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Message: <blockquote data-quote="Conaill" data-source="post: 1575951" data-attributes="member: 1264">The "only when you need a 20" may be true for two attackers with equal stats, but for multiple attackers or with one better attacker, the dpicture can become very different.Oh what the heck, let me just do the math myself. I think the following is correct even if you include crit damage (because it can just be represented as an extra % to average damage):AT1, D1: attack bonus and avg damage for flunkiesAT2, D2: attack bonus and avg damage for ally being aided (typically, AT2>AT1, or D2>D1)AC: armor class, duhN: number of flunkies using Aide Another (so total number of attackers is N+1)P(X) = 1/20 if X<1, 19/20 if X>19, X/20 otherwise  (accounts for automatic hits and misses)dmg on individual attacks = N P(AT1-AC+21) D1 + P(AT2-AC+21) D2dmg using Aide Another  = P(AT2+2NP(AT1+11)-AC+21) D2 Obviously you want to be in the regime where the aided ally has a total attack bonus which is between 1 and 19 away from the AC to hit, so the second equation is within the linear domain of P(X). In that case, the second equation simplifies to:dmg using Aide Another  = [AT2+2NP(AT1+11)-AC+21] D2 / 20Now, if the ally receiving the Aide Another's has a 1/20 or better chance of hitting without a crit even before being aided, this simplifies dramatically. In that case, Aide Another will almost *always* be beneficial, because each flunkie is essentially trading off (1/20th of) one attack against armor class AC with damage D1, for two attacks agains armor class 10 with damage D2, i.e. 2 P(AT1+11) D2 > P(AT1-AC+21) D1For the case where the ally couldn't have hit the target AC without a crit threat before receiving Aide (and assuming AT1<=AT2), the first equation simplifies to:dmg on individual attacks = (N D1 + D2) / 20so Aide Another will be beneficial when[AT2+2NP(AT1+11)-AC+21] D2 > N D1 + D2i.e., when the probability of hitting after AA is greater than (N D1 + D2) / 20D2; or if D1=D2,  if the probability after Aide is greater than the total number of attackers divided by 20.</blockquote>

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