How long can you expect to maintain Concentration during combat? [closed]

Question

Every concentration spell has a listed duration in the spell description, like:

Concentration, up to 1 minute.

Spells that require concentration might end prematurely if the caster takes damage and fails the Constitution save to maintain concentration (Player's Basic Rules, p. 80, italic emphasis mine):

• Taking damage. Whenever you take damage while you are concentrating on a spell, you must make a Constitution saving throw to maintain your concentration. The DC equals 10 or half the damage you take, whichever number is higher. If you take damage from multiple sources, such as an arrow and a dragon’s breath, you make a separate saving throw for each source of damage.

Assuming a combat encounter proceeds in a typical fashion (where the answer must define those assumptions), what is the probabilistic expected duration of a concentration spell?

Answer 1

Rounds

This would be the more important metric, but cannot be answered, as it depends on unanswerable things:

• How many times are you attacked per round, and how high is your AC compared to the enemy's hit bonus
• How many spells you are targeted with per round, and how high are your saves compared to the DC, and if those spells deal damage on a save

Hits

An average can be calculated if we assume you do not get any one damage instance above 21. So think of this as a best case scenario.

Save chance

Without Warcaster, your chance to save is $$\frac{11 + ConSave}{20}$$
As a table:

\begin{array}{r|r} \text{Con Save} & \text{Success %} \\ \hline 0 & 0.55 \\ 1 & 0.6 \\ 2 & 0.65 \\ 3 & 0.70 \\ 4 & 0.75 \\ 5 & 0.80 \\ 6 & 0.85 \\ 7 & 0.90 \\ 8 & 0.95 \\ 9+ & 1.00 \\ \end{array}

So a 17th level Sorcerer with 14 Con has 95%, and a common, 1st level Wizard with the same Con has 65% to succeed. Let's calculate with the latter one.

Accumulation

You have a 65% chance to succeed, so 35% chance to fail the 1st Concentration save.
To fail exactly the 2nd save, you have 22.8% ( 0.35 * 0.65 ) chance.
To fail exactly the 3rd save, you have 14.8% ( 0.35 * 0.65 * 0.65 ) chance.
To fail exactly the 10th save, you have 0.7% ( 0.35 * 0.65 ^ 9 ) chance, and so on.

To find the average, you have to multiply these chances with the number of failures they belong to. The sum of the results is the average, so 1 * 0.35 + 2 * 0.228 + 3 * 0.148 and so on, to infinity.

You can generalize this:
Chance to fail on the Nth save:
$${FailChance * (1-FailChance)^{N-1}}$$.
The SUM is:
$$\sum {(N * FailChance * (1-FailChance)^{N-1})}$$
This is a tough one to crack, but can be changed into this:
$$\frac {FailChance}{(1-FailChance)} *\sum {(N * (1-FailChance)^N)}$$ WolframAlpha says the SUM can be replaced:
$$\frac {FailChance}{(1-FailChance)} *\frac {1-FailChance}{(FailChance)^2}$$ The end result is surprisingly simple: $$\frac {1}{FailChance}$$

Check it with two values
Con save +8: You have a 5% chance to fail, only on a 1.
1 / 0.05 = 20, just what you would expect, on average every 20th save will fail.

Con save +2: As written above, you have a 35% chance to fail.
1 / 0.35 = 2.857. If you start doing the multiplications numerically, the sum of the Product column goes to 2.857 as expected.

\begin{array}{r|r|r} \text{Fail} & \text{Chance} & \text{Product} \\ \hline 1 & 0.350 & 0.350 \\ 2 & 0.228 & 0.455 \\ 3 & 0.148 & 0.444 \\ 4 & 0.096 & 0.384 \\ 5 & 0.062 & 0.312 \\ 6 & 0.041 & 0.244 \\ 7 & 0.026 & 0.185 \\ 8 & 0.017 & 0.137 \\ 9 & 0.011 & 0.100 \\ 10 & 0.007 & 0.072 \\ \end{array}

Results

The average number of hits it takes to stop your concentration by Con save:

\begin{array}{r|r} \text{Con Save} & \text{Hits} \\ \hline 0 & 2.222 \\ 1 & 2.500 \\ 2 & 2.857 \\ 3 & 3.333 \\ 4 & 4.000 \\ 5 & 5.000 \\ 6 & 6.667 \\ 7 & 10.000 \\ 8 & 20.000 \\ 9+ & \infty \\ \end{array}

Consequences

If you are a frontliner, do not expect your Concentration to last without a proficiency bonus or Warcaster unless you have spectacular AC.

A +9 Con save will give you indefinite Concentration, up to the spell's duration.