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You find 2 dollars in your pocket and decide to go gambling. Fortunately, the game you're playing has very favourable odds: each time you play, you gain 1 dollar with probability 3/4 and lose $1 with probability 1/4.

Suppose you continue playing so long as you have money in your pocket. If you lose your first two bets, you go broke and go home after only two rounds; but if you win forever, you'll play forever.

What's the probability you'll eventually go broke?

I think I am overthinking this:

To go broke, you have to end on two Losses, which have p(L) = 1/4

Before the two losses there has to have been an even number of losses and win to have a balance of 2 dollars before the final two losses.

$P(Broke)=left( dfrac {1}{4}right) ^{2}sum ^{infty }_{i=0}left( dfrac {1}{4}right) ^{i}left( dfrac {3}{4}right) ^{i}N_{i}$

But then for each summand I need to multiply by the number of ways it is possible to get to an equal number of losses is without going broke beforehand. so for i = 1 LW and WL, are allowed, for i = 2 WLLW , WLWL, LWWL, LWLW, LLWW are allowed.

So my final calculatin is $P(Broke) = left( dfrac {1}{4}right) ^{2}sum ^{infty }_{i=0}left( dfrac {1}{4}right) ^{i}left( dfrac {3}{4}right) ^{i}left( begin{pmatrix} 2i \ i end{pmatrix}-sum ^{i-1}_{k=1}begin{pmatrix} 2k \ n-1 end{pmatrix}right) $

But this doesn't equal anything and seems rather complicated.

Let $p(n)$ be the probability of going broke starting with $n$ dollars. On the one hand, we have
$$
p(n)=p(1)^ntag1
$$
because moving from $$n$ to $$0$ is the same as moving from $$k$ to $$k-1$ a total of $n$ times, for $k=n,n-1,n-2,dots,1$. On the other hand, we must have
$$
p(1)=frac34 p(2)+frac14qquad (nge1)tag2
$$
Substituting $(1)$ into $(2)$, and setting $n=1$, you get $p(1)=frac34 p(1)^{2}+frac14$, the solution to which is $p(1)=1/3$ or $p(1)=1$. If we had $p(1)=1$, then we would have $p(n)=1$ for all $n$, so that you certainly go broke no matter how much money you start with, which is clearly incorrect.$^*$ Therefore,
$$
bbox[3pt,border: 1.5pt black solid]{p(n)=(1/3)^n.}
$$
$^*$A rigorous proof of this certainly exists, perhaps by applying Stirling's approximation to an exact combinatorial expression for $p(n)$, or with some central limit theorem argument.

Edit: For a more convincing argument, suppose that the process does not stop when you hit zero. If $p(1)=1$, you would certainly hit $$0$, and from there certainly hit $$(-1)$, and from there $$(-2)$, and so on. This means you would drift arbitrarily far down. This contradicts the law of large numbers, which says $S_n/n$ converges to $1/2$ almost surely as $ntoinfty$, where $S_n$ is your winnings after $n$ plays.

As another possible method. Consider a slightly modified game where the game can end if the money reaches $0$ or $N$ dollars (we can take the limit later). Let $P_i$ be the probability that you win starting with $i$ dollars. Suppose the probability of winning a single game is $p$ and $q = 1- p$ is the probability of losing a single game. We can observe the relation
$$ P_i = p P_{i+1} + q P_{i-1}. tag{1} label{1}$$
This is a linear difference equation which can easily be solved for roots (via $P_i = z^i$). The result is
$$P_i = begin{cases} A + B left(frac{q}{p} right)^i & text{if } p neq q, \ A + B i & text{if } p = q = 1/2. end{cases}$$
Then noting the boundary conditions $P_0 = 0$ and $P_N = 1$, we obtain,
$$P_i = begin{cases} frac{1-left(frac{q}{p} right)^i}{1- left(frac{q}{p} right)^N} & text{if } p neq q, \ frac{i}{N} & text{if } p = q = 1/2. end{cases}$$
Taking the limit as $N to infty$, yields the desired result.

Unfortuneately i don't have a very intuitive understanding of the problem, but i thought about it before and that's what i came up with:

Denote $S_n$ be the your money in step $n$ and $S_0=x$ the money you start with. Let $y$ be the (for the time) finite amount of money the bank starts with.
Let $p$ be the probabilty you win $1$$ and $(1-p)$ the probabily that you lose $1$$.

Notice that $left(frac{1-p}{p}right)^{S_n}$ is a martingale, which means
$$mathbb{E}left[left(frac{1-p}{p}right)^{S_n} huge|normalsize left(frac{1-p}{p}right)^{S_k} right] = left(frac{1-p}{p}right)^{S_k}$$
for $kleq n$.

Let $tau = operatorname{inf}{k: S_ntext{ is constant for all }ngeq k }$, that is let $tau$ be the time the game ends (you won all the money from your opponent or you lost all your money).

By the optional stopping theorem,

$$mathbb{P}left(S_tau = 0 right) left(frac{1-p}{p}right)^{0} + left(1- mathbb{P}left(S_tau = 0 right) right)left(frac{1-p}{p}right)^{x+y} = mathbb{E}left[left(frac{1-p}{p}right)^{S_tau}right] = mathbb{E}left[left(frac{1-p}{p}right)^{S_0}right] = left(frac{1-p}{p}right)^{x},$$
so
$$mathbb{P}left(S_tau = 0 right) = frac{left(frac{1-p}{p}right)^{x} - left(frac{1-p}{p}right)^{x+y}}{1 - left(frac{1-p}{p}right)^{x+y}}.$$

For $(1-p) < p$ we see, by letting $yrightarrow infty$, that for a bank with unlimited resources, we get
$$mathbb{P}left(S_tau = 0 right) = left(frac{1-p}{p}right)^{x}.$$

Let $p(n)$ be the probability of ever reaching $n-1$ coins (dollars) starting with $n$ coins. We have that

$$p(n) = 1/4 + 3/4 cdot p(n+1) cdot p(n)$$

Since we can either reach $n-1$ coins directly by losing or win one coin then reach $n$ coins with probability $p(n+1)$ and then reach $n-1$ coins with probability $p(n)$.

But we can see that $p(n+1) = p(n)$, because it's the same the probability of eventually losing one coin, it doesn't matter if we have $n$ or $n+1$ coins to begin with. Substituting, we have

$$p(n) = 1/4 + 3/4 cdot p(n)^2$$

Solving this quadratic equation we have $p(n) = 1/3$ or $p(n) = 1$. We can ignore the case where $p(n) = 1$ since the probability of losing is not 1.

Finally, the probability of losing is

$$p(2) * p(1) = (1/3)^2 = 1/9$$