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If my character drops to 0HP but is not killed instantly by massive damage, what is the probability that I will stabilize via death saves unassisted/unhindered?

As a refresher, here are some of the rules on death saves:

Death Saving Throws. Whenever you start your turn with 0 hit points, you must make a special saving throw, called a death saving throw....

Roll a d20. If the roll is 10 or higher, you succeed. Otherwise, you fail.... On your third success, you become stable....

Rolling a 1 or 20. When you make a death saving throw and roll a 1 on the d20, it counts as two failures. If you roll a 20 on the d20, you regain 1 hit point. (PHB p.197, "Dropping to 0 Hit Points")


Note: this character is not Lucky, neither halfling- nor feat-based. Also no portent (divination wizard).

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There's a 59.5125% chance of survival.

Naively, we might have thought there'd be a 55% chance of survival as 55% of the roll results are good. But the 20 is a slightly better result than the 1 is a bad one, so that pushes up the probability a bit. Let's see how.

The approach

The simplest way to tackle this is to look at the probabilities of surviving in exclusive ways, then combine those probabilities. Remember the rules of combining probabilities of multiple events:

  • If we want to know the probability of (A or B), when A and B don't overlap, we sum the probability of A with the probability of B. (We'll be constructing all of our scenarios below without overlap.)
  • If we want to know the probability of (A and B), we multiply the probability of A with the probability of B.

Notation

We'll use a number or a range of numbers in brackets to indicate the probability of that result on a d20. That is, \$[4]=\frac 1 {20}\$, \$[7]=\frac 1 {20}\$, and \$[10-20]=\frac {11} {20}\$. In this manner we'll generally be concerned with four different results: \$[1]\$ (\$\frac 1 {20}\$ chance), \$[2-9]\$ (\$\frac 8 {20}\$), \$[10-19]\$ (\$\frac {10} {20}\$), and \$[20]\$ (\$\frac 1 {20}\$ again).

When we want to indicate a particular sequence of rolls, we'll list them in order: \$[1][1-9]\$ would be read as "the probability of rolling a 1, then some single-digit number."

When we want to indicate a particular selection of rolls, but don't care about their order* we'll group the unordered results in curly-brackets. Thus \$\{[2-9][10-19][2-9]\}[20]\$ would be read as "the probability of rolling two simple failures and a simple success in any order, followed by a 20."

Ways to stabilize after...

The probabilities of stabilizing on the first, second, third, &c. rolls are exclusive: one can stabilize on the first or on the second, but not both. So we'll determine each of these probabilities and sum them up according to our "or-rule" from above.

1 roll:

$$ [20]=\frac 1 {20} $$

2 rolls:

$$ [1-19][20]= \frac{19}{20} \times \frac 1 {20}= \frac{19}{400} $$

3 rolls:

$$ \begin{align} [10-19][10-19][10-19] = \frac {10} {20} \times \frac{10}{20} \times \frac{10}{20} &= \frac 1 8 \\ \text{or} \quad \{[1][10-19]\}[20] = \left\{2 \times \left(\frac 1 {20} \times \frac{10}{20}\right)\right\} \times \frac 1 {20} &= \frac 1 {400} \\ \text{or} \quad [2-19][2-19][20] = \frac{18}{20} \times \frac{18}{20} \times \frac 1{20} &= \frac{81}{2000} \end{align} $$

$$ \left(\text{total: } \frac 1 8 + \frac 1 {400} + \frac{81}{2000} = \frac{21}{125}\right) $$

4 rolls:

$$ \begin{align} \{[10-19][10-19][1-9]\}[10-20]=\left\{3 \times \left( \frac{10}{20} \times \frac{10}{20} \times \frac 9 {20}\right)\right\} \times \frac{11}{20} &= \frac {297}{1600}\\ \text{or} \quad \{[2-9][2-9][10-19]\}[20]=\left\{3 \times \left(\frac{8}{20} \times \frac{8}{20} \times \frac {10}{20}\right)\right\} \times \frac {1}{20} &= \frac{3}{250} \end{align} $$

$$ \left(\text{total: } \frac {297}{1600} + \frac{3}{250} = \frac{1581}{8000}\right) $$

5 rolls:

$$ \{[2-9][2-9][10-19][10-19]\}[10-20]=\left\{6 \times \left(\frac{8}{20} \times \frac{8}{20} \times \frac{10}{20} \times \frac{10}{20}\right)\right\} \times \frac{11}{20}=\frac{33}{250} $$

Summing it all up:

$$ \frac 1 {20} + \frac{19}{400} + \frac{21}{125} + \frac{1581}{8000} + \frac{33}{250} = \frac{4761}{8000} = 0.595125 $$

Bonus for reading this far: expected time to recovery

With these probabilities in hand it's easy to give an expected time-of-stabilization for those that do stabilize:

$$ t_\text{stable} = \dfrac{ \left( \begin{align} 1 & \times \frac 1 {20} \\ +2 & \times \frac{19}{400} \\ +3 & \times \frac{21}{125} \\ +4 & \times \frac{1581}{8000} \\ +5 & \times \frac{33}{250} \end{align} \right) }{ \frac{4761}{8000} } = 3.5278 \, \text{rounds, or about 21 sec} $$


* - Order is really important. Sometimes. For instance, if we want to know the probability of the first two rolls being a success and a failure, we have to count all the ways that we could get a success then a failure, and add on all the ways we could get a failure then a success.

But some of our rolls are indistinguishable events, and that's where it gets fun. If we want to count the ways of our first three rolls being two failures and a success we only have three orders we can consider (FFS, FSF, SFF), not the six one might expect (ABC, ACB, BAC, BCA, CAB, CBA).

The count of possible arrangements of n objects with multiplicities \$m_1, m_2, \dots\$ is \$N = \dfrac{n!}{m_1! \times m_2! \times \dots}\$. See if you can spot the three times this pops up!

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Consider the number of successes and failures you've had so far against this table, and you can see how much a chance you have of surviving:

\begin{array}{r|lll} & \rlap{\text{number of failures} \rightarrow} \\ \text{number of successes} \; \downarrow & 0 & 1 & 2 \\ \hline 0 & 59.51\% & 41.75\% & 21.25\%\\ 1 & 73.50\% & 56.50\% & 32.50\%\\ 2 & 88.55\% & 77.00\% & 55.00\%\\ \end{array}

If you've had two failures and no successes, you have a 21.25% chance of survival.

I calculated these by means of a recursive algorithm which simulated the rules as-written.

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  • 2
    \$\begingroup\$ This fits nicely with @nitsua60's answer, which describes how (0,0) is actually calculated, by showing the evolving probabilities after a series of rolls. \$\endgroup\$ – chepner Mar 11 '16 at 19:22
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If we use a Markov Chain then we can not only answer this question but also all the other questions that may arise around death saves, like how long it takes to recover and what is you chance of surviving if you have failed once and saved once already.

There are 12 states in this particular Markov process - the initial state where you have made no saving throws yet (State 00), the intermediate states where you have passed X and failed Y saving throws (State XY) and the three terminal states (Conscious with 1hp, Stable and Dead).

The transition matrix is:

$$ \begin{array}{c|cccccccccccc} & 00 & 01 & 02 & 10 &11 & 12 & 20 & 21 & 22 &1\text{hp} & \text{Stable} & \text{Dead}\\ \hline 00 & 0 & 0.40 & 0.05 & 0.50 & 0 & 0 & 0 & 0 & 0 & 0.05 & 0 & 0 \\ 01 & 0 & 0 & 0.40 & 0 & 0.50 & 0 & 0 & 0 & 0 & 0.05 & 0 & 0.05 \\ 02 & 0 & 0 & 0 & 0 & 0 & 0.50 & 0 & 0 & 0 & 0.05 & 0 & 0.45 \\ 10 & 0 & 0 & 0 & 0 & 0.40 & 0.05 & 0.50 & 0 & 0 & 0.05 & 0 & 0 \\ 11 & 0 & 0 & 0 & 0 & 0 & 0.40 & 0 & 0.50 & 0 & 0.05 & 0 & 0.05 \\ 12 &0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.50 & 0.05 & 0 & 0.45 \\ 20 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.40 & 0.05 & 0.05 & 0.50 & 0 \\ 21 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.40 & 0.05 & 0.50 & 0.05 \\ 22 &0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.05 & 0.50 & 0.45 \\ 1\text{hp} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1.00 & 0 & 0 \\ \text{Stable} &0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1.00 & 0 \\ \text{Dead} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1.00 \\ \end{array} $$

Starting from the initial state its easy to derive that your chance of dying is 0.405, your chance of being stable is 0.414 and your chance of being conscious is 0.181 (to 3 significant figures). You can also tell that you can be dead in as little as 2 rounds (0.043) or up in as few as 1 (0.050).

For those who want to play around, here is a spreadsheet - it has scenarios set up for all of the states so you can tell that if you pass your first save, your chance of surviving went up from 0.595 to 0.735.

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  • 5
    \$\begingroup\$ Glad to see a markov chain used in a statistics answer. \$\endgroup\$ – user27327 Apr 11 '17 at 17:34

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