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General Tabletop Discussion
D&D Older Editions, OSR, & D&D Variants
The Quadratic Problem—Speculations on 4e
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<blockquote data-quote="Wulf Ratbane" data-source="post: 3792123" data-attributes="member: 94"><p>This thread has not been particularly active, but my brain has been. And it's starting to hurt. Time for a braindump.</p><p></p><p>I'm still unable to completely (ok, let's be frank, <em>remotely</em>) incorporate Lanchester's Laws into the "Encounter Building" model.</p><p></p><p>Let us say that for any given party of four PCs, we can challenge them equally with either four equally-matched brutes, or eight lesser mooks. Lanchester says that an N-fold increase in numbers must be countered with an N-squared increase in quality. From this we can simply deduce that a single PC, or a single brute, must have <em>four times</em> the quality of a single mook. </p><p></p><p>We also know that the combat/encounter value of four brutes is the same as eight mooks-- both of these encounters are equal in difficulty. If the Encounter Design system says that you may buy either four brutes or eight mooks from the same pool of 1000 xp, for example, we can deduce that a brute costs 250 xp and a mook costs 125 xp. </p><p></p><p>So it would seem elementary that you could use your 1000 xp to build an encounter with TWO brutes (250x2) and FOUR mooks (125x4), right?</p><p></p><p>Not exactly. According to Lanchester's Laws, the combat value of this heterogenous group is NOT equivalent to either 4 brutes or 8 mooks-- it's actually a little bit stronger than either of the homogenous groups.</p><p></p><p>If a mook has a "combat quality" rating of 1, then 8 mooks have a "combat quality" of 8^2 (unit size) x 1 (their average quality) = 64. Similarly, four brutes, each with an individual "combat quality" of 4, have a total combat quality of 4^2 (unit size) x 4 (their average quality) = 64.</p><p></p><p>However, two brutes and four mooks (6 units) would have a total combat quality of 6^2 x [(4+4+1+1+1+1)/6] = 72, and increase in "difficulty" of 12.5%.</p><p></p><p>The main takeaway from Lanchester's Laws is that numbers count for more than quality. To say that five PCs can handle twenty goblins (a fourfold increase in numbers) is to say, in effect, that PCs are <em>sixteen times</em> as effective in combat as goblins. And if orcs are the brutes to our goblin mooks, what must an orc statblock look like? </p><p></p><p>Again, I have a hard time envisioning any kind of statblock for goblins, orcs, or the PCs that has any resemblance to the D&D that I know.</p><p></p><p>And that is before I have even begun to consider that a single "boss" monster designed to be on an even footing with five PCs must have <em>twenty-five</em> times their power.</p><p></p><p>To say that I am eager to see how WoTC solves this problem would be an understatement.</p><p></p><p>There's a very good paper describing Lanchester's Laws at the link below:</p><p></p><p><a href="http://arxiv.org/PS_cache/math/pdf/0606/0606300v1.pdf" target="_blank">http://arxiv.org/PS_cache/math/pdf/0606/0606300v1.pdf</a></p><p></p><p>Be warned that it is sometimes heavy reading (especially if you are already intimidated by this thread, which at times I have been myself). However, equations aside, it does have some very readable and very enjoyable explanations including the terms <em>"a constant bloody melee"</em> and <em>"Never split the party."</em></p></blockquote><p></p>
[QUOTE="Wulf Ratbane, post: 3792123, member: 94"] This thread has not been particularly active, but my brain has been. And it's starting to hurt. Time for a braindump. I'm still unable to completely (ok, let's be frank, [i]remotely[/i]) incorporate Lanchester's Laws into the "Encounter Building" model. Let us say that for any given party of four PCs, we can challenge them equally with either four equally-matched brutes, or eight lesser mooks. Lanchester says that an N-fold increase in numbers must be countered with an N-squared increase in quality. From this we can simply deduce that a single PC, or a single brute, must have [i]four times[/i] the quality of a single mook. We also know that the combat/encounter value of four brutes is the same as eight mooks-- both of these encounters are equal in difficulty. If the Encounter Design system says that you may buy either four brutes or eight mooks from the same pool of 1000 xp, for example, we can deduce that a brute costs 250 xp and a mook costs 125 xp. So it would seem elementary that you could use your 1000 xp to build an encounter with TWO brutes (250x2) and FOUR mooks (125x4), right? Not exactly. According to Lanchester's Laws, the combat value of this heterogenous group is NOT equivalent to either 4 brutes or 8 mooks-- it's actually a little bit stronger than either of the homogenous groups. If a mook has a "combat quality" rating of 1, then 8 mooks have a "combat quality" of 8^2 (unit size) x 1 (their average quality) = 64. Similarly, four brutes, each with an individual "combat quality" of 4, have a total combat quality of 4^2 (unit size) x 4 (their average quality) = 64. However, two brutes and four mooks (6 units) would have a total combat quality of 6^2 x [(4+4+1+1+1+1)/6] = 72, and increase in "difficulty" of 12.5%. The main takeaway from Lanchester's Laws is that numbers count for more than quality. To say that five PCs can handle twenty goblins (a fourfold increase in numbers) is to say, in effect, that PCs are [i]sixteen times[/i] as effective in combat as goblins. And if orcs are the brutes to our goblin mooks, what must an orc statblock look like? Again, I have a hard time envisioning any kind of statblock for goblins, orcs, or the PCs that has any resemblance to the D&D that I know. And that is before I have even begun to consider that a single "boss" monster designed to be on an even footing with five PCs must have [i]twenty-five[/i] times their power. To say that I am eager to see how WoTC solves this problem would be an understatement. There's a very good paper describing Lanchester's Laws at the link below: [url]http://arxiv.org/PS_cache/math/pdf/0606/0606300v1.pdf[/url] Be warned that it is sometimes heavy reading (especially if you are already intimidated by this thread, which at times I have been myself). However, equations aside, it does have some very readable and very enjoyable explanations including the terms [i]"a constant bloody melee"[/i] and [i]"Never split the party."[/i] [/QUOTE]
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