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Message: <blockquote data-quote="Elder-Basilisk" data-source="post: 986559" data-attributes="member: 3146">The difficulty of accounting for cleave is the reason that people don't do it. For average damage/round calculations all you need to know is the character's offense relevant statistics and equipment and the AC of the target. In order to include Cleave, you need to also know the hit points of the target and assign a probability that the character is in an appropriate position to cleave.For creatures like bugbears or zombies and low-level fighters, it's fairly simple. (Zombies are a case where it's often a good idea to power attack in damage/round calculations and cleave serves to multiply the usefulness of power attack; Bugbears are an instance in which it may be advantageous to power attack even though ordinary damage/round calculations show it to be disadvantageous--because cleave nets more attacks).Someone did some calculations with a matrix which was the part of math I never really understood how to use in High School (the last time I studied any math). My very rough concept math didn't deal with damage/round in hit points but rather in monsters and went something like this:Fighter (lvl 4, Str 16, +1 greatsword, weapon specialization= attack bonus +9, damage=2d6+7 (average 13)), fighting lots of AC 17, 15 hp opponent (bugbear IIRC)--assuming that he is always in a position to cleave.W/out power attack, he hits 65% of the time but takes two average hits to drop the bugbear. On the second hit, he will drop the bugbear and cleave into another. Hitting 65% of the time, he therefore drops 11 bugbears in 20 rounds.If he power attacks for 3, OTOH, he deals an average of 16 points of damage 50% of the time. Consequently, he's likely to drop 15 bugbears over 20 rounds.Clearly, my math is somewhat lacking because I assume that every hit will deal average damage. In reality, the non-power attacker will single-shot bugbears on a roll of 10 or better (about 7/36 chance of that) and the power attacker will need a second hit to drop the bugbear on a roll of 6 or lower (about 16/36 chance of that). Nor does it account for critical hits which single-shot the bugbears whether or not the character is power attacking. So the math should come out to much less than the apparent 50% increase in bugbear killing predicted by my average damage assumption.The math is more complex when you run into a situation like large or huge zombies or some animals where power attacking means the difference between killing the creature in two blows or killing it in 3. (Especially when a character has multiple attacks). As I said, I'm not knowledgable enough in matrix theory to make a managable formula for that.</blockquote>

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