The acceleration equation needs to include force terms for air drag $F_a$, and runner friction $F_f$ in addition to the gravity term $g \sin(\theta)$ where $\theta$ is the slope of the luge run and $g$ is gravitation.
As Singh pointed out, gravitation exerts a force proportional to mass, so the total acceleration is
$$a = g \sin(\theta) - \frac{F_f + F_a}{m}$$
Runner friction is more or less proportional to mass, so we can replace it with a constant, i.e. the mass of the rider is not a consideration, giving
$$a = g \sin(\theta) - K - \frac{F_a}{m} $$
Air drag depends on frontal surface area, $A$ and the square of velocity $v^2$, so
$$F_a = DAv^2$$ where D is a drag coefficent to account for the rider's aerodynamic smoothness (or lack thereof).
User11865's answer points out that the rider's surface area is proportional to $\sqrt{m}$ and this is what gives heavier riders an advantage, especially at higher speeds. Let's wrap the density and shape of the human body into a constant, say, $B$, and the acceleration equation now looks like
$$a = g \sin(\theta) - K - \frac{DBv^2}{\sqrt{m}}$$
The acceleration lost to air drag is the only term that depends on mass and having more mass makes it smaller.