The author of my textbook gives the problem:
A gambler repeatedly bets \$1 that a coin will come up heads when tossed. Each time the coin comes up heads, the gambler wins \$1; each time it comes up tails, he loses \$1. Let $P_n$ be the probability that the gambler is ruined if he begins playing with \$ $n$, then
$P_{k-1}=\frac{1}{2}P_{k}+\frac12P_{k-2}$
This follows from the fact that if the gambler has \$$(k − 1)$, then he has an equal chance of winning \$1 or losing \$1, and if he wins \$1, then his chance of being ruined is $P_k$ , whereas if he loses \$1, then his chance of being ruined is $P_{k−2}$.)
I fail to understand this equation. It's not recurrence relation - current term depends on one ahead. Also why does the author multiply the probabilities by $\frac12$? This problem is given as an example in the Expected Value section. How does it relate to Expected Value?
Later the author transforms that equation to $P_k=2P_{k-1}-P_{k-2}$ which doesn't make sense to me (not the algebra but the relation between the terms itself - why current term depends on 2 $P_{k-1}$ and 1 $P_{k-2}$) too since it's based on the reasoning for the original one.
For the solution of the difference equation it's better to write $1 = \frac{1}{2} + \frac{1}{2}$ because that is the term of $P_{k-1}$ in the first equation. Then construct difference from each two consecutive terms, obtain the boundary term and then sum on both sides using telescoping summation.
Keep in mind each term/state is a function of $n$, starting fortune, hence $P_n$ is a limiting probability, i.e probability that the gambler will go bankrupt eventually.