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Power Attack, Two Handed Weapons and Crit
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<blockquote data-quote="frisbeet" data-source="post: 2025644" data-attributes="member: 10287"><p>The restriction to the said formula matters from the following derivation, which I will politely color orange for those less inclined to drool through algebra:</p><p></p><p><span style="color: DarkOrange">Suppose you’re fighting badguy X. It turns out you need, say, a 15 or better to hit X. You threaten a crit on an 18 since it turns out you’re wielding a Falchion. X is vulnerable to crits (aren’t we all).</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">The probability of hitting for crit damage = </span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">3/20*6/20</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">An 18, 19, or 20 threatens out of 20 possible d20 outcomes, so P(threat) = 3/20. There are 6 ways of confirming the crit among 20 possible d20 outcomes, so P(confirm) = 6/20. One law of probability requires that when two things are needed for another thing to happen, the P(another thing happening) = P(thing1 happening)*P(thing2 happening). Ergo, 3/20 * 6/20.</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">The probability of hitting for normal damage = </span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">3/20 + 3/20*14/20</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">That’s 3/20, the chance of rolling a 15, 16, or 17, plus 3/20*14/20, the chance of rolling an 18, 19, or 20, then missing the crit confirmation. That’s another law of probability: P(thing1 or thing2) = P(thing1) + P(thing2).</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Now, generalize. Let N = # of d20 outcomes which hit-but-don’t-threat. This number can be 0, but not less than 0. Let C = threat range. For Falchion, C = 3.</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Then P(hitting for crit) = </span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">C/20*(C + N)/20</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">And P(hitting for normal) =</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">N/20 + C/20*(20 – N – C)/20</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Ok, so what is an “automatic” hit? Isn’t P(hitting) = 1 for an automatic hit? Yes, but it doesn’t help us when we think about critical hits. Sometimes, those automatic hits land for a crit—but not always, of course. The proper way of thinking about it is: When I’ve tabulated all the hits I’ve made with a Falchion in my career, what % of the time were hits crits? The analytic answer to that is simply</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">P(hitting for crit) / {P(hitting for normal) + P(hitting for crit)}</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Or</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">C/20*(C+N)/20 / {N/20 + C/20*(20 – N – C)/20 + C/20*(C+N)/20}</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Algebra:</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">= (C^2 + C*N)/400 / (20*N + 20*C – C*N – C^2 + C^2 + C*N )/400</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">= (C^2 + CN) / (20N + 20C)</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">= C(C+N) / 20 (C+N)</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">= C/20</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Holy schnikees, that seems obvious! Among automatic hits, the % of the time it’s a crit is the threat range / 20.</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Deriving the rest of the formula I first gave:</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Average damage/hit = P(hit not crit) * (D + db) + P(crit) * (wd + db) * M, where M is the weapon crit multiplier, wd the average weapon damage, and db the damage bonus.</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Since we’re talking automatic hits, P(hit not crit) is = 1 – P(crit), not the P(hitting for normal) formula given above. This = 1 – C/20. Then</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Average damage/hit = (1 – C/20)*(wd + db) + C/20*M*(wd+ db).</span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">= (1 – C/20 + C/20*M)*(wd + db)</span></p><p> <span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange"><strong>= (1 + C/20*(M-1))*(wd + db)</strong></span></p><p><span style="color: DarkOrange"></span></p><p><span style="color: DarkOrange">Which is what I offered in my second post.</span></p><p></p><p>Great, answer the question frisbeet. What condition did I require in this derivation? Answer: N, the # of d20 outcomes which hit-but-don’t-threat, must be >=0. This formula is exactly wrong when badguy X’s AC is high enough such that only a 20 or 19 or (generalizing) 20-C+2 or better hits. A very subtle point. Gotcha! for reading this.</p><p></p><p>Which is why, I think, it’s clearer to just do the calculations against several AC taking into account a whole bunch of things yada yada yada click sig.</p></blockquote><p></p>
[QUOTE="frisbeet, post: 2025644, member: 10287"] The restriction to the said formula matters from the following derivation, which I will politely color orange for those less inclined to drool through algebra: [COLOR=DarkOrange]Suppose you’re fighting badguy X. It turns out you need, say, a 15 or better to hit X. You threaten a crit on an 18 since it turns out you’re wielding a Falchion. X is vulnerable to crits (aren’t we all). The probability of hitting for crit damage = 3/20*6/20 An 18, 19, or 20 threatens out of 20 possible d20 outcomes, so P(threat) = 3/20. There are 6 ways of confirming the crit among 20 possible d20 outcomes, so P(confirm) = 6/20. One law of probability requires that when two things are needed for another thing to happen, the P(another thing happening) = P(thing1 happening)*P(thing2 happening). Ergo, 3/20 * 6/20. The probability of hitting for normal damage = 3/20 + 3/20*14/20 That’s 3/20, the chance of rolling a 15, 16, or 17, plus 3/20*14/20, the chance of rolling an 18, 19, or 20, then missing the crit confirmation. That’s another law of probability: P(thing1 or thing2) = P(thing1) + P(thing2). Now, generalize. Let N = # of d20 outcomes which hit-but-don’t-threat. This number can be 0, but not less than 0. Let C = threat range. For Falchion, C = 3. Then P(hitting for crit) = C/20*(C + N)/20 And P(hitting for normal) = N/20 + C/20*(20 – N – C)/20 Ok, so what is an “automatic” hit? Isn’t P(hitting) = 1 for an automatic hit? Yes, but it doesn’t help us when we think about critical hits. Sometimes, those automatic hits land for a crit—but not always, of course. The proper way of thinking about it is: When I’ve tabulated all the hits I’ve made with a Falchion in my career, what % of the time were hits crits? The analytic answer to that is simply P(hitting for crit) / {P(hitting for normal) + P(hitting for crit)} Or C/20*(C+N)/20 / {N/20 + C/20*(20 – N – C)/20 + C/20*(C+N)/20} Algebra: = (C^2 + C*N)/400 / (20*N + 20*C – C*N – C^2 + C^2 + C*N )/400 = (C^2 + CN) / (20N + 20C) = C(C+N) / 20 (C+N) = C/20 Holy schnikees, that seems obvious! Among automatic hits, the % of the time it’s a crit is the threat range / 20. Deriving the rest of the formula I first gave: Average damage/hit = P(hit not crit) * (D + db) + P(crit) * (wd + db) * M, where M is the weapon crit multiplier, wd the average weapon damage, and db the damage bonus. Since we’re talking automatic hits, P(hit not crit) is = 1 – P(crit), not the P(hitting for normal) formula given above. This = 1 – C/20. Then Average damage/hit = (1 – C/20)*(wd + db) + C/20*M*(wd+ db). = (1 – C/20 + C/20*M)*(wd + db) [B]= (1 + C/20*(M-1))*(wd + db)[/B] Which is what I offered in my second post.[/COLOR] Great, answer the question frisbeet. What condition did I require in this derivation? Answer: N, the # of d20 outcomes which hit-but-don’t-threat, must be >=0. This formula is exactly wrong when badguy X’s AC is high enough such that only a 20 or 19 or (generalizing) 20-C+2 or better hits. A very subtle point. Gotcha! for reading this. Which is why, I think, it’s clearer to just do the calculations against several AC taking into account a whole bunch of things yada yada yada click sig. [/QUOTE]
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