= Exact rank-sum tail one above the minimum
{title2=$\Pr\{W\le m(m+1)/2+1\}=2/\binom{m+n}{m}$}
For two independent continuous samples of positive sizes $m,n$ under a common-distribution null, each $m$-subset of ranks $1,\ldots,m+n$ is equally likely. The smallest rank sum is $m(m+1)/2$, attained only by $\{1,\ldots,m\}$. The only subset with rank sum one larger is $\{1,\ldots,m-1,m+1\}$. Hence the exact lower tail at that threshold is $2/\binom{m+n}{m}$. This follows directly by observing that any other change increases the sum by at least two.
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