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In basic, symmetric random walk with $P(Y_{n}=1)=\frac{1}{2}$, $S_{0}=0$, calculate: $$P(S_{1}>0,...,S_{2n-1}>0, S_{2n}=0)$$

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  • $\begingroup$ What if you take the first and last step and bring down the path. You get like some dick path. no? $\endgroup$ – Phicar Apr 8 at 19:14
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You need to count the number of paths $(S_0,S_1,\dots,S_{2n-1},S_{2n})$ such that $S_0=S_{2n}=0$, $S_i>0$ for $0<i<2n$, and consecutive $S_i$'s differ by $\pm1$. The path $(S_1,S_2,\dots,S_{2n-1})$ constitutes a Dyck path of length $2(n-1)$, and these are counted by the Catalan numbers.

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