I have $50$ dollars and I’m gambling on a series of coin flips. For each head I win $2$ dollars and for each tail I lose $1$ dollar. What’s the probability that I will run out of money?
Hint: Suppose we have $x$ dollars, then the probability of ruin satisfies the recursive equation
$$p(x+2) - p(x) = p(x) - p(x-1)$$
Find function $p(x)$.
As indicated, the following recursion holds $$p(n+2)-2p(n)+p(n-1)=0.$$ It only has a solutions on the form $r^n$, where $r$ is a root of the characteristic polynomial $$r^3-2r^1+r^0=0.$$ Finding the roots one has that a general solution of $p$ is given by $$p(n)=A1^n+B\left(\frac{-1+\sqrt{5}}{2}\right)^n+C\left(\frac{-1-\sqrt{5}}{2}\right)^n.$$ Since the underlying random walk has increments with positive mean, $p(n)\rightarrow 0$, as $n\rightarrow\infty$. This gives $A=C=0$. Further, $p(0)=1$, so $B=1$. In conclusion, $$p(n)=\left(\frac{\sqrt{5}-1}{2}\right)^n.$$ $p(50)$ is now easy to calculate.
p=1 solution can be excluded in a rigorous way?
– Johannes Gerer
Feb 26 '17 at 21:16
We only need find the probability $r$ that you lose 1 dollar since the probability that you lose 50 dollars is $r^{50}$ since you must lose 1 dollar 50 times. We can write $$ r = 0.5 + 0.5r^3 $$
since half the time we lose on the first flip, and half the time we win on the first flip, thereby increasing our 1 dollar bankroll to 3 dollars, at which point our risk of ruin becomes $r^3$ since we must now lose 1 dollar 3 times. This has solutions $r = (\sqrt{5}-1)/2$ and $r = 1$. We obviously want the first of these since we have the advantage, and our opponent's $r$ is 1. So $$ P(lose \: 50) = \left(\frac{\sqrt{5}-1}{2}\right)^{50} $$ or about $3.55e^-11$.
Here's an alternative approach (which appears on the surface to be very similar to the recurrence + characteristic polynomial approach) using Martingales.
Let $S_i=X_1+\dots+X_i$ be your money at time $i$; here the $X_j$'s are random variables for the profit (or loss) you make at time $j$. We will find $c$ such that $c^{S_i}$ is a martingale, and then use the Optional Stopping Theorem. Note we need
$$\mathbb{E}[c^{S_{i+1}}|M_1,\dots,M_i]=c^{S_i}\implies \mathbb{E}[c^{X_i}]=\frac{1}{2}c^2+\frac{1}{2}c^{-1}=1\implies c^3-2c+1=0$$ We will take the root $c=\frac{\sqrt{5}-1}{2}$ as it is positive and strictly less than 1, thus for stopping time $T$ representing the first time we hit $0$, $|M_{\min(T,t)}|\leq 1$ for all $t$ is strictly bounded by a finite constant, and thus we can apply Optional Stopping Theorem:
$$\mathbb{E}[M_T]=p\cdot c^0+(1-p)\cdot 0 = M_0=c^{50}\implies p=c^{50}$$
@Johannes Gerer:
I think that a short answer is: If root 1 is not excluded (by setting A=0), $p(n)$ doesn't decrease to zero as $n\to\infty$.
Now why $p(n)$ should decrease to zero as $n\to\infty$.
$\mathbb P$(lose after starting with $0)=1$, since having 0 means loosing by the definition. Also notice that $\forall n>0:p(n)<1$, since there is a chance of not loosing ever. Now since $\mathbb P$(get $+\$2)=\mathbb P($get $-\$1)=0.5$, hence after having 1 we either get 0 (with $\mathbb P=0.5$) or get 3 (with $\mathbb P=0.5$),
So $\mathbb P$(loose after starting with 1)$=:p(1)=$ $0.5p(0)+0.5p(3)=$ $0.5+0.5p(3)>$ $0.5p(3)+0.5p(3)=p(3)$, so $p(1)>p(3)$.
By induction we can show that $p(n)$ is decreasing with $n$. We can see that $p(n)\rightarrow0,n\to\infty$.
We can also see it from the following. We are equally likely moving up and down. But steps up are twice bigger then down, so our game (which is a random walk) has a positive drift, and further we are from 0 more likely that we will never have enough steps down to lose.
