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Is Strength-Focused Archery Viable?
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<blockquote data-quote="Someone" data-source="post: 2867164" data-attributes="member: 5656"><p>Another way to do it would be this:</p><p></p><p>We´ll call probability of hitting P and damage D. P will be a number between 1 and 20 (we´ll ignore auto misses and crits for this simplified demonstration).</p><p></p><p>Therefore, average damage per attack will be: (P/20)*D</p><p></p><p>Now, let´s suppose both P and D start very low. In fact, we´ll suppose they are both 0, and you have a limited amount of points, X, to spread between both of them. Therefore:</p><p></p><p>P+D=X</p><p></p><p>Substituting in the preceding formula:</p><p></p><p>Average damage=(P/20)*(X-P)</p><p></p><p>We want to maximize average damage per attack. If you know or remember anything about calculus, that´s when d/dP(Average damage)=0. Therefore:</p><p></p><p>(X/20)-(P/10)=0 ; P=X/2</p><p></p><p>If P=X/2, then P=D.</p><p></p><p>Maximum damage, supposing you can transfer points between attack and damage (like when you´re power attacking, or making the character) happens when both probability and average damage are equal, unless you´d hit rolling less than a 2, in which case is always advantageous to shift points because you always fail with a 1 (and it´s impossible to roll less than that <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" />). </p><p></p><p>Notice that in game terms “probability” means the number of die sides that would get you a hit. If you have a +10 to hit and are attacking AC 25, you need to roll a 15,16,17,18,19 or 20. That´s 6 valid results, therefore P is 6.</p><p></p><p>Also, notice how if your average damage is 20 or higher (no too difficult to get with a not so high level 2 handed weapon fighter), power attacking lowers your average damage per attack, unless you need less than a 2 to hit.</p></blockquote><p></p>
[QUOTE="Someone, post: 2867164, member: 5656"] Another way to do it would be this: We´ll call probability of hitting P and damage D. P will be a number between 1 and 20 (we´ll ignore auto misses and crits for this simplified demonstration). Therefore, average damage per attack will be: (P/20)*D Now, let´s suppose both P and D start very low. In fact, we´ll suppose they are both 0, and you have a limited amount of points, X, to spread between both of them. Therefore: P+D=X Substituting in the preceding formula: Average damage=(P/20)*(X-P) We want to maximize average damage per attack. If you know or remember anything about calculus, that´s when d/dP(Average damage)=0. Therefore: (X/20)-(P/10)=0 ; P=X/2 If P=X/2, then P=D. Maximum damage, supposing you can transfer points between attack and damage (like when you´re power attacking, or making the character) happens when both probability and average damage are equal, unless you´d hit rolling less than a 2, in which case is always advantageous to shift points because you always fail with a 1 (and it´s impossible to roll less than that :)). Notice that in game terms “probability” means the number of die sides that would get you a hit. If you have a +10 to hit and are attacking AC 25, you need to roll a 15,16,17,18,19 or 20. That´s 6 valid results, therefore P is 6. Also, notice how if your average damage is 20 or higher (no too difficult to get with a not so high level 2 handed weapon fighter), power attacking lowers your average damage per attack, unless you need less than a 2 to hit. [/QUOTE]
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Is Strength-Focused Archery Viable?
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