What is the probability that you never lose this hypothetical dice game? You have a four sided die numbered 1-4 and are playing a game. On the first round, you roll the die once. If you get a 1 you lose the die. If you get a  2 you keep the die. If you get a 3 or 4 you get another, identical die. On the second round you roll each die that you have, and the same thing happens with each die. Once you finish rolling all your dice you move on to the third round and so on. If you have zero dice you lose the game. What is the probability that you never lose the game over infinite rounds?
Let $p$ be the probability of losing a die and all its "spawns" (the dice you got with it by rolling $3$ or $4$ and their spawns and so on). So we have

$$ p = \frac{1}{4} + \frac{1}{4} p + \frac{1}{2}p^2$$

This comes from the following reasoning: You can either lose the die immediately or keep the situation as is (you have the same $p$) or then you get a spawn and have to get rid of that too (hence the $p^2$).

Solving gives $p=\frac{1}{2}$ or $p=1$. Which one should we choose?

EDIT Other solutions have given resources for choosing the correct value, but for completion's sake I'll add some reasoning to my answer (also since I've learned something new myself and to reinforce it for my self :D).

I'll prove we should choose the smaller of the solutions (so in this case $p = \frac{1}{2}$).

Define

$X_n = $ "The number of dice we have after round $n$"

$A_n = \{X_n = 0\} =$ "We have lost all dice after round $n$"

$p_n = \mathbb{P}(A_n)$

We have $A_{n+1} \subset A_n$, so $p=\lim_{n\to\infty} p_n$.

Then define

$$f(x) = \frac{1}{4} + \frac{1}{4}x + \frac{1}{2}x^2.$$

Lemma 1

$$p_{n} = f(p_{n-1}).$$

Proof of Lemma 1: Notice that $f$ is the probability generating function (pgf) of $X_1$ and let's define $f_n$ to be the pgf of $X_n$. Then we have

$$f_n = f \circ f_{n-1}$$

This comes from the fact that if we have $k$ dice, they each "transform as" $f(x)$ and are independent so they produce a term $f(x)^k$ . So it goes like that: we keep composing f with itself. Now, lemma 1 follows easily, since $p_n = f_n(0)$. $$\tag*{$\blacksquare$}$$

Let $t$ be the smallest value on $[0,1]$ for which we have $t=f(t)$ (so $t=\frac{1}{2}$ in this case, but let's do this a bit more generally). Now, if we prove $p_n \leq t$ for all $n$ we are done, since $p$ is their limit. Let's do this inductively. The base case: $p_0=0 \leq t$. Inductive step: assume $p_n \leq t$. Since $f$ is increasing on $[0, 1]$,

$$p_{n+1} = f(p_n) \leq f(t) = t$$

and this finishes the proof.

Notice also, that $f'(1) = \mathbb{E}(X_1)$ is the slope of $f$ at $1$ and since $f'' \geq 0$, $f$ is convex so whether $\mathbb{E}(X_1)$ is bigger than $1$ or not decides whether the smallest solution will be $&lt;1$ or $=1$. Because $f$ is a pgf, $f(1) = 1$ so $1$ will be the "default" solution if there arent' any smaller ones.
there's always the chance that with n dice you will roll all of them and get 1's, it's just a really small probability but it might happen.
Generally on each round you are expected to increase the number of dice you have by 25%.

EDIT: To answer the question more specifically, with n dice your chance of losing is $\frac1{4^n}$. With each round the expected increase of amount of dice you have is 25%, so over infinite rounds you will expect to have $n\to \infty$ which means $\frac1{4^n}\to0$