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For $k\in \{0,1,\dots,n\}$ denote by $\xi_k$ the random variable which equals 1 if $(k,k)$ lies on the path and 0 otherwise. Then the number of crossings of the diagonal equals $\sum \xi_k$. By linearity of expectation, we see that the expected number of crossings equals $\sum \mathbb{E}\xi_k=\binom{2n}n^{-1}\sum_k \binom{2k}k\binom{2(n-k)}{n-k}=4^n/\binom{2n}n$ by a well-known identity $\sum \binom{2k}k\binom{2(n-k)}{n-k}=4^n$ (see, for example, Stanley's Enumerative Combinatorics vol.1 exercise I.3(c).)