickier to calculate. I basically used a spreadsheet

For the damage from rolling 2 six-sided dice and taking the max, you can find an exact formula, rather than using a spreadsheet to enumerate all possible outcomes. This kind of probability distribution is called an "order statistic." Take a look at the following Wikipedia article, especially the section for discrete random variables:

Order Statistics

You apply these formulas to this particular case as follows:

• n = 2 dice
  
• iid X₁,X₂ ~ DU(6,1,1) each die is discrete uniform distribution with probability range {1,...,6}
  
• p₁ = (x-1)/6 is the probability of rolling less than x on one die, x in {1,...,6} of course
  
• p₂ = 1/6 is the probability of rolling x on 1d6

Order statistics are just the random variables in the sample sorted in non-decreasing order from minimum to maximum. Therefore, when n = 2, X₍₂₎ = max {X₁, X₂} is the second order statistic or maximum of the two random variables. Notice the parentheses around the numeral two in the subscript: this differentiates the second (unsorted) random variable X₂ in the sample from the (sorted) second order statistic X₍₂₎.

The CDF P(X₍₂₎ <= x) and PMF P(X₍₂₎ = x) of the second order statistic can now be computed by plugging-and-chugging p₁ and p₂ into the formulas in the encyclopedia article:

Specifically, the probability mass function (PMF) is
P(X₍₂₎ = x) = (-1 + 2 x)/36 for x in {1,...,6}, 0 otherwise

Therefore the mean (expected value of X₍₂₎) is
E(X₍₂₎) = Sum[x P(X₍₂₎ = x), {x,1,6}] = Sum[x (-1 + 2 x)/36, {x,1,6}] = 161/36 (exact) = 4.472222 (2 repeats forever)

You don't even have to compute this sum yourself: Just plug it into WolframAlpha.

Notice the value of the mean 4.472 from the formula is the same value of the mean you got from your spreadsheet. But now you also have a nice expression for the probability of getting any value of x from 1 to 6, namely (-1 + 2 x)/36, without using a spreadsheet.

[Harbinger]

Thanks for that aher!

I was never a huge fan of stats (more a calculus guy). Since I've restarted playing D&D (quit when I was 12) I have been reteaching myself using first principles. I'll add your formula to my bag of tricks, but I'll need to derive it myself to truly grok it.

[kent]

It might not be obvious to folks from the above that X2 is the higher of the two dice rolls.

[Zenopus]

Feb 14, 2012 14:07:14 GMT -6 Harbinger said:

To compare these two techniques you can calculate damage per round:

dpr = % chance to hit x average damage

Let's assume you're an OD&D 4th level fighter needing a 10 to hit leather (AC 7). That's 55% chance. Now if you roll two dice and take the best, that's the same as saying 'What is my chance of hitting at least once on two dice?'.

This is easier to calculate if you reverse the question: 'What is the chance that I not hit with two dice?'. This is %miss x %miss = 45% x 45% = 20.25%

So your chance to hit is 1 - 0.2025 = 0.7975

And average damage on a d6 is 3.5 so:

DPR for roll atk twice = 0.7975 x 3.5 = 2.79125 dmg per round

Not sure I'm following all this, but average damage per round is increasing from 1.9 (.55 x 3.5) to 2.79 (.7975 x 3.5)? Isn't that the same as giving the fighter almost a +5 to hit? That seems like a large to-hit bonus for OD&D.

[aher]

Feb 15, 2012 6:38:34 GMT -6 kent said:

It might not be obvious to folks from the above that X2 is the higher of the two dice rolls.

Yes, thanks for pointing this out. I didn't explain the notation in my original post. So here's more detail about the notation used in order statistics:

• X₁,...,X_n are the unordered/unsorted random variables (also called variates)
  
• x₁,...,x_n are the unordered/unsorted observations. Notice upper-case for variates. Lower-case letters for observations.
  
• X₍₁₎ <= ... <= X_(n) are the ordered/sorted variates, called the "order statistics." Notice the parentheses around the subscripts.
  
• x₍₁₎ <= ... <= x_(n) are the ordered observations. Again, observations are lower-case. Random variables (or variates) are upper case. Parentheses for order.
  

Some authors prefer this notation for the order statistics: X_1:n <= ... <= X_n:n, called "extensive form." No parentheses in the subscripts, but the sample size "n" is shown explicitly.

As harbinger said, this stuff is trickier. Hope this makes it a little clearer.

[Harbinger]

Feb 15, 2012 22:05:25 GMT -6 Zenopus said:

Not sure I'm following all this, but average damage per round is increasing from 1.9 (.55 x 3.5) to 2.79 (.7975 x 3.5)? Isn't that the same as giving the fighter almost a +5 to hit? That seems like a large to-hit bonus for OD&D.

Good point. This is why AD&D two-weapon fighting is given a -2 on the main weapon and -4 on the off-hand weapon (I think those were the numbers). I've read somewhere that evens out the bonus. Let's see with a d8 sword and a d6 short sword.

.55 x 4.5 = 2.475

.45 x 4.5 + .35 x 3.5 = 3.25

Nope, it's still a +3 ~ +4

-6 and -4 works:

.35 x 4.5 + .25 x 3.5 = 2.45

Of course, it's still a curve (not linear), so this only matches at this particular character level.

[darkling]

Feb 17, 2012 16:15:29 GMT -6 Harbinger said:

Of course, it's still a curve (not linear), so this only matches at this particular character level.

Mmhmm! I managed to create the same graph you made earlier with the basic single weapon DPR in as a base-line. Though it occurs to me that I have no idea how to upload it as an attachment, maybe I should just upload it to a separate imagehost.

I think I am actually a big fan of how this is falling out so that there are strong situational drives to certain fighting styles. I think that once it is more fleshed out it will only get better. Remember that while using that second hand for either a second weapon or a double grip on a single weapon will always net a basic DPR advantage it can hinder your character in other ways.

Of course I tend to lean heavily towards tracking things like weight and logistics, so it becomes a pertinent questions in a number of situations whether you have a free hand for balance as you fight on a cliff face, will you have to drop your weapon to save yourself when a trap goes off, do you really want the extra encumbrance of lugging around another sword, etc. E.g. in my last session my party ran into a classic trapdoor that threatened to dump them two levels down into a lake housing a small shrine with a water elemental. Those with free hands got a chance to grab onto something and save themselves, those with full hands had to choose between taking the plunge and dropping whatever they were holding into the water below, and once they hit the water we began looking at just how much weight was dragging them to a watery grave and so wound up with characters desperately dropping their weapons and packs to try to swim.

And this only becomes more interesting if you introduce weapons with damages beyond 1d6. I might try to run those numbers.

[darkling]

Okay, so I have been running more numbers and I stumbled across something really interesting!

As we all know the average result of rolling 1d6 is 3.5.
Likewise the average result on 1d4 is 2.5 and on 1d8 it is 4.5.
So here's the thing: the average when rolling 2d6 and keeping the highest is 4.47 or ~4.5. And the average result when rolling 2d6 and keeping the lowest is 2.52.
So basically, averaged over time, you get almost identical results from forcing a player to roll 1d6 twice and take the max or min as you would from scaling the die type. Which I think is cool.

[Harbinger]

Feb 20, 2012 9:20:21 GMT -6 darkling said:

Okay, so I have been running more numbers and I stumbled across something really interesting!

As we all know the average result of rolling 1d6 is 3.5.
Likewise the average result on 1d4 is 2.5 and on 1d8 it is 4.5.
So here's the thing: the average when rolling 2d6 and keeping the highest is 4.47 or ~4.5. And the average result when rolling 2d6 and keeping the lowest is 2.52.
So basically, averaged over time, you get almost identical results from forcing a player to roll 1d6 twice and take the max or min as you would from scaling the die type. Which I think is cool.

Welcome to the world of neat'o mechanics!

I've always liked using:
- Magic User - take min of 2d6
- Cleric - d6
- Fighter - take max of 2d6

Or you could say 'light', 'medium', 'martial' weapon.

[aher]

Feb 20, 2012 9:20:21 GMT -6 darkling said:

So basically, averaged over time, you get almost identical results from forcing a player to roll 1d6 twice and take the max or min as you would from scaling the die type.

In the long run, for large samples, the differences between the sample means of d8 and max{d6,d6} may be negligible. However, in the short run, for small samples, the differences between these two distributions could be critical.

Let's say I get one last shot at a foe before I'm paralyzed. I make the hit. Now I'm about to roll for damage.

• Case (a) I need to roll 7 or more, or I'm doomed for sure. My chances are 0.25 on d8, but zero on max{d6,d6}.
  
• Case (b) I need 5 or more. The probability of rolling 5 or more on d8 is 0.5. The probability of rolling 5 or more on max{d6,d6} is 0.555 (5 repeats forever), giving me a slight edge (0.0555...) over d8.
  
• Case (c) The only thing I can't roll is a "1". The probability of rolling a "1" is 1/8 on a d8 but only 1/36 on a max{d6,d6}. So the chance of rolling "1" on a d8 is 4.5 times higher on a d8 than on max{d6,d6}.

The word "average" is somewhat ambiguous. While it commonly refers to "mean", it could mean...

• mean: expected value E(X), i.e., sum(x f(x) for each outcome x)
  
• median: middle value m; F(m) >= 1/2 and (1 - F(m) + f(m)) >= 1/2
  
• mode: most frequent value; x for which f(x) attains its maximum
  (where f is the pmf and F is the cdf).

With this in mind, the "averages" of these two distributions -- d8 and max{d6,d6} -- are somewhat different:

• mean of d8 = 4.5 ≈ mean of max{d6,d6} = 4.47222 (2 repeats forever)
  
• median d8 = 4.5, but median max{d6,d6} = 5
  
• d8 is modeless, but mode of max{d6,d6} is 6

Furthermore, the spreads of these two distributions are quite different:

• stddev d8 ≈ 2.29129
  
• stddev max{d6,d6} ≈ 1.40408355

This means a random sample drawn from max{d6,d6} will be clustered together more closely around the mean, than will a random sample drawn from d8.

Here's a summary of the properties of all the distributions under consideration in this thread:

d4
mean = E(d4) = 5/2 = 2.5
median = 2.5
mode = undefined
variance = var(d4) = 5/4 = 1.25
stddev = sqrt(5)/2 ≈ 1.11803
pmf = P(d4=x) = 1/4 for x in {1,2,3,4}; 0 otherwise
cdf = P(d4<=x) = floor(x)/4 for x in [0,4] N.B. This is a non-decreasing, right-continuous step function

min{d6,d6}
mean = 91/36 = 2.52778
median = 2
mode = 1
var = 2555/1296 = 1.9714506172839506172839 (506172839 repeating)
stddev = sqrt(2555)/36 ≈ 1.40408355067779191
pmf = P(min{d6,d6}=x) = (13-2x)/36 for x in {1,2,3,4,5,6}; 0 otherwise
cdf = P(min{d6,d6}<=x) = (12x - x^2)/36 after step(6) where "after" means function composition and "step(6)" is a step function, defined below.

d6
mean = 7/2 = 3.5
median = 3.5
mode = undefined
var = 35/12 = 2.91666 (6 repeats forever)
stddev ≈ 1.707825
pmf = P(d6=x) = 1/6 for x in {1,2,3,4,5,6}; 0 otherwise
cdf = P(d6<=x) = floor(x)/6 for x in [0,6]

max{d6,d6}
mean = 161/36 = 4.472222 (2 repeats forever)
median = 5
mode = 6
var = 2555/1296 = 1.9714506172839506172839 (506172839 repeating)
stddev = sqrt(2555)/36 ≈ 1.40408355067779191
pmf = P(min{d6,d6}=x) = (2x-1)/36 for x in {1,2,3,4,5,6}; 0 otherwise
cdf = P(min{d6,d6}<=x) = (x^2)/36 after step(6)

d8
mean = 9/2 = 4.5
median = 4.5
mode = undefined
var = 21/4 = 5.25
stddev = sqrt(21)/2 ≈ 2.29129
pmf = P(d8=x) = 1/8 for x in {1,2,3,4,5,6,7,8}; 0 otherwise
cdf = P(d8<=x) = floor(x)/8 for x in [0,8]

Where...

step(a) : Reals -> {0,1,2,...,a} is a step function, defined as
step(a) = 0 χ_(-∞,1) + sum(b χ_[b,b+1) for b in {1,...,a-1}) + a χ_[a,∞)

χ_A : Reals -> {0,1} is the indicator function of the set A, defined as
χ_A(x) = 1 if x in A; 0 otherwise