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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?

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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.

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  • $\begingroup$ Ah, yes! I somehow wrote $p$ instead of $1$ for the first root. And to think I was checking and rechecking the subsequent calculations, without so much as a glance back! Lesson learnt, thank you! :) $\endgroup$ – Train Heartnet Feb 5 '18 at 14:47
  • $\begingroup$ But even then, with the now correct $p_i = \dfrac{p^2-q^2}{p^2} + \dfrac{q^4-p^2q^2}{p^4}(\dfrac{q}{p})^i$ (assuming $p \neq q$), I'm getting $p_0 = \dfrac{p^4+q^4}{p^4}$, which is incorrect. $\endgroup$ – Train Heartnet Feb 5 '18 at 15:19
  • $\begingroup$ @Heartnet You made a sign error in simplifying. $\endgroup$ – Ian Feb 5 '18 at 15:22
  • $\begingroup$ Ah, darn it. Okay, assuming the $p_i$ I wrote is correct, now I have $p_0 = \dfrac{(p^2-q^2)^2}{p^4}$, which still is incorrect. Or did you mean I made a sign error elsewhere, while obtaining $p_i$? I'm really sorry to be a bother, but I just want to nail this problem. $\endgroup$ – Train Heartnet Feb 5 '18 at 15:34
  • $\begingroup$ @TrainHeartnet There was definitely a sign error there but I think there was also a mistake further back. I usually prefer to do this with linear algebra, which in this case means solving the system $a_1+(q/p)^2a_2=1,a_1+(q/p)^{-2}a_2=0$. This solution to that is not all that complicated; the only ugly thing is the division by the determinant. $\endgroup$ – Ian Feb 5 '18 at 15:38

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