For a trading interview assessment I was presented with this problem:
3 traders have a payoff at the end of the day that is uniformly distributed between 0 and 100$.
What is the probability that the Payoff of A is bigger than B and the payoff of B is bigger than C?
MY attempt:
The probability of any value in the uniform distribution is $\frac{1}{101}$.
From experience with past problem I think that the probability that A is bigger than B is $\frac{A+1}{101}$ and B bigger than C is $\frac{B+1}{101}$ which can be written as $\frac{A+1 +1}{101}$
So I thought the correct answer was to multiply those and get $\frac{A^2+3A+2}{101}$
since A should be 1 the probability is $\frac{6}{101}$
Visualize the $1\times 1\times 1$ cube. We want the volume of the region s.t. $z>y>x$. So $$I=\int_{z>y>x}dzdydx=\int_{[0,1]}\int_{[0,z]}\int_{[0,y]}dzdydx=\int_{[0,1]}\int_{[0,z]}ydzdy=\int_{[0,1]}\frac{z^2}{2}dz=\frac{1}{6}$$