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Rules for "multiplication" in D&D and Pathfinder are unusual.

This is what Pathfinder rules say about multiplication:

When you are asked to apply more than one multiplier to a roll, the multipliers are not multiplied by one another. Instead, you combine them into a single multiplier, with each extra multiple adding 1 less than its value to the first multiple.

...and this is what D&D 3.5e rules say:

When two or more multipliers apply to any abstract value (such as a modifier or a die roll), however, combine them into a single multiple, with each extra multiple adding 1 less than its value to the first multiple.

I know the D&D rule for multiplication does not apply to real-world values, only to abstract ones, and I have no idea if the problem I'm describing here ever happens in D&D. This is not true for Pathfinder, where the provided rules are true for every out-of-game, rule-induced multiplication, and the problem arises (see below).


So if you double \$A\$, then triple it, you get \$A \times \left[2+\left(3-1\right)\right] = 4A\$ instead of the expected \$6A\$ that a true multiplication would give us.

How does this work when dividing?
(e.g. when halving creation times because of two different features or feats in Pathfinder. I'm not sure halving twice ever applies to abstract values in D&D.)

  • Do I just divide like in the real world, since this rule is only valid for multipliers?

    $$ A \times \frac{1}{2} \times \frac{1}{2} = A \times \frac{1}{4} $$

  • Do I apply the multiply rules, since division is just multiplication with fractions?

    $$ A \times \frac{1}{2} \times \frac{1}{2} = A \times \left[ \frac{1}{2} + \left( \frac{1}{2} - 1 \right) \right] = A \times 0 $$

    I guess not. As you can see the formula breaks for any second or later multiplier that is 1 or lower (which should not happen, with numbers > 1 the formula works equally well whatever the order of the multipliers).

  • Do I find the inverse of the operation, so that when I multiply the values back it gives me the explected value??

    $$ A \times \frac{1}{2} \times \frac{1}{2} = A \times \frac{1}{2 + \left(2 - 1\right)} = A \times \frac{1}{3} = B $$

    $$ \text{so that } B \times 2 \times 2 = \left(A \times \frac{1}{3}\right) \times \left[2 + \left(2 - 1\right)\right] = A $$

I don't really expect the authors to have thought about this, but who knows?
Maybe you know.
...or at least I'm hoping so.

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    \$\begingroup\$ I think I see what you're getting at, but can you give us a concrete example of a situation where you'd need to know? That would be a lot clearer. \$\endgroup\$ – SirTechSpec Jun 26 '16 at 12:50
  • \$\begingroup\$ @SirTechSpec rpg.stackexchange.com/q/83184/4173 this is what spawned the question in my mind. I'm editing the question accordingly. \$\endgroup\$ – Zachiel Jun 26 '16 at 13:18
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Multiplication of "xN" multipliers on a roll is a special case with a special rule in the game (http://www.d20pfsrd.com/basics-ability-scores/glossary#TOC-Multiplying). It doesn't even apply to all multiplication in the game, just that on rolls that use a multiplier. Division is not lined out as a special case, it works like normal math (plus rounding down of course), which should be clear from all the rules that discuss the progression from 1/2 to 1/4 (greater evasion, movement rates, etc.). They say halved and they mean halved, not "times 1/2" - mathematically equivalent in math class, but not the way the rules work.

Note that the answer to the question you link was incorrect (it has been corrected) under any interpretation of whether 'division is multiplication by a fractional numerator' because the multiplication rule strictly states it only applies to a roll that has a multiplier on it - "When you are asked to apply more than one multiplier to a roll" - not times, not distances, not things that aren't a roll. It is very clearly scoped.

Note also that many effects - like size change - work in moving up and down categories, not actual doubling or halving.

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  • \$\begingroup\$ I wrote my question here before the answers in the linked question came, in order to provide a starting point for the answers there. \$\endgroup\$ – Zachiel Jul 10 '16 at 9:08
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There's no explicit rule for that, so that falls in GM discretion lands by default. But we are trying to figure out a rule to manage this special case.

Let's leave here, for reference, the multiplying rule:

Multiplying: When you are asked to apply more than one multiplier to a roll, the multipliers are not multiplied by one another. Instead, you combine them into a single multiplier, with each extra multiple adding 1 less than its value to the first multiple. For example, if you are asked to apply a \$\times2\$ multiplier twice, the result would be \$\times3\$, not \$\times4\$.

Let's remember this rule applies to rolls, not everything. And the rule does not even apply to all rolls, there are some exceptions, use common sense. Determine if the rules apply to the current case if we were multiplying.

Now, i'd go with finding the inverse of the operation, 'so that when [we] multiply the values back it gives [us] the expected value', one of your suggestions.

$$ a = \frac{\left(\frac{A}{2}\right)}{2} $$

$$ a = A \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) $$

$$ a = A \left[\frac{1}{2 + \left(2 - 1\right)}\right] $$

$$ a = \frac{A}{3} $$

Why?

First: because it keeps the balance between both operations, in simple scenarios. That simple. So they can be mathematically countered. If not, division becomes stronger than multiplication for the same cases, breaking the balance of the game, meaning people would better learn spells that halve your numbers.

Second: because it works similar to the multiply rule, not as an exception of the exception.

Third: because this way we avoid division by zero, negative dividers or multipliers and so on.

The general rule would be:

Dividing: When you are asked to apply more than one divider to a roll, the dividers are not multiplied by one another. Instead, you combine them into a single divider, with each extra divider adding 1 less than its value to the first divider. For example, if you are asked to apply a \$\div3\$ divider twice, the result would be \$\div5\$, not \$\div9\$.

In math terms:

$$ a = A\left(\frac{1}{B_1}\right)\left(\frac{1}{B_2}\right) \dotsm \left(\frac{1}{B_n}\right) $$

$$ a = A \left[\frac{1}{1 + \left(B_1 - 1\right) + \left(B_2 - 1\right) + \dotsb + \left(B_n - 1\right)}\right] $$

The big con is: This rule does not behave properly when combining both rules. Suppose we are asked to multiply a roll by \$\times2\$ twice, then by \$\times4\$, and also halve it twice, we would expect for the result to be \$\times4\$, right? but no. Look:

$$ r = R \times 2 \times 2 \times 4 \div 2 \div 2 $$

$$ r = R \frac{1 + \left(2-1\right) + \left(2-1\right) + \left(4-1\right)}{1 + \left(2-1\right) + \left(2-1\right)} $$

$$ r = R \frac{1 + 1 + 1 + 3}{1 + 1 + 1} $$

$$ r = R \left(\frac{6}{3}\right) $$

$$ r = R \left(2\right) $$

The result would be \$2\$, not even \$3\$. The workaround would be to cancel opposite multiplier/dividers before making the operation... something easy at a table but not if you are programming a game or app with the rules in mind, and have to consider effects duration, area of effects and so on.

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