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20130118, 01:51 AM (ISO 8601)
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How EXACTLY do you calculate an Empowered spell result with an odd number of dice?
This is based on a D&D question but isn't terribly systemspecific
Let's start with a rough approximation: 1d4 becomes 1d6, 5d8 becomes 5d12, and so on. However, that's not quite right! If, instead of multiplying by 1.5, we merely double, it's more obvious; twice 1d4 is definitely not the same as 1d8, because it has a different mean, different standard deviation, different minimum, and so on.
So how about adding half dice? For example, 3d4 becomes 3d4+3d2, 11d6 becomes 11d6+11d3, and so on. Well, obviously those are rather tedious, since in many cases you have to roll the regular die size and carefully count the results, then halve them. Also, it's still not quite correct; it has too high a minimum and too high a mean, and its standard deviation looks a bit low, though I can't say for sure.
Then suppose we attempt to adjust for this by rolling the very minimum number of halfsized dice and folding the others back into the regular die size: 7d6 becomes 10d6+1d3, 9d8 becomes 13d8+1d4, and so on. Checking on AnyDice shows that this is very close, but still a bit too high.
Speaking of AnyDice, there's an algorithm in there for multiplying dice, but it seems to merely multiply the results. In any case, it seems to return results with slightly too low a mean, and with gaps between possible numbers.
A theoretical solution for empowering 3d4 should have a minimum of 4.5, a maximum of 18, and a mean of 11.25. Here's a summary of die schemes I've tried, including two I didn't mention, and none of which are quite right. If any brave soul wants to calculate the desired standard deviation, or can think of a better approximation that isn't entirely impractical to perform, I'd be endlessly grateful.Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130118, 02:11 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Roll the normal damage, multiply the total by 1.5.

20130118, 02:46 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
Anyone knows blue is for sarcas'ing in · "Take 10 SAN damage from Dark Orchid" · Use of gray may indicate nitpicking · Green is sincerity

20130118, 02:54 AM (ISO 8601)
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20130118, 03:17 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
AnyDice rounds noninteger values for a roll, and my guess is that it uses these artificiallyskewed numbers to calculate the deviation. You may find that this makes it impossible to find what you're looking for. Actually, looking at the means it gives, I'm certain that it uses artificiallyskewed numbers to calculate mean and std. dev. This is likely what causes the slightlylow means you noticed as well, as fractions appear to be rounded down.
Also, is there a reason you didn't do "output 3d4+3d4/2"? It does have a different graph than those you used, though the differentlyshaped bell curve probably means it doesn't fit. By the shape of the curve alone, I'd call "empowered multiplication" best.

20130118, 03:52 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Remember that "Empower" as the D&D feat operates under the D&D maths rules.
Half of 1d4 is not 1d2
Consider the possible results:
1 > 0.5 round down to 0*
2 > 1
3 > 1.5 round down to 1
4 > 2
*Now the 1 > 0 is a special case. If this was a base damage roll then the "minimum 1" rule applies and 0 > 1, but this only applies in aggregate (so gives different probability distributions depending on how many dice are being rolled as it only acts when they all come up on a 1). However Empower is a bonus to existing damage so the "minimum 1" rule never applies to the empower damage.
In fact, with D&D you are supposed to just multiply the original dice by 50% to get the empower total, the only time you don't is when "Maximise" is also applied when you have to roll the dice solely for the Empower figure.
Now for other time when you want "half a dX" you may be able to use equivalences, but D&D's Empower is special circumstances and it just does not work. (Oh the roll NdX + half(NdX) method works, it's just not quite correct by the rules.)
Edit: the distributions are even more messed up than I said above  rounding only works on the total so the result distributions actually depend on the number of dice, they don't scale properly.Last edited by Khedrac; 20130118 at 03:55 AM.

20130118, 05:37 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Being unable to get, say 8 damage from an Empowered Burning Hands at CL 5 (but being able to get 7 or 9 damage) is weird and unnatural. Any discontinuity in the damage roll curves indicates a quantum jump in the spell's effectiveness, beyond that necessary for abstraction: it indicates something else is going on.
Of course, inworld, it would be difficult for anyone to track, even with centuries of magical research into CL, HD, and so forth. However, difficult is not the same as impossible, and with enough persistence it could be discovered.
Hmm. That may be true. (I'm already aware that I may have already mapped out the best schemes that are practical.)
Also, is there a reason you didn't do "output 3d4+3d4/2"? It does have a different graph than those you used, though the differentlyshaped bell curve probably means it doesn't fit. By the shape of the curve alone, I'd call "empowered multiplication" best.
I hadn't really taken the time to figure out why the halfsized dice were too high, but that makes sense.
In fact, with D&D you are supposed to just multiply the original dice by 50% to get the empower total, the only time you don't is when "Maximise" is also applied when you have to roll the dice solely for the Empower figure.
Now for other time when you want "half a dX" you may be able to use equivalences, but D&D's Empower is special circumstances and it just does not work. (Oh the roll NdX + half(NdX) method works, it's just not quite correct by the rules.)
Edit: the distributions are even more messed up than I said above  rounding only works on the total so the result distributions actually depend on the number of dice, they don't scale properly.
Anyway, I adjusted the program to add a few more schemes . Tell me what you think. I'm beginning to take a hankering to "minimal partial multiplication", myself.Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130118, 06:24 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
You must really have a lot of time to waste
I'm usually very interested in dice, math and probability but this issue never came to my mind.
I understand your point, but there are some problems that work against a smooth distribution, like the rules on rounding and the fact that dice give inevitably discrete results.
The problem is more evident with a small number of small dice, and you rightly choose 3d4 for your tests.
I do not have time to do the math behind this, but I have the feeling that you can go near the result you are searching by rolling 2 sets and halve the second.
I just noticed that you already did: 3d4+(3d4/2) named "partial multiplication". This should give the right distribution (except the fact that 3d4/2 is rounded) and avoid 'holes' in distribution.
Didn't it work?

20130118, 07:45 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
It was just something sparked by another thread that was discussing the best spells to Empower, and why; it seemed that spells that give low numbers of d4s are best, which led to the realization that, at those ranges, it's nontrivial to calculate exactly what the result is. (Take time stop, or enervation, for example.)
Also I am, in general, highly dedicated to best practices.
I understand your point, but there are some problems that work against a smooth distribution, like the rules on rounding and the fact that dice give inevitably discrete results.
I just noticed that you already did: 3d4+(3d4/2) named "partial multiplication". This should give the right distribution (except the fact that 3d4/2 is rounded) and avoid 'holes' in distribution.
Didn't it work?
OTOH, there's only really one other serious candidate: minimal partial multiplication, which is a shortcut to reduce the amount of division you have to do at the table. But here's a shortlist graph anyway.Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130121, 06:27 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Hi, I have some time to return on the question.
I don't understand the part on the rule reading.
I have only access to the D20 SRD, and it doesn't say much. What is strictly allowed by the rules?
Before rereading this thread, I thought that the rules stated: "roll normal damage, then add 1/2 of it (or multiply by 1.5, or use 150%, etc)".
Only now I understand that you interpret: "roll all the dice + half the dice". And this can be done simply with a 6d6 fireball, that becomes 9d6, but not with odd numbers of dice (look, it's in the thread title ). This make me think that it's not the right way to do it.
To my reading, the strictly allowed method is the multiplication (x1.5) above. An acceptable ruling/houserule can be two rolls: 1 full + 1/2 of the second. This only for the sake of smoothness of distribution. (And I know players who would be strongly against it, as in this way rolling the maximum it's more difficult)

20130121, 08:17 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
In the D&D 3.5 Players Handbook under Empower Spell it states:
Originally Posted by PHB

20130121, 09:07 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
One moment, does not it state all variable, numeric effects? I thought it was 1d4 x 1,5 and then +1.
Not a big difference on Magic Missile, but maybe higher in other cases.

20130121, 09:18 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
π = 4
Consider a 5' radius blast: this affects 4 squares which have a circumference of 40' — Actually it's worse than that.
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20130121, 11:04 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Yep. Shockingly enough you are expected to add 50% to the damage or other variable numeric effect by adding 50% to the damage or other variable numeric effect.
Not by rolling twice and taking half of one, and not by rolling 9d6 for a 6d6 spell.
There's nothing whatsoever in the rules to justify any interpretation except "add 50% to the result" and when doing so you should be using the usual D&D rounding rules, which does result in gaps in the distribution.
Who cares? Repeat after me, HP are an abstraction anyway.
Any discussion of whether you should roll 6d6+(6d6)/2 or 9d6 and of how you roll a half die is a discussion of someone's house rule. Strangly, the people discussing this house rule seem to think they're discussing an ambiguity in RAW, when RAW are completely clear and even provide helpful examples.

20130121, 11:09 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
It does, but this is an odd one  the FAQ gives a little more details but here goes (as I understand it).
If a spell does 1dX+(N per level) the variable numeric effect (vne) is 1dX
Because Magic Missile does (1d4+1) × Y where Y is level dependent the vne is "1d4+1" and the empower is applied after the +1.
There are other spells in this category but they are quite rare.
Anyway it does make the distribution interesting  for one missile:
Dice Result > Damage
1 > (1+1)*3/2 = 3
2 > (2+1)*3/2 = 4
3 > (3+1)*3/2 = 6
4 > (4+1)*3/2 = 7

20130121, 11:47 AM (ISO 8601)
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20130121, 09:54 PM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Personally I've always added up the damage, then multiplied by 1.5, which is how it's written in the feat. Yes, it's not perfect abstraction allowing for any value between minimum and maximum, but it is how the feat is written by RAW. While I'm sure we could write out a better version, we must first acknowledge that creating a random set was not the intent of the feat, and it actually is working as intended (One of the few things in 3.5 that does in all cases where it is employed :p)
Still, don't let that discourage you from finding a better way, and posting it up in Homebrew :3

20130122, 05:36 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Fair enough. I find this unsatisfactory for reasons of consistency as well as statistics, but I suppose I'll have to accept that I'm diverging.
Ehhh, not sure it's worth putting in homebrew per se, it's just a slightly smoother fit that should otherwise be mechanically identical.
Although, one area there could be a substantial change, now that I think about it, is if you apply one of the few "extra points of damage per die" abilities. That's ... unfortunate, and to be fair may be enough of a reason to use the inferior WotC algorithm. (OTOH, Empower Spell isn't especially overpowered, nor are Warmages or War Mages, so....)
Incidentally, perhaps for that reason, Empower Power is even clearer in the SRD; it mentions that you should "multiply 1½ times the damage total of the augmented power", among other things.Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130122, 07:10 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Actually I'm not sure that I'm all that consistent about how I do it — but then both methods yield the same result and with similar distributions — so it probably doesn't matter. I think I changed over when I started to see Empower + Maximise.
π = 4
Consider a 5' radius blast: this affects 4 squares which have a circumference of 40' — Actually it's worse than that.
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20130122, 07:37 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Black text is for sarcasm, also sincerity. You'll just have to read between the lines and infer from context like an animal

20130122, 09:46 PM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
I already addressed this in the post you quoted, going on to say "beyond that necessary for abstraction". Inworld, the existence of HP could be deduced, which I find unfortunate but essentially unavoidable; the existence of larger jumps with Empowered spells is just as unfortunate and far more avoidable.
Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130124, 11:15 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Speaking mathematically, multiplication by 1.5 is not "onto" (or "surjective"  nor injective either, for that matter) from the natural numbers (1,2,3,...) to the natural numbers. This is inherent to multiplication by 1.5, and rounding down doesn't change it. Basically, it may be weird, but it is natural.
There's also no way to "fix" it, barring picking a set of natural numbers that you feel is enough and writing a custom function that is 11 ("injective"). That's because as you get into higher numbers, there are bigger gaps.
tl;dr: as long as you're multiplying by 1.5, there will be gaps.Last edited by rockdeworld; 20130125 at 07:44 AM.
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20130125, 06:21 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
OK but... that's assuming you're starting with natural numbers in the first place, which I find strange. Sure, eventually damage has to be expressed as a whole number, but why would you apply rounding earlier in the process than you need to?
Or, in programming terms, float > int, not int > int.
Honestly, I think that's the problem I have with a lot of WotC math: they made assumptions that are questionable, and that I disagree with ("damage is quantized in essence, not merely as a convenience", "truncating works for everything, including averages", and so forth).Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130125, 07:42 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Last edited by rockdeworld; 20130125 at 09:17 AM.
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20130125, 07:56 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
It indicates something else going on, yeah: lazy math.
Insetting that jump does not matter. Since the simple 1,5 multiplication, compared to a more accurate method, does not break the boundaries of the effect nor changes the distribution of possibilities, it does not matter. Trying for a more accurate method will slow down your game for no gain.Last edited by Andreaz; 20130125 at 07:57 AM.

20130125, 09:12 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Black text is for sarcasm, also sincerity. You'll just have to read between the lines and infer from context like an animal

20130125, 08:29 PM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
Well, I think this has pretty much wound down, with the result that two schemes are sensible:
 (3d4) * 1.5 — the RAW way to do it
 4d4 + (1d4/2) — a more precise approximation that minimizes the number of halved dice
Everything else is less accurate and/or less efficient.
Conceptually, I consider dice a way to generate natural numbers that are representative of the actual continuum of damage possibilities, which is not, or should not be, inherently composed of natural numbers.
Unfortunately, it's possible by following the rules strictly for a character to deduce the existence of these jumps in damage possibilities; it's not possible for something to do 3.5 damage, obviously, never mind 3.272492347 damage.
What I object to, though, is in extending this any further than strictly essential: the resolution should never be coarser than natural numbers if it is at all avoidable, so being unable to generate 8 damage is a bad sign. Conveniently, it is avoidable, and it's actually not terribly hard to do so either.
I don't actually think it will slow things down significantly; the difference between (3d4)*1.5 and 4d4+(1d4*0.5) or 3d4+(3d4*0.5) is not substantial. (Although admittedly proper rounding may be a bit sticky, but consistency there is practical and useful.)
Also, different die schemas demonstrably do change the distribution, not only in probability but in some cases minimum, maximum, and obviously spots in between.
Well, I consider a onetime examination worth it if it costs relatively little to use in game (as mentioned above). I'm not sure what you mean about the problem of "giving attention to the abstractions".Projects: Homebrew, Gentlemen's Agreement, DMPCs, Forbidden Knowledge safety, and Top Ten Worst. Also, Quotes and RACSD are good.
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20130126, 12:08 AM (ISO 8601)
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Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice
You also seem to be most worried about the small end. I see the RAW method as significantly favorable on the higher end, when you're rolling 20orso dice already (or more!). There, you just roll 20 dice, for instance, and take the total * 1.5, versus your method that would have you rolling 30 dice. Ten extra dice to roll and count, that's a significant hassle to me.