Pick's theorem concerns "simple" polygons with vertices in $\mathbb Z^2.$ "Simple" means they have neither self-intersections nor holes. It says $$ A = I + \frac B 2 -1 $$ where $A$ is the area of the polygon, $I$ is the number of points in $\mathbb Z^2$ in the interior, and $B$ is the number of points in $\mathbb Z^2$ on the boundary.
Area is an additive set function: If you chop a polygon into two polygons then its area is the sum of their areas.
In order to make this function additive, we had to subtract $1.$
The gamma function is $$ \Gamma(\alpha) = \int _0^\infty x^{\alpha-1} e^{-x} \, dx, $$ so the measure $f_\alpha(x)\, dx = \dfrac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)} \, dx$ for $x>0$ is a probability measure.
This convolution identity holds: $$ (f_\alpha*f_\beta)(x) = \int_0^x f_\alpha(u) f_\beta(x-u)\, du = f_{\alpha+\beta}(x). $$ In order to make the index in this convolution equal to the sum of the two separate indices, we had to subtract $1$ from $\alpha$ and $\beta$ in the exponent.
Are these two instances of subtraction of $1$ being just the thing needed in order to achieve additivity both instances of some common general proposition?