The potion of poison (DMG p.188) is an uncommon magic item that deals poison damage over time to anyone who drinks it. However, the way it deals damage is rather complicated, with multiple saves and reducing damage.
How much damage would you expect a potion of poison to deal (that is, what is the average damage)? You can consider the case where the victim fails the initial saving throw separately to the case where the result of the initial save is unknown (because passing the initial save deals just 3d6).
Double bonus points if you can provide more detailed information than just the mean damage (like a statistical distribution of damage).
For reference, the potion of poison functions as follows:
If you drink it, you take 3d6 poison damage, and you must succeed on a DC 13 Constitution saving throw or be poisoned. At the start of each of your turns while you are poisoned in this way, you take 3d6 poison damage. At the end of each of your turns, you can repeat the saving throw. On a successful save, the poison damage you take on your subsequent turns decreases by 1d6. The poison ends when the damage decreases to 0.
Best Answer
We have to build this case-by-case. First, we will build the basic progression of the damage dice assumming we fail the first two saves, then build the full case backwards from there.
Here's is the order of events:
We will first calculate the average from this point forward. There are three stages to ridding ourselves of this poison: a 3d6 stage, a 2d6 stage, and a 1d6 stage. The average damage for each stage is modeled by a geometric distrubtion with probability of success \$p\$ and probability of failure \$(1-p)\$. A geometric distribution, when counting the number of failures before success, has an expected value of \$\displaystyle{\frac{1-p}{p}}\$. The average damage of the 1d6 stage is \$\displaystyle{\frac{1d6(1-p)}{p}+0d6}\$: on a failed saving throw we take another 1d6 damage, on a success we take 0 damage.
However, the 2d6 stage is a bit different. On a failed save, we take 2d6 damage, but on a successful save we take 1d6 damage before entering the 1d6 stage. What I mean here is that the 1d6 stage is defined as "saving throw then resulting damage". If we succeed on a save in the 2d6 stage, we take 1d6 damage prior to making any saves in the 1d6 stage. So the expected damage from the 2d6 stage is: \$\displaystyle{\frac{2d6(1-p)}{p}+1d6}\$: we take 2d6 damage for an average of \$\displaystyle{\frac{1-p}{p}}\$ trials, and then a guaranteed 1d6 damage before entering the 1d6 stage. So the average damage for the 2d6 and 1d6 stages together is:
$$\frac{2d6(1-p)}{p}+1d6+\frac{1d6(1-p)}{p}+0d6.$$
Extending this to the 3d6 stage, we similarly take 2d6 upon a successful save prior to entering the 2d6 stage. So the damage from all three stages is given by:
$$\frac{3d6(1-p)}{p}+2d6+\frac{2d6(1-p)}{p}+1d6+\frac{1d6(1-p)}{p}+0d6$$
The interesting thing here (and the thing that took me much too long to realize) is that the 2d6 and 1d6 between phases is guaranteed damage once you fail the initial saving throw for the potion.
From here, it is easy to adjust this last expression to present the full expected damage from drinking the potion. The first 3d6 is guaranteed, and we only take subsequent damage if we fail the initial save, with probability \$(1-p)\$, so the average damage is:
$$3d6+(1-p)\displaystyle{\left(\frac{3d6(1-p)}{p}+2d6+\frac{2d6(1-p)}{p}+1d6+\frac{1d6(1-p)}{p}+0d6\right)},$$
with some algebra, using \$\displaystyle{1+\frac{1-p}{p}=\frac{1}{p}}\$, we obtain:
$$3d6+(1-p)\displaystyle{\left(\frac{3d6(1-p)}{p}+\frac{2d6}{p}+\frac{1d6}{p}\right)}.$$
Things change if we consume the potion on someone else's turn, because we would not get to make the save to reduce the damage before entering the 3d6 stage. The order of events is:
In this scenario, if you fail the initial save, another 3d6 is guaranteed. So the average damage in this case would be:
$$3d6+(1-p)\displaystyle{\left(3d6+\frac{3d6(1-p)}{p}+2d6+\frac{2d6(1-p)}{p}+1d6+\frac{1d6(1-p)}{p}+0d6\right)},$$
and again, with some algebra, we obtain:
$$3d6+(1-p)\displaystyle{\left(\frac{3d6}{p}+\frac{2d6}{p}+\frac{1d6}{p}\right)}.$$
Here is a table comparing the two cases with real values:
And here is a chart showcasing the magnitude of the difference between disadvantage and advantage for non-negative CON save bonuses:
As you can see, having disadvantage on the saves against the poison can be very perilous when your CON save bonus is low.
Acknowledgements: I would like to thank Eddymage for their significant assistance in working through the math here. Without their help, I would have been confused for a few minutes before moving on without writing up an answer. If this were a paper, they would be a coauthor.