Gambling game: What is the probability of eventually going broke

Question

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.

Answer 1

Let $p(n)$ be the probability of going broke starting with $n$ dollars. On the one hand, we have $$ p(n)=p(1)^n\tag1 $$ 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 $$ p(1)=\frac34 p(2)+\frac14\qquad (n\ge1)\tag2 $$ 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 $n\to\infty$, where $S_n$ is your winnings after $n$ plays.

Answer 2

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.

Answer 3

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 $k\leq n$.

Let $\tau = \operatorname{inf}\{k: S_n\text{ is constant for all }n\geq k \}$, 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 $y\rightarrow \infty$, that for a bank with unlimited resources, we get $$\mathbb{P}\left(S_\tau = 0 \right) = \left(\frac{1-p}{p}\right)^{x}.$$

Answer 4

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

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

Answer 5

It was still bugging me that we have 2 roots of the quadratic equation in the solution and I wanted a rigorous proof that probability of going broke is not 1, no matter how intuitive it seems.

I think I came up with the bound:

Suppose that we start with 1 dollar. To go broke we obviously need odd number of bets, suppose it is 2n+1. To go broke after 2n+1 bets we need n winning bets and n+1 losing bets. According to binomial distribution probability of going broke after 2n+1 bets:

P_broke(2n+1) <= C(2n+1, n) * (3/4)^n * (1/4)^(n+1)

(C() denotes binomial coefficient)

<= because sequences of bets where you go broke before 2n+1 are not valid (cannot go into debt). In fact, for any n > 0, <= will actually be < because we exclude the sequences where you go broke before 2n+1 bets.

Total probability of going broke will be bounded by summation of the above with n from 0 to infinity.

It is easy to show that C(2(n+1)+1,n+1) = C(2n+1,n)2(2n+3)/(n+2) < 4*C(2n+1,n) from which it is easy to show by induction that C(2n+1,n) < 4^n. Thus we have:

P_broke(2n+1) < 1/4 * (3/4)^n

Total probability of going broke P_broke = sum(0, inf) P_broke(2n+1), thus

P_broke < 1/4 sum(0, inf) (3/4)^n = 1/4 * 1 / (1 - 3/4) = 1

Thus P_broke < 1.

This is the best i could come up with so far, i m sure that for given probabilities a tighter bound can be proven, please post if you know.