In a fair gambler's ruin problem, where the gambler starts with k dollars, wins \$1 with probability 1/2 and loses \$1 with probability 1/2, and stops when he/she reaches \$n or \$0. 

In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p\_k$ be defined as the probability of reaching \$n with \$k in one's inventory. Then they use the fact that $p\_k - p\_{k-1} = p\_{k-1} - p\_{k-2} = ... = p\_1 - p\_0 = p\_1$. 

Intuitively this means the probability of reaching \$n with \$k minus the probability of reaching \$n with \$k-1 is equivalent to the probability of reaching \$n with only \$1. 

Is there an intuitive reason why this is the case?