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<blockquote data-quote="The Shadow" data-source="post: 6002585" data-attributes="member: 16760">I looked at this some yesterday.&nbsp; Let's say you do die-type + bonus in damage, abbreviated d and b respectively.Then the average damage you do is d/2 + b + 1/2.&nbsp; (Technically, it'd be 19/20 times that plus 1/20 times (d + b), but the difference is negligible for the amount of effort it adds. Do note that including crits would inevitably make Battlerager just a tiny bit worse, though.)If the chance of hitting an opponent is p, then the chance of hitting with disadvantage is p^2.So with Battlerager, your average damage is (d + b)*p^2.Whereas without the feat, you deal maximum damage with a probability of 1/20 + 19/(20*d).&nbsp; Or just 1/20 if you're only interested in crits.So for Battlerager to make a difference for you, you'd need:(d + b)*p &gt; d/2 + b + 1/2Unfortunately, there are three solutions, depending on whether p is &gt;, &lt;, or = to 0.5:p=0.5:&nbsp; b &lt; -1.&nbsp; In other words, if you have even odds to hit, Battlerager gives you a net benefit only if your bonus is negative.&nbsp; Since this is extremely unlikely for a melee combatant who takes this feat, Battlerager is a net loss under realistic circumstances.p&lt;0.5:&nbsp; d &lt; [1 + 2(1-p)b]/[(2p-1)].&nbsp; Assuming b is positive, d has to be a negative number, which is impossible.&nbsp; So for this case also, Battlerager is a net loss.p&gt;0.5:&nbsp; d &gt; [1 + 2(1-p)b]/[(2p-1)].&nbsp; Here at last we have a case where Battlerager has a net benefit.&nbsp; Let's take the worst-case scenario of p=0.55 and assume a bonus b of +3.Then d has to be greater than 37, which is impossible in D&amp;D.Now take the best-case scenario, with p=0.95.Then d has to be greater than 0.072, which is true of all D&amp;D weapons.Where's the breakeven point?&nbsp; If d is 12 (a greataxe, say), then p has to be 0.633, or in other words, you need to hit on a 7.&nbsp; This doesn't change significantly if your damage bonus is 4.If d is 8, you have to hit on a 6.In short, maximum-damage Battlerager is a losing proposition on average, except for easy-to-hit opponents.&nbsp; One could argue that this is realistic - that berserking is a losing proposition on the whole against armored opponents - but it doesn't make the feat look all that attractive.However, Battlerager gets a lot more attractive if your bonus goes negative - in other words, if you've been debuffed.&nbsp; That's something to take into consideration.&nbsp; And of course, if you're at a disadvantage for any reason, the feat suddenly gets completely awesome.</blockquote>

[QUOTE="The Shadow, post: 6002585, member: 16760"] I looked at this some yesterday. Let's say you do die-type + bonus in damage, abbreviated d and b respectively. Then the average damage you do is d/2 + b + 1/2. (Technically, it'd be 19/20 times that plus 1/20 times (d + b), but the difference is negligible for the amount of effort it adds. Do note that including crits would inevitably make Battlerager just a tiny bit worse, though.) If the chance of hitting an opponent is p, then the chance of hitting with disadvantage is p^2. So with Battlerager, your average damage is (d + b)*p^2. Whereas without the feat, you deal maximum damage with a probability of 1/20 + 19/(20*d). Or just 1/20 if you're only interested in crits. So for Battlerager to make a difference for you, you'd need: (d + b)*p > d/2 + b + 1/2 Unfortunately, there are three solutions, depending on whether p is >, <, or = to 0.5: p=0.5: b < -1. In other words, if you have even odds to hit, Battlerager gives you a net benefit only if your bonus is negative. Since this is extremely unlikely for a melee combatant who takes this feat, Battlerager is a net loss under realistic circumstances. p<0.5: d < [1 + 2(1-p)b]/[(2p-1)]. Assuming b is positive, d has to be a negative number, which is impossible. So for this case also, Battlerager is a net loss. p>0.5: d > [1 + 2(1-p)b]/[(2p-1)]. Here at last we have a case where Battlerager has a net benefit. Let's take the worst-case scenario of p=0.55 and assume a bonus b of +3. Then d has to be greater than 37, which is impossible in D&D. Now take the best-case scenario, with p=0.95. Then d has to be greater than 0.072, which is true of all D&D weapons. Where's the breakeven point? If d is 12 (a greataxe, say), then p has to be 0.633, or in other words, you need to hit on a 7. This doesn't change significantly if your damage bonus is 4. If d is 8, you have to hit on a 6. In short, maximum-damage Battlerager is a losing proposition on average, except for easy-to-hit opponents. One could argue that this is realistic - that berserking is a losing proposition on the whole against armored opponents - but it doesn't make the feat look all that attractive. However, Battlerager gets a lot more attractive if your bonus goes negative - in other words, if you've been debuffed. That's something to take into consideration. And of course, if you're at a disadvantage for any reason, the feat suddenly gets completely awesome. [/QUOTE]

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