Is it okay to roll multiple attacks that all have advantage in one cluster?

I am planning to make my first Ranger and going Hunter to eventually get Volley. If I get the preemptive strike from an unseen position, then RAW I get advantage on attack rolls.

Let's say I am moving stealthily through a forest and spot 6 orcs marching down in a caravan, all within 10 ft of each other. I need to make 6 attack rolls, all with advantage.

Because of the advantage do I need to do them one by one (or use pairs of color coded dice)? Or is it considered fair to roll 12 x d20 at once, then take the highest 6 rolls?

• Are you asking whether the six highest of 12d20 are differently distributed than the results of normal six rolls with advantage? – kviiri Jul 7 '19 at 18:14

Rolling 12d20 and taking the 6 highest is not equivalent

Taking the 6 highest rolls out of 12d20 is not equivalent to 6 attacks with advantage. It is significantly better. I could lay out the probability argument, but it's easier to cherry-pick an example that demonstrates the difference. Let's say you roll your 6 attacks with advantage. On the first 3 attacks, you roll double 1s. On the last 3 attacks, you roll double 20s. That means that 3 out of 6 attacks hit. If you instead pooled those 12 rolls all together and took the 6 highest, all 6 attacks would hit, because you rolled 6 20s. With a little work, it's not too hard to show that taking the top 6 dice is always guaranteed to give a result at least as good as rolling with advantage normally, and will usually give a better result (i.e. more hits and/or more crits).

To make it fast and fair: decide before rolling how to pair the dice

Obviously, it would be nice to resolve all 6 attacks at once with a single roll of 12d20, and you can do so in a fair way, even with 12 identical dice, by defining a rule ahead of time that dictates how you will pair up the dice. A simple rule is to pair the dice up from left to right. Roll all the dice, then take the two leftmost dice and pair them up. Then the next two leftmost after that, and so on, until you have 6 pairs. Finally, once you have paired up the dice, start resolving attacks. Make sure to tell the DM what rule you are using so they can check that you are following it instead of cherry-picking pairs.

• If you are doing this digitally, dice output is easy to pair up like this. Physically, it's much less so due to random scatter. You can instead pick 6 sets of 2 d20s with similar colors to pair up - but only if you have that many d20s that match. – Zibbobz Jul 8 '19 at 13:09
• @Zibbobz it can be quite easy if you roll on paper with lines. Then you can easily pick the dice from left to right, since the lines act as an accurate ruler. – Falco Jul 8 '19 at 13:31
• @Zibbobz You could also go for different sized dice and pair them up that way. Might be a little easier to find 6 large d20s and then 6 small ones, versus color coding. Then you can say something like "large dice are for regular rolls, small ones are for advantage rolls" – D.Spetz Jul 8 '19 at 13:43
• @Zibbobz if they are on the same line, then just take em top to bottom, since they are on the same line, one will be clearly top and the other bottom. – Falco Jul 8 '19 at 13:50
• Yes, I've tried to give a rule that is easy and quick to apply in the worst case (all identical dice). If your d20s are easy to tell apart, you can use a better pairing system. – Ryan C. Thompson Jul 8 '19 at 15:21

I would say, if I was DM, that you would have to use separate rolls or some differentiation of dice pairs.

For instance, if you rolled at advantage for six attacks, you could get a result on the dice of 20 and 12 on the first attack, 18 and 14 on the second, 1 and 4 on the third, 12 and 7 on the fourth, 20 and 1 on the fifth, and 15 and 17 on the sixth. Considering each pair in order, you'd have a result of 20, 18, 4, 12, 20, and 17. However, if you rolled 12 identical dice at once, you'd get to pick all the highest dice for a corresponding series of results: 20, 20, 18, 17, 15, and 14. This could make for a major difference in the outcome of the battle. Rolling 12 dice at once would allow you to essentially swap a low roll against one Orc for a higher roll against a different Orc.

If you did this roll in a group of undifferentiated dice, you could pick the lower of the two dice intended for your attack roll against Orc A and instead use it on Orc B, when your two attack rolls against B were both lower than that value (and might otherwise have caused you to miss B). This would be essentially cheating, like changing the face-up face of the dice by sleight of hand after the dice stopped rolling.

In other words no, it would not be "fair".

• I've attempted to clarify your phrasing a bit; please check to make sure it matches what you were trying to say. – V2Blast Jul 8 '19 at 2:46
• Wow, exactly what I was trying to convey but couldn't phrase as I saw it in my head. I hate disgraphia... Thank you. – Karæthon Jul 8 '19 at 2:52
• No problem, glad to help! :) – V2Blast Jul 8 '19 at 2:53
• Just read through the edit history, nice job on the edit (and then V2 shows the usual helpful edit as only V2 can do). +1. Thanks for making the improvement/clarification. – KorvinStarmast Jul 8 '19 at 3:16

I ran some calculations to show what this might look like, the first set of numbers is what I called "Pooled" Advantage where you take the 6 highest dice of 12d20, the second is "Standard" Advantage where you take the highest values from 6 pairs of d20s.

I used python to do 100 million rolls of each variant. The results are without modifiers, with critical misses, and with crits counting 1.5x (labeled "w/crits") against ACs from 1-20. The results are surprisingly close to one another with less than 3% more natural 20s.

$$\begin{array}{|l|l|l|l|l|l|l|} \hline \text{AC} & \text{Pooled} & \text{w/crits} & \text{Standard} & \text{w/crits} & \text{Delta} & \text{w/crits} \\ \hline 1 & 5.999999 & 6.300019 & 5.985012 & 6.277496 & 0.014988 & 0.022522 \\ \hline 2 & 5.999999 & 6.300019 & 5.985012 & 6.277496 & 0.014988 & 0.022522 \\ \hline 3 & 5.999945 & 6.299964 & 5.94003 & 6.232515 & 0.059915 & 0.06745 \\ \hline 4 & 5.99925 & 6.299269 & 5.865021 & 6.157506 & 0.134229 & 0.141764 \\ \hline 5 & 5.995451 & 6.29547 & 5.760039 & 6.052523 & 0.235412 & 0.242947 \\ \hline 6 & 5.982554 & 6.282574 & 5.624994 & 5.917478 & 0.357561 & 0.365095 \\ \hline 7 & 5.950025 & 6.250044 & 5.459995 & 5.75248 & 0.49003 & 0.497564 \\ \hline 8 & 5.883357 & 6.183376 & 5.264972 & 5.557456 & 0.618385 & 0.62592 \\ \hline 9 & 5.766096 & 6.066115 & 5.039953 & 5.332437 & 0.726144 & 0.733678 \\ \hline 10 & 5.582928 & 5.882947 & 4.784862 & 5.077347 & 0.798066 & 0.805601 \\ \hline 11 & 5.323168 & 5.623187 & 4.499758 & 4.792243 & 0.82341 & 0.830944 \\ \hline 12 & 4.982936 & 5.282955 & 4.184749 & 4.477234 & 0.798187 & 0.805722 \\ \hline 13 & 4.566012 & 4.866031 & 3.839827 & 4.132311 & 0.726185 & 0.73372 \\ \hline 14 & 4.083328 & 4.383347 & 3.464796 & 3.757281 & 0.618532 & 0.626067 \\ \hline 15 & 3.549994 & 3.850013 & 3.059882 & 3.352366 & 0.490112 & 0.497647 \\ \hline 16 & 2.982406 & 3.282426 & 2.624903 & 2.917388 & 0.357503 & 0.365038 \\ \hline 17 & 2.395416 & 2.695435 & 2.159914 & 2.452399 & 0.235501 & 0.243036 \\ \hline 18 & 1.799203 & 2.099222 & 1.664884 & 1.957368 & 0.134319 & 0.141853 \\ \hline 19 & 1.199884 & 1.499903 & 1.139971 & 1.432456 & 0.059913 & 0.067447 \\ \hline 20 & 0.600038 & 0.900058 & 0.584969 & 0.877454 & 0.015069 & 0.022604 \\ \hline \end{array}$$

Here is a plot of the average number of hits for Pooled vs. Standard (w/crits):

Here is a plot of how many more hits pooled gets you vs standard:

The delta is not as high as I expected it to be. There are 0.83/0.82 or 17.3%/18.3% (with/without crits) more hits at an AC of 11, over the whole range the average is 0.39/0.38 or 9.2%/9.8% more hits (with/without crits).

Note, the earlier version of this table was generated with a script that rolled dice with values from 1-21, not 1-20.

One more method: reroll

One rapid method I've used as a DM (or Druid handling eight wolves), using average damage:

• Roll a single d20 for all attacks, then reroll misses.

While this fails to calculate the possibly of extra crits on the initial hits, that too can be accommodated (if desired) by rerolling the misses off to the left, and rerolling the initial (non-crit) hits off to the right (to check for extra crits).