Let $X_1, X_2, Y_1, Y_2, Z_1, Z_2$ be non-negative random variables which have following constraints.
$X_1+Y_1+Z_1=1$
$X_2+Y_2+Z_2=1$
$(X_i,Y_i,Z_i)$ is uniformly distributed. It means, the point of $(X_i,Y_i,Z_i)$ is uniformly distributed on the surface of the triangle with the vertex of (1,0,0)(0,1,0)(0,0,1).
Surely $(X_1,Y_1,Z_1)$ and $(X_2,Y_2,Z_2)$ are independent.
let: $X=X_1 + X_2 $, $Y=Y_1 + Y_2 $, $Z=Z_1 + Z_2 $
In this case, the probability of $\{X$ is bigger than $Y$ and $Z$ both$\}$ would be $\dfrac{1}{3}$.
My question is:
What is the probability of "$c+X$ is bigger than $Y$ and $Z$" when $c$ is a constant"?
For example: what is $\mathbb{P}\left[0.2+X >\max\{Y,Z\}\right]$?
Best Answer
Letting $f(y_1,z_1,y_2,z_2)={\large\chi}_{\ c+(1-y_1-z_1)+(1-y_2-z_2)\ > \ \max(y_1+y_2,z_1+z_2)}\ $, the probability is $$ 4 \int_{y_1=0}^1\int_{z_1=0}^{1-y_1} \int_{y_2=0}^1\int_{z_2=0}^{1-y_2} f(y_1,z_1,y_2,z_2)\ dy_1\ dz_1 \ dy_2 \ dz_2 $$ The $4$ arises because the probability measures are $2\ dy_1 \ dz_1$ and $2\ dy_2 \ dz_2$ to make $\int_{y_1=0}^1\int_{z_1=0}^{1-y_1} 2\ dy_1 \ dz_1= \int_{y_2=0}^1\int_{z_2=0}^{1-y_2} 2\ dy_2 \ dz_2=1$.
According to Mathematica, the result is $$\frac{36+64c+24c^2-24c^3-c^4}{108}\ \text{ if }\ 0<c<1$$ and for $c=1/5$ this is $30979/67500$.