Exact rank-sum tail one above the minimum ($\mathrm{Pr}\{W\le m(m+1)/2+1\}=2/\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$)

For two independent continuous samples of positive sizes $m,n$ under a common-distribution null, each $m$-subset of ranks $1,\dots ,m+n$ is equally likely. The smallest rank sum is $m(m+1)/2$, attained only by $\{1,\dots ,m\}$. The only subset with rank sum one larger is $\{1,\dots ,m-1,m+1\}$. Hence the exact lower tail at that threshold is $2/\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$. This follows directly by observing that any other change increases the sum by at least two.