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For a target corner $(\alpha ,\beta )$ in the Greene–Nijenhuis–Wilf hook walk, put $A_i=h(i,\beta )-1$ and $B_j=h(\alpha ,j)-1$. The probability of terminating there from $(a,b)$ factors as $F_a{G}_b$, where $F_{\alpha }=G_{\beta }=1$, $F_a=A_a^{-1}{\prod }_{i=a+1}^{\alpha -1}(1+A_i^{-1})$ for $a<\alpha$, and $G_b=B_b^{-1}{\prod }_{j=b+1}^{\beta -1}(1+B_j^{-1})$ for $b<\beta$. It is zero when the start is outside the target's northwest rectangle. The factorization uses $h(i,j)-1=A_i+B_j$ and the first-step recurrence.

For a cell $x=(i,j)$ in a Young diagram, its hook of a Young diagram consists of $x$, all cells to its right in row $i$, and all cells below it in column $j$. If ${\lambda }^{\prime }$ is the conjugate partition, its hook length isThe hook-length formula is

In the Greene–Nijenhuis–Wilf hook walk, a walk at a noncorner cell chooses uniformly one of the other $h(i,j)-1$ cells in its hook, and stops at a corner. Moves are strictly rightwards or downwards, so termination is certain. A terminal cell must be a Removable node of a Young diagram. For a chosen corner $(\alpha ,\beta )$, its probability is zero unless the start satisfies $a\le \alpha$ and $b\le \beta$.

For a start in that northwest rectangle, defineThe corner conditions are ${\lambda }_{\alpha }=\beta$ and ${\lambda }_{\beta }^{\prime }=\alpha$, so $A_{\alpha }=B_{\beta }=0$, and all earlier $A_i,B_j$ are positive. DefineandThen the closed conditional probability isIn particular, if both inequalities are strict, this isThe separate definitions of $F_{\alpha },G_{\beta }$ handle starts in the terminal row, in the terminal column, or at the corner itself without division by zero.

To prove the formula, the product definitions give the identitieswith the terminal zero identities interpreted as empty sums. For example, summing the $F$ recurrence backwards gives ${\sum }_{r=i+1}^{\alpha }{F}_r={\prod }_{r=i+1}^{\alpha -1}(1+A_r^{-1})$. AlsoThe first-step equation for the hook walk terminal probability isJumps beyond the target rectangle contribute zero. Substituting $p_{i,j}=F_i{G}_j$ proves this equation, and $p_{\alpha ,\beta }=1$ gives its boundary value. Since every step increases a coordinate, backward recursion determines the probability uniquely. This proves the claimed expression.