If you do not spend any actions to deal with it, persistent damage ends if you succeed on a DC 15 flat check.
What is the expected duration? (as calculated from the chance of ending it)
If you do not spend any actions to deal with it, persistent damage ends if you succeed on a DC 15 flat check.
What is the expected duration? (as calculated from the chance of ending it)
On your first turn, you always take damage (\$p=1\$). The probability of failing a DC 15 flat check is \$0.7\$. So on any turn after the first, you have \$0.7\$ chance of taking damage, if you took damage the last turn. So that makes the overall expected value:
$$S = 1 + 0.7 + 0.7^2 + 0.7^3 + … \text{(ad infinitum)}$$
An easy way to calculate that is to observe that:
$$S \times 0.7 = 0.7 + 0.7^2 + 0.7^3 + 0.7^4 + … = S - 1,$$
from which (by subtracting \$S\$ from both sides and multiplying by \$-1\$) we get:
$$S \times 0.3 = S \times \frac{3}{10} = 1$$
and thus:
$$S = \frac{10}{3} = 3 + \frac13.$$
Based on geometric distribution, the average number is 1/p, where p the chance of success.
You have 30% chance to succeed on a DC 15 flat check (6 in 20). So you have to try 20/6 times on average to succeed, 3.333 rounds.
You do not take the next damage when you succeed, so we we need to subtract 1, but we also add 1 as you take damage even before the first check.