In the book probability theory and random processes by Grimmet, I don't understand the last line of the proof of the probability that the first time that a random walk hits the number $b$.
I don't understand the last step where he shows that the line above is equals to $\frac{b}{n}P(S_n=b)$. the result is the same as the probability of getting to $b$ without revisiting the origin.
Here is the definition of $M_n$
We have $$ \mathbb{P} (S_k=x) = \binom{ k}{ \frac {k+x}{2} }p^\frac {k+x}{2}q^{k-\frac {k+x}{2}}, $$ so \begin{alignat*}{2} f_b(n) & = \binom{n-1}{\frac{n-1+b-1}{2}}p^{\frac{n-1+b-1}{2}+1}q^{\frac{n-1-b+1}{2}}-\binom{n-1}{\frac{n-1+b+1}{2}}p^{\frac{n-1+b+1}{2}}q^{\frac{n-1-b-1}{2}+1}\\ & = \left[\binom{n-1}{\frac{n+b}{2}-1}-\binom{n-1}{\frac{n+b}{2}}\right]p^{\frac{n+b}{2}}q^{\frac{n-b}{2}}\\ & = \left[ \frac{(n-1)!}{\left(\frac{n+b}{2}-1\right)!\left(\frac{n-b}{2}\right)!} - \frac{(n-1)!}{\left(\frac{n+b}{2}\right)!\left(\frac{n-b}{2}-1\right)!} \right]p^{\frac{n+b}{2}}q^{\frac{n-b}{2}}\\ & = \frac{(n-1)!}{\left(\frac{n-b}{2}\right)!\left(\frac{n+b}{2}\right)!}\left[\frac{n+b}{2}-\frac{n-b}{2}\right]p^{\frac{n+b}{2}}q^{\frac{n-b}{2}}\\ & = \frac{b}{n}\mathbb{P}(S_n = b). \end{alignat*}