On an unrelated question, user Yakk shared the following bit of advice in a comment to keep combat moving quickly for mass numbers where advantage/disadvantage might apply:

Mass attacks with disadvantage: work out target number. Roll # of goblin d20. Count how many hit. Now roll those again, and they only hit if they hit the second time. Nothing crits (the odds of it happening where 1 in 400 to start, so we'll say screw it). For advantage, reroll misses, treat both 19s and 20s as 20s on the first roll, and 20s as 19s on the reroll. (The difference between this and doing it manually is tiny).

How does this affect the odds of:

Bonus question, what is the total hit/miss/crit rate for the entire group of goblins?

(I am aware there are other ways of handling mass combat and I know the general probabilities for advantage/disadvantage (those are easy to figure out on my own). On this question I am specifically concerned with the probabilities as described in the quoted text.)

Hitting and missing is unchanged. A miss with the lower of two die with disadvantage is a hit, and a hit with the higher of two die is a hit, and this system replicates that without performing any redundant rolls.

What this system does modify is critical hits, although only slightly. The chance of a critical hit with disadvantage goes from 1/400 to 0 (or 0/400). The chance of a critical hit with advantage goes from 39/400 to 2/20 (or 40/400).

A critical miss is still just a miss, and this system doesn't seem to account for them, so they'd remain unchanged.

• I guess it's a little simpler than I thought. I'll admit I hadn't thought much about how it might work out before I posted the question but now I see it actually works out quite well without much change to the odds at all. – LegendaryDude Jan 31 '17 at 21:59
• If the fight is massive enough, you might plausibly want to care about the 1/400 chance of a crit with disadvantage. (With 20 attackers, the probability of one of them critting is roughly 1/20.) In any case, if you do care, it's easy enough to handle: before rerolling the successes, set any nat 20s aside and reroll them separately. If any of them rolls a nat 20 again, that's a crit. – Ilmari Karonen Jan 31 '17 at 22:14

## Single hit probability

$M$: A miss with a single attack

$H$: A hit with a single attack

$C$: A critical hit with a single attack

It is obvious that $M$ and $H$ are complimentary events. That is, if you miss you don't hit and vice versa, so: $P(M) = 1 - P(H)$.

It is also obvious that $C$ is a sub-event of $H$. That is, you can't get a critical hit unless you hit, so: $P(H\cap C) = P(C)$.

5e does not have any such thing as a critical miss so I am not going to deal with this.

We also define the target number $t$, as the number needed to roll or above to get a hit. $t$ must be between 2 and 20 inclusive, since a 1 always misses and a 20 always hits.

Normal attacks

\begin{align} P(H) = h & = {21 - t \over 20} \\ P(M) = m & = {t - 1 \over 20} \\ P(C) & = {1 \over 20} \\ P(C|H) = c & = {P(C\cap H) \over P(H)} \\ & = {P(C) \over P(H)} \\ & = {1 \over 21-t} \\ \end{align}

\begin{align} P(H) = h & = 1 - \left({t - 1 \over 20}\right)^2 \\ & = {400 - (t-1)^2 \over 400} \\ P(M) = m & = \left({t - 1 \over 20}\right)^2 \\ P(C) &= 1 - \left({19 \over 20}\right)^2 \\ & = {39 \over 400} \\ P(C|H) = c & = {39 \over {400 - (t-1)^2}}{} \\ \end{align}

\begin{align} P(H) = h & = {(21 - t)^2 \over 400} \\ P(M) = m & = {400 - (21 - t)^2 \over 400} \\ P(C) &= \left({1 \over 20}\right)^2 \\ & = {1 \over 400} \\ P(C|H) = c & = {1 \over (21 - t)^2} \\ \end{align}

Now, you can pick your target number and get numerical solutions to these but I will continue to treat them algebraically. The important point to note is that the chance of critical hit given a hit was achieved is dependent on the target number so the quote that dismisses it as inconsequential in not valid. For example, if $t$ = 20 then every hit is a critical hit irrespective of advantage/disadvantage.

## Multiple attack probability

This is a straightforward application of the Binomial Distribution.

Specifically, for $n$ attacks, the probability of $k$ hits is:

$$P(X=k) = \binom{n}{k}h^km^{1-k}$$

And the probability that of these $k$, $j$ will be critical hits is:

$$P(Y=j|X=k)= \binom{n}{k}h^km^{1-k}\binom{k}{j}c^j(1-c)^{1-j}$$

The mean number of hits is simply $nh$ and the mean number of critical hits is $nhc$.

## Proposed alternative

Normal attacks

There is no difference.

This gives a straight chance of a critical hit of ${1 \over {10}} {({40\over 400})}$ where the actual chance is $39 \over 400$.

I won't take this any further because you can actually do this strictly in accordance with the rules with no additional work:

• Roll all the dice,
• reroll all the misses,
• look at all the dice now on the table (originals and rerrolls), 20s are critical hits and all the other hits are just hits.

This obviously reduces the chance of a critical hit, from $1 \over 400$ to 0.

You can actually make this in accordance with RAW with one tiny additional step:

• Roll all the dice,
• all the misses are misses,
• reroll all the 20s away from the other hits, if these miss they miss, if they hit they hit and if you get a 20 they are criticals,
• reroll all the other hits, if they hit they hit and if they miss they miss - these cannot be criticals.
• Your proposal for advantaged attacks does not take into account that a reroll on a hit could result in a crit. – inthemanual Feb 1 '17 at 0:03
• You could take this calculated probability for the chance to hit and add it to the general damage question rpg.stackexchange.com/questions/93212/…, or consult the table in rpg.stackexchange.com/questions/14690/… for the "C," chance to hit. However, my discreet math is at its limit....it would be a separate question I guess, but how to make the equations simulate x rounds of combat...taking into account creatures dying each round? – Ἄρτεμις Feb 1 '17 at 0:16
• Also, not sure what the h and m is in your equations....and not sure what use the "probability of k hits" is in practice. You just want the avg number of hits. Do those equations simplify more when actually used? – Ἄρτεμις Feb 1 '17 at 0:24
• @Ἄρτεμις h, m & c are defined for each of normal, advantage and disadvantage in the first part. The probability of k hits is the probability of k hits - for example, if n=5 you can use this to work out the probability of k=0,1,2,3,4 or 5 hits. – Dale M Feb 1 '17 at 0:26
• The lowercase through me off. Well, I guess that is what the OP was asking, but not sure if that is what he meant to ask, lol. – Ἄρτεμις Feb 1 '17 at 0:28