# Markov chain application Gambler's Ruin Problem

Could anyone please explain me why you must assume to be win at i+1 and lose at i-1 in order to get to the highlighted lines? Thanks a lot. .

A player $A$ plays a sequence of independent games against an opponent $B$. At each game either the player wins say \$1 with probability$p$or loses \$1 with probability $q=1-p$. We assume that player $A$ initially has an amount of \$k and his opponent$B$has \$(N-k). That is, the total amount in the game is \$N, and no inflow or outflow of money is allowed. The game continues until either of the players get ruined. If$X_{n}$denotes the accumulated amount with the player$A$after$n$games, then the sequence$\{X_{0},X_{1},X_{2},\cdots,\}$represents a time homogeneous finite state Markov chain with the state space$S=\{0,1,2,3,\cdots,N\}$. The states,$0$and$N$being absorbing states. We are interested in computing the probability that player$A$wins all the amount in the game. Let$P_{k}$denote the probability that player$A$wins all the amount in the game, given that he has an initial capital of \$k. \begin{equation*} P_{k}=P\{X_{n}=N, \mbox{ for some n }\vert X_{0}=k\} \end{equation*} To compute the above probability, we observe that either he wins the first game, with probability $p$, and go on to win all the amount starting with capital \$(k+1), OR, he lose the first game, with probability$q=1-p$and go on to win all the amount starting with capital \$(k-1). This give the difference equation, \begin{equation*} P_{k}=pP_{k+1}+qP_{k-1}, \quad 0<k<N. \end{equation*}