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    Default 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 system-specific

    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 half-sized 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.
    Quote Originally Posted by Water_Bear View Post
    That's RAW for you; 100% Rules-Legal, 110% silly.
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    Default 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.

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by kieza View Post
    Roll the normal damage, multiply the total by 1.5.
    That's essentially AnyDice's method. As mentioned in passing, that leaves gaps where some totals aren't possible.
    Quote Originally Posted by Water_Bear View Post
    That's RAW for you; 100% Rules-Legal, 110% silly.
    Quote Originally Posted by hamishspence View Post
    "Common sense" and "RAW" are not exactly on speaking terms
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    Anyone knows blue is for sarcas'ing in "Take 10 SAN damage from Dark Orchid" Use of gray may indicate nitpicking Green is sincerity

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by tuggyne View Post
    That's essentially AnyDice's method. As mentioned in passing, that leaves gaps where some totals aren't possible.
    How is this a problem?

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    AnyDice rounds non-integer values for a roll, and my guess is that it uses these artificially-skewed 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 artificially-skewed numbers to calculate mean and std. dev. This is likely what causes the slightly-low 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 differently-shaped bell curve probably means it doesn't fit. By the shape of the curve alone, I'd call "empowered multiplication" best.

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    Default 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; 2013-01-18 at 03:55 AM.

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by Zeful View Post
    How is this a problem?
    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, in-world, 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.

    Quote Originally Posted by NecroRebel View Post
    AnyDice rounds non-integer values for a roll, and my guess is that it uses these artificially-skewed 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 artificially-skewed numbers to calculate mean and std. dev. This is likely what causes the slightly-low means you noticed as well, as fractions appear to be rounded down.
    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 differently-shaped bell curve probably means it doesn't fit. By the shape of the curve alone, I'd call "empowered multiplication" best.
    No, there isn't, I just didn't think of it.

    Quote Originally Posted by Khedrac View Post
    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
    I hadn't really taken the time to figure out why the half-sized 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.
    This is probably not as clear as you'd like it to be; do you mean "multiply the number of original dice by 50%, and then roll those", or "multiply the original die roll by 50%"? Because usually D&D roll multiplication operates on the former principle (critical hits, for example).

    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.)
    Hmm. Ideally I'd want something that is statistically sensible, or as near as can be managed, while still at least paying lip service to D&D's weird rounding rules and what-not.

    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.
    Yeah, that makes sense. So rounding on each die would work better, you think?

    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.
    Quote Originally Posted by Water_Bear View Post
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    Default 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?

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by CalamaroJoe View Post
    I'm usually very interested in dice, math and probability but this issue never came to my mind.
    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 non-trivial 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.
    Yeah, D&D's rounding rules are messed up, and die quantization is unavoidable, but I believe it's possible to get a really good match, even if it's not perfect.

    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?
    It sort of does. It avoids gaps, certainly, and it closely matches what I suspect the ideal distribution to be, but it's not quite right, probably due to rounding. It's also not unambiguously correct: a strict reading of the rules may indicate it's disallowed, and it's not the only distribution that is similarly close.

    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.
    Quote Originally Posted by Water_Bear View Post
    That's RAW for you; 100% Rules-Legal, 110% silly.
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    Default 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 re-reading 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)

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    Default 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:
    Quote Originally Posted by PHB
    For example, an empowered magic missile deals 1-1/2 times its normal damage (roll 1d4+1 and multiply the result by 1-1/2 for each missile).
    This is pretty clear about the mechanic one is expected to use for the adding 50%.

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    Default 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.

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by NecroRebel View Post
    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 differently-shaped bell curve probably means it doesn't fit. By the shape of the curve alone, I'd call "empowered multiplication" best.
    This is how I do it it's a bit slow but it's easy enough.
    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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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by Khedrac View Post
    In the D&D 3.5 Players Handbook under Empower Spell it states:
    This is pretty clear about the mechanic one is expected to use for the adding 50%.
    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.

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by CalamaroJoe View Post
    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.
    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

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by Doug Lampert View Post
    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.
    Thank you for enlightening us, then.

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    Default 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

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by Khedrac View Post
    In the D&D 3.5 Players Handbook under Empower Spell it states:
    This is pretty clear about the mechanic one is expected to use for the adding 50%.
    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.

    Quote Originally Posted by Acanous View Post
    Still, don't let that discourage you from finding a better way, and posting it up in Homebrew :3
    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 1 times the damage total of the augmented power", among other things.
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    Default 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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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by tuggyne View Post
    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,
    No moreso than being able to take 0 points of damage and 1 point of damage but not 0.1 or 0.5 points of damage. Consider that a commoner has 4 hp, 1 point of damage is 25% points from death.
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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by Mastikator View Post
    No moreso than being able to take 0 points of damage and 1 point of damage but not 0.1 or 0.5 points of damage. Consider that a commoner has 4 hp, 1 point of damage is 25% points from death.
    I already addressed this in the post you quoted, going on to say "beyond that necessary for abstraction". In-world, 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.
    Quote Originally Posted by Water_Bear View Post
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    Quote Originally Posted by hamishspence View Post
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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by tuggyne View Post
    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.
    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 1-1 ("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; 2013-01-25 at 07:44 AM.
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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by rockdeworld View Post
    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 wierd, 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 1-1 ("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.
    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).
    Quote Originally Posted by Water_Bear View Post
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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by tuggyne View Post
    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?
    adx always results in a natural number for any a and x.

    f:A\N->N given by f(x)=1.5x, where A contains non-natural numbers, will almost never give you a natural number output, so I don't understand your question.
    Last edited by rockdeworld; 2013-01-25 at 09:17 AM.
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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by tuggyne View Post
    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.
    It indicates something else going on, yeah: lazy math.
    In-setting 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; 2013-01-25 at 07:57 AM.

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    Default Re: How EXACTLY do you calculate an Empowered spell result with an odd number of dice

    Quote Originally Posted by tuggyne View Post
    I already addressed this in the post you quoted, going on to say "beyond that necessary for abstraction". In-world, 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.
    Is it really worth the effort though? Giving attention to the abstractions may have a bigger impact than obscure quantum jumps in possible damage outcomes from empowered spells.
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    Default 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.

    Quote Originally Posted by rockdeworld View Post
    adx always results in a natural number for any a and x.

    f:A\N->N given by f(x)=1.5x, where A contains non-natural numbers, will almost never give you a natural number output, so I don't understand your question.
    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.

    Quote Originally Posted by Andreaz View Post
    It indicates something else going on, yeah: lazy math.
    In-setting 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.
    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.

    Quote Originally Posted by Mastikator View Post
    Is it really worth the effort though? Giving attention to the abstractions may have a bigger impact than obscure quantum jumps in possible damage outcomes from empowered spells.
    Well, I consider a one-time 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".
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    Default 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 20-or-so 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.

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