The probability of a mere human surviving a series of unmodified death saving throws has been established as 59.5125%. But Halfling Luck applies to death saving throws in a dramatic way: A roll of 1 on a death save normally counts as two failures, but halflings are much less likely to suffer this result.

Previous work in the field of calculating the effects of Halfling Luck does not apply directly to death saving throws, because of this asymmetry between races concerning how terrible it is to roll a 1.

How much more likely is it that a halfling will survive a series of death saving throws as opposed to one of those poor unfortunate nonhalflings?

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    \$\begingroup\$ Besides following naturally from existing questions, this issue has real-world applications as I try to figure out how much of a discount halflings should get on their resurrection insurance premiums. \$\endgroup\$ Jun 2 '17 at 4:13
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    \$\begingroup\$ You don't even have to be a halfling--the Lucky feat would seriously wreak havoc with the attached question's numbers =) \$\endgroup\$
    – nitsua60
    Jun 2 '17 at 4:22

A Halfling has a 67.877607328% to survive their death saves

Refer to this question for calculations of the distribution.

In brief, to re-roll at all, you need to roll a 1 first. The chances of that are \$\frac{1}{20}\$ out of the gate. Once you do re-roll, you have a \$\frac{1}{20}\$ chance of re-rolling any specific number. So the chance of re-rolling and getting any specific number is \$\frac{1^2}{20^2} = \frac{1}{400} = 0.25\%\$.

To roll a 1, you must re-roll and get a 1 on that re-roll. The chances of that is already given, \$0.25\%\$. To roll anything other than a 1, you can get there by two ways: you can roll it the first try (\$5\%\$ chance), or roll a 1 on the first try and get it on the re-roll (\$0.25\%\$ chance). Together, both give you a \$5.25\%\$ chance to roll anything not a 1.


To get the answer, you need only to manipulate the probabilities nitsua60 wrote down in this question.

You survive your death saves if:

  • You roll a 20 the first time;

$$Pr(X=20) = 5.25\%$$

  • You roll below a 20 the first roll, and a 20 the second roll;

$$Pr(X<20) \times Pr(X=20) = (1 - Pr(X=20)) \times Pr(X=20) = 4.974375\%$$

  • You roll between a 10 and 19 on all three rolls;

$$Pr(10<X<19)^3 = (0.525)^3 = 14.4703125\%$$

  • You roll a 1 the first roll, between a 10 and 19 the second, and a 20 the third roll (and count this twice for symmetry of the first two rolls);

$$2 \times Pr(X=1) \times Pr(10<X<19) \times Pr(X=20)$$

$$2 \times 0.0025 \times 0.525 \times 0.0525 = 0.01378125\%$$

  • You roll between a 2 and 19 the first two rolls, but get a 20 the third roll

$$Pr(2<X<19)^2 \times Pr(X=20) = 0.945^2 \times 0.0525 = 4.68838125\%$$

  • You roll between a 10 and 19 on the first two rolls, you roll between a 1 and 9 on the third, and between a 10 and 20 on the fourth (count this thrice for symmetry);

$$3 \times Pr(10<X<19)^2 \times Pr(1<X<9) \times Pr(10<X<20)$$

$$3 \times (0.525)^2 \times 0.4225 \times 0.5775 = 20.175233203\%$$

  • You roll between a 2 and 9 on the first two rolls, between a 10 and 19 on the third, and a 20 on the fourth one (count this thrice for symmetry);

$$3 \times Pr(2<X<9)^2 \times Pr(10<X<19) \times Pr(X=20)$$

$$3 \times (0.42)^2 \times 0.525 \times 0.0525 = 1.4586075\%$$

  • You roll between a 2 and 9 the first two rolls, between a 10 and 19 the next two rolls, and between a 10 and 20 on the fifth roll (and count this six times for symmetry);

$$6 \times Pr(2<X<9)^2 \times Pr(10<X<19)^2 \times Pr(10<X<20)$$

$$6 \times (0.42)^2 \times (0.525)^2 \times (0.5775) = 16.846916625\%$$

Summing up.

$$5.25\% + 4.974375\% + 14.4703125\% + 0.01378125\% + 4.68838125\% + 20.175233203\% + 1.4586075\% + 16.846916625\% = 67.877607328\%$$

Halfling vs Non-Halfing

As a Halfling, your overall chance to survive jumps from ~60% to ~68%. But this massive increase in probability to survive comes from the fact that there are so many ways to survive a death save already, and almost all of those ways to survive (specifically, the ones that don't involve you rolling a Nat 1) gain some small boost in probability. All those small boosts turn out to add up to a large increase.

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    \$\begingroup\$ The "+2 to death saves" bit has me curious. Are you just approximating the 10% difference in outcome-probability by the flat +2, or did you actually run the probabilities and find that (d20+2) worked out close to (d20 reroll 1)? \$\endgroup\$
    – nitsua60
    Jul 13 '17 at 12:08
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    \$\begingroup\$ @nitsua60 I approximated. That is a good point, a true +2 would be an entirely different calculation. \$\endgroup\$
    – user27327
    Jul 13 '17 at 12:18
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    \$\begingroup\$ I think I've spotted a tiny typo. \$Pr(1<X<9) = 0.4225\$, not \$0.475\$, giving \$20.175233203124996\%\$ not \$22.682214843\%\$ and a final answer of \$67.877607328125\%\$ \$\endgroup\$ Jun 2 '20 at 4:43

Starting with 0 successes and 0 failures, the chance of a halfling surviving is 67.9%. You can compute this figure using this recursive function in python.

def chance_survival(f, s):
    chance_crit_fail = 1/20 * 1/20
    chance_norm_fail = 1/20 * 8/20 + 8/20
    chance_norm_succ = 1/20 * 10/20 + 10/20
    chance_crit_succ = 1/20 * 1/20 + 1/20
    if f >= 3:
        return 0
    elif s >= 3:
        return 1
        return (
            chance_crit_fail * chance_survival(f + 2, s) +
            chance_norm_fail * chance_survival(f + 1, s) +
            chance_norm_succ * chance_survival(f, s + 1) +
            chance_crit_succ * chance_survival(f, s + 3)

For reference, using similar code, you can compute that the chance of someone without the lucky trait surviving is 59.5%. With advantage on all rolls, the chance is 94.7%, and with disadvantage on all rolls the chance is 15.1%.

  • \$\begingroup\$ Any thoughts on why this gives a result ten percentage-points below the other answer? \$\endgroup\$
    – nitsua60
    Mar 14 at 3:11
  • \$\begingroup\$ @nitsua60 I haven't checked the other answer completely, but it's likely the other answer double counts some ways to survive dropping to 0 HP, since the first few paragraphs seem correct. \$\endgroup\$
    – Vaelus
    Mar 14 at 3:17
  • \$\begingroup\$ @nitsua60 Turns out I had a typo in my code when I initially computed it. (The code in the answer was correct, but the numbers were wrong). The answer comes out to the same as jonny's comment on the other answer. \$\endgroup\$
    – Vaelus
    Mar 14 at 3:34

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