Full Answer
Answering these questions:
- What is the expected number of hits using the strategy outlined in the question:
- without including the expected number of hits from the BA attack
- with including the expected number of hits from the BA attack
- What is the expected damage using the strategy outlined in the question:
- without including the expected damage from the BA attack
- with including the expected damage from the BA attack
- What is the probability to trigger the bonus attack?
Q1 and Q2 are split, unlike the intent in OP question since, due to how critical damage and disadvantage interact, we can't simply get the expected damage from the expected number of hits.
Assumptions
- Let the average damage from a single non-critical attack on a hit be in the form: \$D + d\$, where \$D\$ is the damage components which are from dice damage (e.g., 1d4 for a dagger \$\rightarrow D=\frac{4+1}{2}=2.5\$, and 2d6 for a greatsword \$\rightarrow D=2\frac{6+1}{2}=7\$), while \$d\$ is the fixed damage (e.g., +3 from your Str or Dex). Note that on a critical hit, the damage will be \$2D+d\$ as only the dice is doubled.
- Let \$p\$ be the probability of an attack to hit on a flat roll. Note that due to a nat 20 always hitting and nat 1 always missing, we have \$0.05 \leq p \leq 0.95\$. For example with a +10 to hit against a 18 AC target, we will hit on a roll of 8-20, so \$p=0.65\$ in this case.
- Let \$p'\$ be the probability of an attack to crit. This is normally 0.05 as crit is only on nat 20, but some class features (e.g., Champion) may modify this (i.e., 0.1 for crit on 19-20).
We will start with explaining the basics for probability to hit, expected number of hits, and expected damage, including critical hits. These will be used later in the final calculation.
Basics of Expected Number of Hits
If we have 1 attack with probability \$p\$ to hit, what is the expected number of hits? This is simply: $$1\cdot p = p$$
If that attack is with disadvantage, the expected number of hits is simply: $$1\cdot p^2 = p^2$$
If there are \$N^\circ\$ attacks with flat roll and \$N^-\$ attacks at disadvantage, the expected number of hits: $$N^\circ\cdot p + N^-\cdot p^2$$
Basics of Expected Damage
For the purpose of explanation, we start from the simplest case and then progressively go closer to our goal.
What is the expected damage from a single attack with probability \$p\$ to hit?
There are two components here:
- Expected damage without accounting for critical hit.
This is simply \$p(D+d)\$
- Expected extra damage on a crit. This is simply \$p'\cdot D\$
The total expected damage, accounting for critical hits, \$E_1(p,p')\$:
$$\fbox{$E_1(p,p') = p(D+d) + p'D$}$$
Example for a dagger with 1d4+3 damage, and 0.65 to hit, 0.05 to crit (only on nat 20):
$$E_1(0.65,0.05) = 0.65\cdot(2.5+3) + 0.05\cdot 2.5 = 3.575 + 0.125 = 3.7$$
Another way to calculate this, if you're unconvinced by the formula above, is to separate the non-crit hits and crit hits:
$$\begin{array}{rl}
E_1(p,p') &= \underbrace{(p-p')(D+d)}_{\text{non-crit avg dmg}} + \underbrace{p'(2D+d)}_{\text{crit avg dmg}}\\
&= p(D+d) - p'(D+d) + p'(D+d) + p'D\\
&= p(D+d) + p'D\end{array}$$
So, the same as the previous one, so we're good.
What is the expected damage from a single attack with Disadvantage?
To hit, we need both rolls to hit, and to crit we need both rolls to crit, so:
$$\fbox{$E_1^-(p,p') = p^2(D+d) + p'^2\cdot D$}$$
Example with the same 1d4+3 dagger:
$$E_1^-(0.65,0.05) = 0.65^2(2.5+3) + 0.05^2\cdot2.5 = 2.32375 + 0.00625 = 2.33$$
What is the expected damage from an attack with disadvantage given it hits?
This is slightly different from the previous one, since this one assumes the attack hits, and then we ask "What's the expected damage from this attack that we know it hits?"
This is \$\displaystyle \underbrace{(D+d)}_{\text{base damage, since we know it hits}}+\underbrace{\frac{p'^2}{p^2}}_{\text{p to crit given it hits}}D = \fbox{$\frac{E_1^-(p,p')}{p^2}$}\$.
What is the expected damage from \$N\$ attacks?
Due to linearity of expectation, simply add them together!
$$E_N(p,p') = N\cdot E_1(p,p')$$
$$E_N^-(p,p') = N\cdot E_1^-(p,p')$$
The Real Question
Now we're ready to answer the real question.
This is basically the same proof as MannerPots' answer, but with critical damage included.
What is the expected damage of the strategy outlined in the question?
For clarity, we outline the strategy here:
- We have \$N\$ main attacks, each with probability \$p\$ to hit, and \$p'\$ to crit.
- If we choose to hit with disadvantage with one of our main attacks and hit, we gain a single bonus attack (BA), with the same probability \$p\$ to hit, and \$p'\$ to crit.
- Keep attacking at disadvantage until we get a hit, then roll normally for all subsequent attacks
- When we gain the BA from hitting with one of our main attacks, we take it.
We consider each of the following cases separately:
- First attack with disadvantage hits, then we roll normally for the rest \$N-1\$ attacks, plus one BA. So we have \$N\$ attacks with flat roll. These attacks have an expected damage of \$E_N(p,p')\$. Together with the first hit, the total expected damage from this case is \$\displaystyle\frac{E_1^-(p,p')}{p^2}+E_N(p,p')\$. This case happens \$p^2\$ of the time, when the first attack hits.
- First attack misses, second attack hits. We have 1 miss at disadvantage (probability \$1-p^2\$), 1 hit at disadvantage (probability \$p^2\$; expected damage \$\displaystyle\frac{E_1^-(p,p')}{p^2}\$), and \$N-1\$ attacks including BA (expected damage \$E_{N-1}(p,p')\$). Total expected damage \$\displaystyle\frac{E_1^-(p,p')}{p^2}+E_{N-1}(p,p')\$. This case happens with probability \$(1-p^2)\cdot p^2\$
- First two attack miss, third attack hits. Similarly, we have expected damage from this case \$\displaystyle\frac{E_1^-(p,p')}{p^2}+E_{N-2}(p,p')\$ with probability \$(1-p^2)^2\cdot p^2\$.
- ...
- The first \$N-1\$ attack miss, last attack hit. Expected damage \$\displaystyle\frac{E_1^-(p,p')}{p^2}+E_1(p,p')\$, probability \$(1-p^2)^{N-1}\cdot p^2\$.
- All attacks miss, and we don't get BA. 0 damage, probability \$(1-p^2)^N\$.
Total expected damage is sum of (probability x expected damage of each case):
$$\begin{array}{rl}
E &= \displaystyle\sum_{i=0}^{N-1}(1-p^2)^i\cdot p^2 \cdot \left(\frac{E_1^-(p,p')}{p^2}+E_{N-i}(p,p')\right) \\
&= \displaystyle \left[E_1^-(p,p')\sum_{i=0}^{N-1} (1-p^2)^i\right] + \left[p^2E_1(p,p')\sum_{i=0}^{N-1}(1-p^2)^i(N-i)\right] \\
&= \displaystyle \left[\left(E_1^-(p,p')+Np^2E_1(p,p')\right)\sum_{i=0}^{N-1}(1-p^2)^i\right] - \left[p^2E_1(p,p')\sum_{i=0}^{N-1}i(1-p^2)^i\right]
\end{array}$$
Note that this is the same form as the original MannerPots' answer, but we have \$E_1^-(p,p')\$ in place of \$p^2d\$, and \$E_1(p,p')\$ in place of \$pd\$.
Now, substituting the expectation and separating \$p\$ from \$p'\$ we have:
$$\begin{array}{rcl}
E &=& \displaystyle \left[\left(E_1^-(p,p')+Np^2E_1(p,p')\right)\sum_{i=0}^{N-1}(1-p^2)^i\right] - \left[p^2E_1(p,p')\sum_{i=0}^{N-1}i(1-p^2)^i\right] \\
&=& \displaystyle (D+d)\left(Np+\frac{(p^2+p-1)(1-(1-p^2)^N)}{p}\right) \\
&&\displaystyle + D\left(Np' + \frac{p'(p^2+p'-1)(1-(1-p^2)^N)}{p^2}\right)
\end{array}$$
The first term is the same as the MannerPots' non-crit version, and the second term gives the additional term from the crit damage. After writing all these, I note that we can replace \$D+d\$ in the first term with "damage when not critting" and \$D\$ in the second term with "extra damage when crit".
Table for D=0, d=1 (equivalent to Number of Hits)
p |
1 attack |
2 attacks |
3 attacks |
4 attacks |
5 attacks |
0.05 |
0.0026 |
0.0054 |
0.0082 |
0.0112 |
0.0143 |
0.10 |
0.0110 |
0.0229 |
0.0357 |
0.0493 |
0.0638 |
0.15 |
0.0259 |
0.0545 |
0.0859 |
0.1200 |
0.1567 |
0.20 |
0.0480 |
0.1021 |
0.1620 |
0.2275 |
0.2984 |
0.25 |
0.0781 |
0.1670 |
0.2659 |
0.3743 |
0.4915 |
0.30 |
0.1170 |
0.2505 |
0.3989 |
0.5610 |
0.7355 |
0.35 |
0.1654 |
0.3534 |
0.5612 |
0.7865 |
1.0270 |
0.40 |
0.2240 |
0.4762 |
0.7520 |
1.0477 |
1.3600 |
0.45 |
0.2936 |
0.6189 |
0.9695 |
1.3401 |
1.7269 |
0.50 |
0.3750 |
0.7813 |
1.2109 |
1.6582 |
2.1187 |
0.55 |
0.4689 |
0.9623 |
1.4728 |
1.9953 |
2.5261 |
0.60 |
0.5760 |
1.1606 |
1.7508 |
2.3445 |
2.9405 |
0.65 |
0.6971 |
1.3743 |
2.0401 |
2.6991 |
3.3544 |
0.70 |
0.8330 |
1.6008 |
2.3354 |
3.0531 |
3.7621 |
0.75 |
0.9844 |
1.8369 |
2.6318 |
3.4014 |
4.1600 |
0.80 |
1.1520 |
2.0787 |
2.9243 |
3.7408 |
4.5467 |
0.85 |
1.3366 |
2.3217 |
3.2091 |
4.0695 |
4.9224 |
0.90 |
1.5390 |
2.5604 |
3.4835 |
4.3879 |
5.2887 |
0.95 |
1.7599 |
2.7888 |
3.7465 |
4.6973 |
5.6474 |
1.00 |
2.0000 |
3.0000 |
4.0000 |
5.0000 |
6.0000 |
For the expected number of hits excluding the BA attack, simply subtract the expected number of hits from BA, which is just probability to trigger BA times expected number of hits from a BA (flat roll).
$$\begin{array}{rcl}
E_{\text{Main}} &=& E - E_{\text{BA}} \\
&=& \displaystyle (D+d)\left(\left(N-1+(1-p^2)^N\right)p+\frac{(p^2+p-1)(1-(1-p^2)^N)}{p}\right) \\
&&\displaystyle + D\left(\left(N-1+(1-p^2)^N\right)p' + \frac{p'(p^2+p'-1)(1-(1-p^2)^N)}{p^2}\right)
\end{array}$$
Table for D=2.5 d=3 (Dagger 1d4+3)
p |
1 attack |
2 attacks |
3 attacks |
4 attacks |
5 attacks |
0.05 |
0.0210 |
0.0429 |
0.0658 |
0.0897 |
0.1145 |
0.10 |
0.0680 |
0.1421 |
0.2221 |
0.3082 |
0.4001 |
0.15 |
0.1514 |
0.3207 |
0.5076 |
0.7117 |
0.9326 |
0.20 |
0.2752 |
0.5885 |
0.9382 |
1.3299 |
1.7413 |
0.25 |
0.4438 |
0.9535 |
1.5252 |
2.1548 |
2.8389 |
0.30 |
0.6610 |
1.4223 |
2.2748 |
3.2103 |
4.2214 |
0.35 |
0.9311 |
1.9993 |
3.1878 |
4.4818 |
5.8684 |
0.40 |
1.2583 |
2.6872 |
4.2595 |
5.9522 |
7.7461 |
0.45 |
1.6465 |
3.4861 |
5.4797 |
7.5960 |
9.8103 |
0.50 |
2.1000 |
4.3938 |
6.8328 |
9.3809 |
12.0106 |
0.55 |
2.6229 |
5.4052 |
8.2988 |
11.2699 |
14.2951 |
0.60 |
3.2193 |
6.5123 |
9.8533 |
13.2244 |
16.6148 |
0.65 |
3.8933 |
7.7049 |
11.4693 |
15.2065 |
18.9280 |
0.70 |
4.6490 |
8.9677 |
13.1180 |
17.1825 |
21.2031 |
0.75 |
5.4906 |
10.2834 |
14.7709 |
19.1248 |
23.4202 |
0.80 |
6.4223 |
11.6303 |
16.4011 |
21.0147 |
25.5715 |
0.85 |
7.4480 |
12.9828 |
17.9867 |
22.8433 |
27.6590 |
0.90 |
8.5720 |
14.3114 |
19.5127 |
24.6117 |
29.6912 |
0.95 |
9.7984 |
15.5821 |
20.9744 |
26.3285 |
31.6789 |
1.00 |
11.1313 |
16.7563 |
22.3813 |
28.0063 |
33.6313 |
For reference, expected damage with just 1 attack flat roll (including crit) is 3.7. So \$p=0.65\$ improves the expected damage, similar conclusion as before.
Probability to trigger bonus attack
For completeness, the probability to trigger the bonus attack is \$1-(1-p^2)^N\$, the full reasoning can be read in MannerPots' answer. Basically we find the probability of not triggering it, then take the opposite.
Old Answer
Continuing MannerPots excellent answer to get a closed form, the average damage is:
$$
\begin{array}{rcl}
\bar{d} &=&\displaystyle \sum_{i=0}^{n-1} p^2(1-p^2)^{i}\left(d + (n-i)pd\right) \\
&=&\displaystyle p^2d\left((1+np)\sum_{i=0}^{n-1}(1-p^2)^i - p\sum_{i=0}^{n-1}i(1-p^2)^i\right)\\
&=&\displaystyle p^2d\left((1+np)\frac{1-(1-p^2)^n}{p^2} - p(1-p^2)\frac{1-(1-p^2)^n-np^2(1-p^2)^{n-1}}{p^4}\right)\\
&=&\displaystyle d\left(np+\frac{(p^2+p-1)(1-(1-p^2)^n)}{p}\right)
\end{array}
$$
Compared to the base case \$dnp\$, there is the extra probability at the end. And it's positive or negative depends on whether \$p^2+p-1\$ is positive or negative. It's positive when \$p > 0.618\$, similar to MannerPots's conclusion, that you need 0.65 chance to hit in order to benefit using this strategy. Note that this doesn't depend on \$n\$, the number of attacks you can make.
Against an 18 AC enemy, this means you need +10 to hit in order to benefit. +9 against 17 AC, +8 against 16 AC, etc. This seems not too different from sharpshooter, where you need a certain chance to hit in order to benefit from it. So I'd also say that this feature is ok.