Why does P(Calvin wins the match from a tie game) = pP(wins the match when ahead by 1 point) + (1 — p)P(Calvin wins the match when behind by 1 point)? I don't understand user694818's answer. His username is gray, so I do not know when this user last visited this website.
Please DON'T moot other solutions. I already perused the author's alternative solution in his publicly downloadable, but downsized, solutions manual.

Here's how I would set up the problem:
Let $T$ be the probability that Calvin wins the match from a tie game.  Let $A$ be the probability that Calvin wins the match when he is ahead by one point, and let $B$ be the probability that Calvin wins the match when he is behind by one point.  Then show that:
$$T=pA+(1-p)B\\A=p+\color{Red}{(1 - p)T}\\B=pT$$
and solve that for $T$.


*

*Why does $T=pA+(1-p)B$? If Calvin ties, then Calvin must win 2 more games. Why doesn't T contain $pp$?


*I understand that if A happens, then Calvin need win merely 1 more game, with probability p. But why $\color{Red}{(1 - p)T}$? $1 - p$ = Pr(Calvin loses). How can Calvin lose, but then be "ahead by one point"?


*Why $B = pT$?  If Calvin is behind by 1 point, he must win 3 more games because he's behind my 1 point and the game postulates "the first player to win two games more than his opponent". So shouldn't B contain $ppp$?
 A: *

*Why does $T=pA+(1-p)B?$
$T$ is the probability that Calvin wins from a tie. Suppose Calvin and Hobbes are tied. With probability $p$ he will win the match, at which point he will have probability $A$ of winning. With probability $1-p$ he will lose the match, at which point he will have probability $A$ of winning.


*Why does $A=p+(1-p)T$?

$A$ is the probability that Calvin wins once he is ahead. Suppose Calvin is ahead. With probability $p$ he can win the match and thus the whole game. With probability $1-p$ he will lose the match and then once again have probability $T$ to win.


*Why does $B=pT$?

$B$ is the probability Calvin will win once he is behind. Suppose Calvin is behind. He must win the match to win the game. He has probability $p$ to win the match, at which point he has probability $T$ to win the game.
You ask why these expressions do not contain $p^2$ or $p^3$ terms. If you like, there is a $p^2$ term that pops out if you plug the expression for $A$ into the expression for $T$. However, the expression for $T$ will then also have terms containing $T$ and $A$. Repeatedly plugging expressions into each other does not lead to a simple computation of $T$. This is analogous to what would happen if you tried to consider all of the infinitely many different possible sequences of wins and losses that might occur, and calculate each of their probabilities. The method suggested by user694818 greatly simplifies this situation.
A: 




*Calvin and Hobbes play a match consisting of a series of games, where Calvin has probability $p$ of winning each game (independently). 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).




Let $T$ be the probability that Calvin wins the match from a tie game.  Let $A$ be the probability that Calvin wins the match when he is ahead by one point, and let $B$ be the probability that Calvin wins the match when he is behind by one point.  Then: $$T=pA+(1-p)B\tag1$$$$A=p+(1 - p)T\tag2$$$$B=pT.\tag3$$


These are recursive equations.


*

*Why does $T=pA+(1-p)B$? If Calvin ties, then Calvin must win 2 more games. Why doesn't T contain $pp$?


Calvin either wins or loses the first game after the tie game; the probability that he eventually wins the match given that he wins this game is precisely the probability $A$ that he eventually wins the match when he is ahead by one point; the probability that he eventually wins the match given that he loses this game is precisely the probability $B$ that he eventually wins the match when he is behind by one point. So, $T=pA+(1-p)B.\tag1$



*I understand that if A happens, then Calvin need win merely 1 more game, with probability p. But why $\color{Red}{(1 - p)T}$? $1 - p$ = Pr(Calvin loses). How can Calvin lose, but then be "ahead by one point"?


Calvin either wins or loses the first game after being ahead by one point; the probability that he eventually wins the match given that he wins this game is $1;$ the probability that he eventually wins the match given that he loses this game is precisely the probability $T$ that he eventually wins the match from a tie game. So, $A=(p)1+(1 - p)T.\tag2$



*Why $B = pT$?  If Calvin is behind by 1 point, he must win 3 more games because he's behind my 1 point and the game postulates "the first player to win two games more than his opponent". So shouldn't B contain $ppp$?


Calvin either wins or loses the first game after being behind by one point; the probability that he eventually wins the match given that he wins this game is precisely the probability $T$ that he eventually wins the match from a tie game; the probability that he eventually wins the match given that he loses this game is $0.$ So, $B=pT+(1-p)0.\tag3$
A: The explanation of the equations has already been given by others for the Markov approach.
I would like to give another (simpler) approach for this question, taking two games at a time.
If Calvin is to win from scratch, (ie Hobbes is not to win),  either Calvin is to win in the next two games with probability $p^2$ or go back to scratch with probability $2pq$
Thus $T = p^2 +2pq*T,\;\;and\;\; T = \dfrac {p^2}{1-2pq}$
