A lot of gamers consistently optimize characters average damage per round. My intuition tells me, though, there are times when your character should be opting for an action that does the highest minimum damage per hit. In other words, choosing the action with the lowest maximum hits to kill a monster.
Although I trust my intuition here, not everyone agrees with me. I would like to have mathematical evidence to back up my gut feeling.
Wizard vs. Goblin
Consider this example. A 3rd level wizard finds himself alone, facing a goblin at close range. (Beyond the goblin lies certain safety.) Out of spell slots and other supplies, the wizard considers whether to use Firebolt or his dagger.
Our wizard has:
- Intelligence: 16
- Dexterity: 16
- AC: 13
- HP: 17
- Crossbow Expert Feat (so
no disadvantage for attacking with Firebolt at close range.)
So, our wizard has a +5 to hit with either the Firebolt or the dagger.
The goblin has AC: 15, 7 Hit Points, attacks at +4, and does 1d6+2 damage. Assume both combatants fight to zero HP.
My gut is telling me that the wizard should reach for his dagger (that does 1d4+3 damage) rather than casting Firebolt (for 1d10+0 damage) even though Firebolt will do slightly higher average damage.
Including calculations for critical hits, Firebolt will do 3.025 average DPR against the goblin, while the dagger does 2.875 average DPR. The D&D 5e tag is there deliberately – the slightly-trickier math around critical hits is part of this..
Counting on bad luck
My thinking is that bad luck on damage rolls won’t be disastrous for the wizard if he attacks with the dagger. Two hits with the dagger will definitely kill the goblin, while the goblin might survive several Firebolts.
Is my intuition correct? Is the wizard more likely to beat the goblin using his dagger than with Firebolt?
I’m hoping someone can provide some math to prove what the wizard should do.
(Note, I think it's obvious that if you have an attack that can guarantee killing a creature in one attack, you'd want to use that attack over one that hits as often but sometimes does less damage. The above example is meant to be more subtle than that, a example where non-mathematically-minded folks might disagree on the best tactic.)
Best Answer
Background Theory
Broadly when choosing to optimise damage in a single round there are five variables you need to take account of:
When you optimise DPR, you are optimising only one/two of these five potential variables (Criting can alter the ADPR).
If you are only rolling one dice, your damage variance on that attack will be relatively high.
To reduce the variance of the attack there are three broad strategies:
In the situation where there is a damage type resistance/vulnerability then the choice between the two becomes clearer as you will be halving/doubling the result from the damage dice which will decrease/increase your DPR.
To answer whether or not you are more likely to beat a specific enemy depends on the HP, and variability in that enemies HP (determined by their hit dice). The same two considerations apply here.
If you are fighting an enemy with a large amount of hit dice, then by the same application of the law of large numbers you are more likely to be fighting an average specemin of the enemy.
Similarly if you are fighting an enemy with a large fixed HP component then you are similarly more likely fighting an average enemy.
Application to your example
The Firebolt has a range of damage values when it hits of 1-10 (with a 10% chance of each)
The dagger has a different range of damage values when it hits 4-7 (with a 25% chance of each).
A Goblin’s HP is calculated by rolling 2d6 (the 7 HP in the Monster Manual is simply the average value of this distribution). This gives the Goblin Monster a range of Hit Point values between 2 & 12 HP (obviously the extremes of 2 and 12 are pretty unlikely).
Both attacks have a 45% chance of a non-critical hit and a 5% chance of a critical hit.
The rest of this analysis will assume that we have hit (that makes the math slightly earier, and we can convert it into number of rounds by using this information later).
As a result, when we hit that translates to a 90% of hits are a non-crit, and 10% are a crit.
I’m also going to assume, for simplicities sake, that we are doubling the result of the dice when we crit, instead of doubling the number of dice we roll.
Vanilla Attacks
1 Hit to Kill
Taking your specific example of a Goblin (7 HP). When our attacks hit
Taking all of this into account, the 1 hit kill percentages are:
3 Hits or more to Kill
At the other extreme, the chance of it taking more than two hits to fell the Goblin.
Ways we can get this total with Firebolt
2 x Crits:
1 x Crit:
0 x Crits:
All of that is:
Which totals to 13.26%
Final Totals
As a result the probabilities of killing the Goblin with a repeated attack type are:
1 hit to kill:
2 hits or less to kill:
3+ hits or more to kill:
Mix & Match
This is of course assuming you don’t mix and match attacks.
If we allow mixing and matching your choices change. For example, if you hit with the Firebolt or the Dagger for 6 damage, it doesn’t matter which attack you hit with next.
Similarly if you hit with the firebolt for 3 or more (80% probability for the hit), using the dagger next hit guarantees the kill.
If you hit with the firebolt for 1 on the first hit (10%) then your probabilities are:
The Dagger’s smaller variance will probably give it enough to pip the Firebolt (but it’s close).
If you hit with the firebolt for 2 damage on the first hit (10%*90% + 10%*10% = 10%) then your probabilities are:
You can see where this is trending…
Consolidated Percentages
1 Hit to Kill
2 Hits or less to Kill
3 Hits or less to Kill
3 Hits or more to Kill
Strategy against a 7 HP Goblin
Hit with Firebolt first, as it gives you a better chance of one-shot killing the Goblin.
With two hits the probabilities of killing the Goblin are: Firebolt + Dagger = 94% Dagger + Dagger = 100%
If you hit with the Firebolt on your first hit, and it doesn't kill the Goblin (and the Goblin has 7HP) then you are better off switching your second attack to be the dagger, as you have a higher (and less variable) probability of killing the Goblin with the Dagger on the second hit.
Given that the increase of kill % with 1 hit on a Firebolt (13%) is greater than the decrease (6%) in % we get over two rounds from not doing Dagger, Dagger, Optimal play is to do Firebolt + Dagger.
On average how many rounds does it take for you to hit the Goblin?
We effectively have repeated trials to success of a iid Bernoulli random variables with p=0.5. The expected waiting time until a success (a hit) for this type of process is given by the expected value of a Geometric distribution, with p=0.5.
E[Rounds to Hit] = 1/0.5 = 2 rounds.
Thus the expected rounds to get 2 hits is 4.
Conclusion
This sort of spread will change depending on the actual HP of the Goblin.