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!
