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?
 A: 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 $<1$ or $=1$. Because $f$ is a pgf, $f(1) = 1$ so $1$ will be the "default" solution if there arent' any smaller ones.
A: This is a classic example of a branching process.  As other answers have pointed out, the extinction prob $p = Prob(lose) = 1$ or $1/2$.  To pick the correct answer, the linked wikipedia article has this to say:

it can be shown that starting with one individual in generation zero, the expected size of generation $n$ equals $\mu^n$ where $\mu$ is the expected number of children of each individual. If $\mu < 1$, then the expected number of individuals goes rapidly to zero, which implies ultimate extinction with probability $1$ by Markov's inequality. Alternatively, if $\mu > 1$, then the probability of ultimate extinction is less than $1$.

In the OP example, $\mu = (0 + 1 + 2 + 2) / 4 = 5/4 > 1$, so if you believe the wikipedia article, then $p < 1$.  So the right answer is:
$$p = \frac12$$
Disclaimer: I personally don't know enough about branching processes to verify what wikipedia said.  The article has some more details (in addition to what's quoted above) but I haven't read it in detail.
A: Let $p$ be the probability that you eventually lose starting from one die.  The key idea is that, since throwing each die is independent, the probability of eventually losing when you hold $n$ dice is $p^n$.  Therefore, by the rule of the game, $$p=\frac14+\frac14p+\frac12p^2\\2p^2-3p+1=0$$
So this gives us $p=0.5$ or $p=1$.  Looking at this problem with other rules I'm inclined to reject $p=1$, as it always turns up, so I will say there is a 50% chance of eventually losing.
A: 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 $\frac1{4^n}\to0$
