I was trying to solve this problem (Strategic Practice Week 3, Homework problem 4 in Harvard's Stat 110 class), by framing it as a gambler's ruin problem:
Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability $p$ of winning each game (independently) and $q = 1-p$. They play with a “win by two” rule: the first player to win two games more than his opponent wins the match. Find the probability that Calvin wins the match (in terms of $p$), by interpreting the problem as a gambler's ruin problem.
Here's how I approached the problem:
Let,
$W:$ Calvin wins the match;
$D_i:$ (Wins by Calvin) $-$ (Wins by Hobbes) $= i$
$p_i:$ $\Pr($W | $D_i$$)$
Now, by conditioning on the first game, and using the law of total probability, we get:
$p_i = p p_{i+1} + qp_{i-1}$, with $p_2 = 1$ (Calvin wins with certainty if the difference is $2$) and $p_{-2} = 0$ (Calvin loses with certainty if the difference is $-2$).
Solving this recurrence relation (which I omit here, since it's mostly algebra gymnastics), we get:
$p_i = \dfrac{p^6}{p^8 - q^4}p^i - \dfrac{p^2q^2}{p^8-q^4}(\dfrac{q}{p})^i$
Since both Calvin and Hobbes start with a $0$ difference in wins, what we need to find is $p_0 = \dfrac{p^2(p^4-q^2)}{p^8-q^4}$.
However, the answer turns out to be $\dfrac{p^2}{p^2+q^2}$, which can be easily obtained by using the law of total probability and conditioning on the number of wins in the first 2 matches (0 win, 1 win or 2 wins, with the number of wins $X \sim $ Bin($2,p$)).
Where am I going wrong with my interpretation of the problem as a gambler's ruin problem?
Your formulation is correct. Your solution to the recurrence relation (the part you omitted) is not. You should find
$$\lambda=\frac{1 \pm \sqrt{1-4pq}}{2p}=\frac{1 \pm (2p-1)}{2p}=1,\frac{q}{p}.$$
Thus if $0<p<1$ then the basis of solutions is given by $\{ 1,(q/p)^i \}$, except when $p=q$ in which case the basis of solutions is given by $\{ 1,i \}$. So in particular the inclusion of $p^i$ in the basis of solutions was not correct.