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<blockquote data-quote="Staffan" data-source="post: 1829765" data-attributes="member: 907">First, mathematical stuff. Basically, what you want to do is to maximize the formula 0.05*(n-P)*(d+P), where N is number of d20 rolls that lead to a hit (i.e. if you need 15 to hit, n=6) without power attack, P is the amount of power attack, and d is your average damage without power attack. Since, in any given situation, n and d are constants, as is the 0.05 factor (for the d20), the problem is basically the same as &quot;If drawing a rectangle with a predetermined sum of the sides, what is the best ratio of width to height in order to get maximum area?&quot; The answer to that question is making a square - in other words, you want to get (n-P) as close as possible to (d+P). The value of P that gives the closest value is the &quot;best&quot; amount of power attack. If you have multiple attacks, use the average value of n.Example: You're fighting something that you need a 3 to hit with your primary and off-hand attacks (and thus an 8 with your secondary). 3 gives an n of 18, and 8 gives n=13. The relevant n is thus (18+18+13)/3 = ~16. Your damage is 1d8+8 (well, +5 with the secondary, but let's ignore that for the moment), for an average of 12.5. Here we have n-d = 4, so split the difference and you'll see that the optimal amount of power attack is 2.So, short version: Make sure (number of results that hit-PA) is as close as possible to (average damage+PA). That'll get you reasonably close.</blockquote>

[QUOTE="Staffan, post: 1829765, member: 907"] First, mathematical stuff. Basically, what you want to do is to maximize the formula 0.05*(n-P)*(d+P), where N is number of d20 rolls that lead to a hit (i.e. if you need 15 to hit, n=6) without power attack, P is the amount of power attack, and d is your average damage without power attack. Since, in any given situation, n and d are constants, as is the 0.05 factor (for the d20), the problem is basically the same as "If drawing a rectangle with a predetermined sum of the sides, what is the best ratio of width to height in order to get maximum area?" The answer to that question is making a square - in other words, you want to get (n-P) as close as possible to (d+P). The value of P that gives the closest value is the "best" amount of power attack. If you have multiple attacks, use the average value of n. Example: You're fighting something that you need a 3 to hit with your primary and off-hand attacks (and thus an 8 with your secondary). 3 gives an n of 18, and 8 gives n=13. The relevant n is thus (18+18+13)/3 = ~16. Your damage is 1d8+8 (well, +5 with the secondary, but let's ignore that for the moment), for an average of 12.5. Here we have n-d = 4, so split the difference and you'll see that the optimal amount of power attack is 2. So, short version: Make sure (number of results that hit-PA) is as close as possible to (average damage+PA). That'll get you reasonably close. [/QUOTE]

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