Problem 16.4 Easy Free 1 variation

Fair Gambler's Ruin

A particle performs a simple symmetric random walk on the integer points $0,1,\dots,10$ , moving to a neighbouring point with probability $1/2$ at each step and stopping as soon as it reaches $0$ or $10$ . Starting from $6$ , find the probability that it is absorbed at $10$ rather than at $0$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. The position is a martingale, so the hitting probability is linear in the starting point.

Let $p_k$ be the probability of absorption at $10$ starting from $k$ . Conditioning on the first step gives

p_k=\tfrac12 p_{k-1}+\tfrac12 p_{k+1}\qquad(1\le k\le 9),\qquad p_0=0,\quad p_{10}=1.

The recurrence says $p_{k+1}-p_k=p_k-p_{k-1}$ . So the increments are constant and $p_k$ is linear in $k$ . The boundary values then force $p_k=k/10$ .

Alternative. Let $X_n$ be the position and $T$ the absorption time.

$T<\infty$ almost surely. From any interior point, ten consecutive steps in the same direction (probability $2^{-10}$ ) force absorption, so $\mathbb{P}(T>10m)\le(1-2^{-10})^m\to0$ .
$X_{n\wedge T}$ is a martingale bounded between $0$ and $10$ .
Optional stopping therefore gives $6=\mathbb{E}[X_T]=10\,\mathbb{P}(X_T=10)$ .

Either way, starting from $6$ :

\boxed{\mathbb{P}(\text{absorbed at }10)=\frac{6}{10}=\frac35}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 16.4a

Alice and Bob start with $15 and $10 respectively and repeatedly bet $1 on independent fair coin flips, each betting on the opposite outcome. If Alice ever goes bankrupt, she may buy back in once for $15 and keep playing against Bob, who retains whatever he holds at that moment; if she goes bankrupt a second time she loses outright. Find the probability that Bob is the one who eventually goes bankrupt.

Work it out, then compare with the answer.

Solution

Alice’s fortune performs a fair walk, so by the main result the probability of ruin is linear in the stakes.

First phase: Alice has $15$ and the total is $25$ . She goes bankrupt first with probability $\frac{10}{25}=\frac25$ , and Bob then holds $25$ .

Second phase: after the rebuy Alice has $15$ against Bob’s $25$ . Because the coin flips are independent, this is a fresh fair game on a total of $40$ . Alice is ruined in it with probability $\frac{25}{40}=\frac58$ .

Alice loses the whole contest only if she is ruined in both phases, which has probability $\frac25\cdot\frac58=\frac14$ .

This is the same as Alice starting with $30$ against $10$ , where Bob’s ruin probability is $\frac{30}{40}$ . Hence

\boxed{\mathbb{P}(\text{Bob goes bankrupt})=\frac34}

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