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The Mathematical Model of the d20 System
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<blockquote data-quote="Wulf Ratbane" data-source="post: 4188364" data-attributes="member: 94"><p>Hey Johnny--</p><p></p><p>I've done a whole lot of work since that PDF. I'll just say for now that I think it's incomplete and I'm now trying to wrangle Lanchester's Square Law into the whole encounter design model. It's not easy... </p><p></p><p>The PDF will serve you just fine in the meantime, but you do need to understand that the meaning of EL5 in the PDF <strong>is not</strong> the same as EL5 as WotC uses the term. You need to be able to throw out your old knowledge before applying the new. "EL5" might as well mean "Orange" for all its applicability to the term as WotC used it.</p><p></p><p></p><p></p><p>They = WotC, or They = Me (Bad Axe)? </p><p></p><p>For the record, the curve is quadratic (to the power of 2), not (strictly speaking) exponential.</p><p></p><p></p><p></p><p>That is correct.</p><p></p><p>What you're looking for is the amount of Offense that the creature can put out per action, multiplied by the Defense (which approximates the number of actions you expect it to stick around for).</p><p></p><p>The simplest and roughest model here (or as you aptly put it, the "identical and spherical horse theory"-- I like that, I'll keep using it!) is Hit Points x Average Attack Damage.</p><p></p><p>Here's a fun little process you might want to try. Pick a monster out of the SRD at random-- but one that doesn't have a lot of unusual abilities. Something straightforward. Multiply its hit points times its main attack average damage. Now divide that result by 25. Now take the square root of that. You'll be in the ballpark of its CR. </p><p></p><p>If you used a typical brute monster and your result ended up higher than expected, there's a good chance that monster is a lot rougher than expected. The CR3 Ogre weighs in at a little over CR4, which I think is probably more accurate. </p><p></p><p>The CR5 Troll, meanwhile, weighs in at an expected 4.89... unless you start taking his full attack routine and Rend into account. Suddenly it jumps way up, and the troll gets a lot tougher-- as anyone who has ever stood toe to toe with a troll for a full attack can attest. </p><p></p><p>Obviously, a lot of things can make your horses less spherical. This is where the "Art vs. Science" of monster design starts to play with our mathematical expectations.</p><p></p><p>Higher than expected AC (on the order of +4 or +5 points higher than expected) is going to give the creature a lot more staying power-- akin to doubling its hit points. An area of effect attack that can catch the whole party in one action is like multiplying the creature's damage output by the size of the party. And a Save-or-Suck attack that can take one PC out of the fight in one action is as good as doubling that creature's combat power.</p><p></p><p>It is a wildly approximate model based on the average of monster abilities (statistical analysis thanks to Ryan Stoughton). So while it holds up very well across the average-- identical and spherical!-- I certainly wouldn't bet it to WIN.</p><p></p><p>Yet. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":)" title="Smile :)" data-smilie="1"data-shortname=":)" /></p><p></p><p>I'm working on it, though.</p></blockquote><p></p>
[QUOTE="Wulf Ratbane, post: 4188364, member: 94"] Hey Johnny-- I've done a whole lot of work since that PDF. I'll just say for now that I think it's incomplete and I'm now trying to wrangle Lanchester's Square Law into the whole encounter design model. It's not easy... The PDF will serve you just fine in the meantime, but you do need to understand that the meaning of EL5 in the PDF [b]is not[/b] the same as EL5 as WotC uses the term. You need to be able to throw out your old knowledge before applying the new. "EL5" might as well mean "Orange" for all its applicability to the term as WotC used it. They = WotC, or They = Me (Bad Axe)? For the record, the curve is quadratic (to the power of 2), not (strictly speaking) exponential. That is correct. What you're looking for is the amount of Offense that the creature can put out per action, multiplied by the Defense (which approximates the number of actions you expect it to stick around for). The simplest and roughest model here (or as you aptly put it, the "identical and spherical horse theory"-- I like that, I'll keep using it!) is Hit Points x Average Attack Damage. Here's a fun little process you might want to try. Pick a monster out of the SRD at random-- but one that doesn't have a lot of unusual abilities. Something straightforward. Multiply its hit points times its main attack average damage. Now divide that result by 25. Now take the square root of that. You'll be in the ballpark of its CR. If you used a typical brute monster and your result ended up higher than expected, there's a good chance that monster is a lot rougher than expected. The CR3 Ogre weighs in at a little over CR4, which I think is probably more accurate. The CR5 Troll, meanwhile, weighs in at an expected 4.89... unless you start taking his full attack routine and Rend into account. Suddenly it jumps way up, and the troll gets a lot tougher-- as anyone who has ever stood toe to toe with a troll for a full attack can attest. Obviously, a lot of things can make your horses less spherical. This is where the "Art vs. Science" of monster design starts to play with our mathematical expectations. Higher than expected AC (on the order of +4 or +5 points higher than expected) is going to give the creature a lot more staying power-- akin to doubling its hit points. An area of effect attack that can catch the whole party in one action is like multiplying the creature's damage output by the size of the party. And a Save-or-Suck attack that can take one PC out of the fight in one action is as good as doubling that creature's combat power. It is a wildly approximate model based on the average of monster abilities (statistical analysis thanks to Ryan Stoughton). So while it holds up very well across the average-- identical and spherical!-- I certainly wouldn't bet it to WIN. Yet. :) I'm working on it, though. [/QUOTE]
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