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Is there an equation to determine the Ability Modifier?

I'm not talking about the chart that lists

  • 1 = −5
  • 2-3 = −4
  • 4-5 = −3
  • 6-7 = −2
  • etc.

I mean a legit equation that, if I somehow misplaced this information, I could determine what the modifier is based on the score?

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5 Answers 5

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The formula is simply:

$$ \frac{\mathrm{ability} - 10}{2} $$

with the result rounded down, if it contains any halves.

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I know this is old but I think the current answers are close but slightly inadequate. I think the formula itself (rather than comments/descriptions) needs to account for rounding somehow, so here is an actual formula (in Google Sheets or Microsoft Excel syntax) that does work, and I feel is human-readable:

=(ROUNDDOWN(A1/2) - 5)

Where A1 is the ability score (or cell).

You could also use truncation, e.g. TRUNC(A1/2,0)-5, casting, e.g. INT(A1/2)-5, or the floor function, e.g. FLOOR(A1/2)-5 to effectively turn the decimals into integers. I feel like ROUNDDOWN is a bit more self-explanatory.

Mathematically, the above formula would be expressed using the floor function notation:

$$\left\lfloor\frac{a}{2}\right\rfloor - 5$$

Where \$a\$ is the ability score.

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  • \$\begingroup\$ I am amused that your formula is nearly identical to my original post. +1, and I am curious if readers will like it more than my current answer. \$\endgroup\$
    – Zhuge
    Oct 31, 2018 at 5:50
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It's quite simple if you do a different equation like this:

$$\left(\frac{\text{score}}{2}\right) - 5=\text{Modifier}$$

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    \$\begingroup\$ This doesn't quite work either; you still need some sort of situational rounding or truncation. \$\endgroup\$
    – Josh
    Oct 28, 2018 at 13:44
  • \$\begingroup\$ -1, because as Josh mentions it doesn't perfectly capture what is needed to calculate the answer, and otherwise doesn't say anything my answer didn't already. \$\endgroup\$
    – Zhuge
    Oct 31, 2018 at 5:51
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The Euclidean quotient of Ability divided by 2, remainder discarded, minus 5

Euclidean division is the process of dividing two integers to produce an integer quotient and an integer remainder. For example, 11 div 2 yields a quotient of 5 and a remainder of 1, where div signifies Euclidean division.

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  • \$\begingroup\$ Similar answers have been stated previously using a variety of descriptions of the division required or processing of the quotient after real number division. Perhaps this is the rigorous answer sought? \$\endgroup\$
    – Tuorg
    Nov 1, 2018 at 1:47
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I think it's already been stated, but the method I use is round the base number down to the nearest even (17->16) divide by half (16/2 = 8) and then subtract 5 (8-5=3)

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