This is heuristic so it may not be what you are asking for. There is a probabilistic argument for the hook length formula. Let $\lambda$ be a partition: then the number of standard tableaux of shape $\lambda$ is $$ \frac{n!}{\prod_{(i,j)\in\lambda}h(i,j)} $$ Here $(i,j)$ runs over the coordinates of the boxes of the Ferrers diagram and $h(i,j)$ is the hook length of the box $(i,j)$.

Then the usual definition of a standard tableau is that it is a filling of the boxes by $1,2,\ldots ,n$ such that the entries increase along each row and down each column. However this is equivalent to saying that for all boxes the entry in $(i,j)$ is the smallest entry in the hook for $(i,j)$.

Now there are $n!$ ways to fill the boxes and the probability of the condition holding for box $(i,j)$ is $\frac1{h(i,j)}$. If these probabilities were independent (which they are not!) this would prove the hook length formula.