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Rolling one attack with advantage is worse than rolling two normal attacks (because we have two attack rolls either way, but only one vs two chances of dealing dmaage).

However, what of the inverse situation? Is it better to roll two attacks with disadvantage, or one normal attack?

Since this will likely depend on target number, assume an unmodified rolls of 8, 12 and 15 are needed to hit.

The main reason I’m asking this question is for a homebrew feat that allows trading one attack for disadvantage on two attacks, but this could also come up in RAW, if you had disadvantage and multi attack, the decision would be between shoving your opponent prone and taking one attack, or taking two with disadvantage.

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    \$\begingroup\$ “Obviously, due to simple logic, rolling one attack with advantage is worse than rolling two normal attacks.” Definitely not obvious, and I’m not even sure it’s always true. \$\endgroup\$ Mar 27, 2022 at 15:21
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    \$\begingroup\$ @ThomasMarkov it is true. Rolling one attack with advantage = two attack rolls with one chance of damage. Two normal attacks = two attack rolls with two chances of damage. \$\endgroup\$ Mar 27, 2022 at 15:26
  • \$\begingroup\$ Hmm. I'm not sure if it would be better for you to formulate your Homebrew and do your own analysis first, then come to us with the questions of where you think that you need to double check or don't understand something. But the details of who the speed is available to who it is supposed to be for are going to be very relevant to determining the balance of this. \$\endgroup\$
    – NotArch
    Mar 27, 2022 at 15:32
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    \$\begingroup\$ Any calculations would need to account for critical hits and misses too. (Don't want to explain more in a comment and I'm too dumb and lazy to write out a full answer! But crits would sway any calculations) \$\endgroup\$
    – PJRZ
    Mar 27, 2022 at 15:33
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    \$\begingroup\$ Related on When should I use the -5 for GWM? \$\endgroup\$
    – NotArch
    Mar 27, 2022 at 16:24

2 Answers 2

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2 attacks at disadvantage is better only when a single roll has more than 50% chance to hit

It's probably easiest to talk about this through calculating average damage based on a 'standard' hit chance. This standard hit chance is the chance to hit with a single roll. To make calculations easy and understandable, let's set the damage dealt on a hit to 1, changing this number does not at all affect the results but it saves me some type work. I'm not taking critical hits into account, they do change the result but only slightly in favor of a single normal attack. Accounting for crits is a bit rough because their influence depends on more variables, like the size of the damage die compared to your damage bonus and any additional features like Sneak Attack or Hunter's Mark that add more dice to your damage rolls. This answer is meant as a general guideline for how these different attacks affect general gameplay and unless some character has a crazy combination of features that heavily benefit from that small chance to crit it holds up pretty well.

On a single standard attack with a potential damage output of 1 your average damage is simply the hit chance, for example if you have 80% chance to hit for 1 damage your average damage will be 0.80x1=0.80.

On a single attack at disadvantage you pick the lowest roll, so in essence both rolls need to 'hit'. If normally you would have a 80% chance to hit, you now have a 0.8x0.8=0.64=64% chance to hit. So you have a 64% chance to hit for 1 damage, your average damage in this case is 0.64.

On two attacks at disadvantage you could do some math to figure out the average hit chance but since we're talking about average damage output this is equivalent to simply doubling the average damage from a single attack at disadvantage. So with a normal 80% chance to hit, brought to 64% due to disadvantage, you have an average damage of 2x(0.8x0.8)=2x(0.64)=1.28.

Now we can generalize this. For a single attack your average damage is P, with P being the standard hit chance for an individual roll. For two attacks at disadvantage this average damage is then 2*(P*P) or 2P^2. Plotting these 2 functions yields:

enter image description here

The straight line is the single straight roll, and as you can see it is worse than attacking two times at disadvantage as long as the normal hit chance is above 50%.

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  • \$\begingroup\$ For those curious, the longer route to calculate the average damage for two attacks at disadvantage is: Average Damage = (P^2 x (1-P^2)) x 1 + ((1-P^2) x P^2) x 1 + (P^2 x P^2) x 2 + ((1-P^2) x (1-P^2) x 0) = (2 x (P^2 x (1-P^2)) + (P^4) x 2) = (2 x P^2) x ((1-P^2)+P^2) = 2P^2 \$\endgroup\$
    – anon
    Mar 27, 2022 at 16:03
  • \$\begingroup\$ @Pepijn I think you also need to change it in the text it still says single attack is worse when to hit is under 50%, while chart shows single attack is better there \$\endgroup\$ Mar 27, 2022 at 16:29
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    \$\begingroup\$ There is a slight wrinkle to this with criticals. I would not expect that to be explained as thoroughly and graphed as for the base comparison. However, it is worth noting, if someone has criticals on 19-20 and/or extra dice on a critical, or simply a lot of their damage is from dice rolls that would double up on a critical, then the trade-off point may be higher than 0.5. \$\endgroup\$ Mar 28, 2022 at 17:52
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If your odds of hitting on a straight attack roll are 50%, then both are the same. If the odds are lower than that, a single attack is better, and if the odds are higher, two attacks with disadvantage are better:

The expected number of hits from a single attack roll with probability p to hit is simply p, while the expected number of hits from two rolls with disadvantage is 2p².

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