Probability of A winning the game

Question

The following question is from problem 30 of the 2021 level 11-12 Mathematical Kangaroo:

A certain game is won when one player gets 3 points ahead. Two players A and B are playing the game and at a particular point, A is 1 point ahead. Each player has an equal probability of winning each point. What is the probability that A wins the game?

Apparently, the solution is 2/3, but I don’t quite know how the solution is gotten. Any help will be much appreciated!

Answer 1

Model this as a Markov Chain, with state $-3, -2, -1, 0,1,2,3$ where the numbers represent the points of $A$ minus the points of $B$. $-3$ and $3$ are the absorbing states.

Let $p(i)$ be the probability that it is absorbed at $3$ if we start the transition at $i$.

$p(-3)=0, p(3)=1$.

We have $p(i) = \frac12 p(i-1)+\frac12p(i+1), -2 \le i \le 2. $

or $p(i+1)-2p(i)+p(i-1)=0, -2 \le i \le 2. $

Its corresponding characteristic polynomial $x^2-2x+1=0$ which has a repeated root $1$.

$p(i)=A+Bi$

We have $p(-3)=0$ and $p(3)=1$.

$$-3B+A=0$$ $$3B+A=1$$

Solve for $A$ and $B$ and you just compute $p(1)$.

Intuitive answer:

It is equally likely to travel left or to travel right and you can get the answer by considering the ratio of distance from $1$ to $3$ and the distance from $1$ to $-3$.

Answer 2

This is a random walk between the points $-3$ and $3$. As each step is $50/50$, and the starting point is $1$, $B$ needs $4$ steps and $A$ only $2$, hence the probability that $A$ wins is $\frac46=\frac23$.