Well, first up, we need to see what we're dealing with.

If you take any damage while
you have 0 hit points, you suffer a death saving throw
failure.

Ok. Now let's look at *Scorching Ray* for an example.

Make a ranged spell attack for each ray. On a hit, the target takes 2d6 fire damage.

You suffer a death saving throw failure when you take damage, and you take damage each time a ray of *Scorching Ray* hits you. So if you're hit by 3 rays, you take 3 death saving throw failures. *Eldritch Blast* uses the same wording.

Now, Extra Attack. Extra Attack says:

Beginning at 5th level, you can attack twice, instead of
once, whenever you take the Attack action on your turn.

So you can make two attacks, and that's all it means. When you make an attack:

- Choose a target. [...]
- Determine modifiers. [...]
- Resolve the attack. You make the attack roll. On a hit,
you roll damage, unless the particular attack has rules
that specify otherwise. Some attacks cause special
effects in addition to or instead of damage.

Each time you make an attack, you roll damage. You don't, for example, make all of your attacks and then roll damage for them all cumulatively. If you use Extra Attack to hit an unconscious creature twice, it takes damage twice, and suffers 2 death saving throw failures.

As additional backup (thanks Adeptus), we have:

If the damage is from a critical hit, you suffer two
failures instead.

A critical hit is something that happens to an individual attack, like one ray of *Scorching Ray* or one attack from Extra Attack, not to the whole of either of them.

As far as brutality is concerned, I would point out that there are good reasons not to attack unconscious creatures: Unconscious creatures aren't a threat. From the perspective of a monster, it would generally make more sense to deal with the adventurers who are still trying to kill them than wasting time finishing off the ones who are already out of the fight.

## Best Answer

## 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:

orB), 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.)andB), 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

oron 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!}