Actually, it doesn't. If a black hole and a star have the same mass, their gravity is the same. As the formula for the attraction (gravity) between two objects is given as $F=\dfrac{GMm}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole/star, $m$ is the mass of the smaller object, and $r$ is the distance between the two. Even though a black hole is tiny and a star is much, much bigger than a black hole, the radius of each object does not matter. It only depends on mass and radiusdistance.
The reason why it seems like the gravitational pull is stronger is because the black hole indeed is much smaller than the star with the same mass. If you had a $1 M_\odot$ star and a black hole of the same mass, at distances less than $1 R_\odot$ would mean that you would be inside the star, and still outside the black hole. Therefore, you would feel gravity from the black hole and accelerate toward it, while (assuming you weren't vaporized by the Sun's internal temperatures, you would experience drag from the hydrogen gas, as well as (even if the gas was essentially dragless) more mass being concentrated in the other direction (away from your vector of falling) as a star is not a point-like particle, its mass is distributed throughout the star.
