Probability $a+b+c>d+e$ for $a,b,c,d,e$ uniform in $[0,1]$ Let $a,b,c,d,e$ be drawn independently and uniformly at random from $[0,1]$. What is the probability that $a+b+c>d+e$?
Considering a unit hypercube, we want the volume "below" the plane $a+b+c-d-e=0$. The corners that are "above" the plane are $$(0,0,0,1,0),(0,0,0,0,1),(0,0,0,1,1),(1,0,0,1,1),(0,1,0,1,1),(0,0,1,1,1).$$ How can we compute the volume of this part of the hypercube?
 A: Notice if $d, e$ are random variables uniform over $[0,1]$, so do $\tilde{d} \stackrel{def}{=} 1 - d\;$ and $\;\tilde{e}\stackrel{def}{=} 1 - e$.
We have
$$a + b + c > d + e \quad\iff\quad a + b + c + \tilde{d} + \tilde{e} > 2$$
For any $p = (x_1,x_2,x_3,x_4,x_5) \in \mathbb{R}^5$, let $\;\ell(p) = x_1 + x_2 + x_3 + x_4 + x_5$.
The probability $\mathcal{P}$ we seek is equal to:
$$\mathcal{P} = \verb/Volume/\big\{\; p \in [0,1]^5 : \ell(p) > 2 \;\big\}
= 1 - \verb/Volume/\big\{\; p \in [0,1]^5 : \ell(p) \le 2 \;\big\}\tag{*1}$$
Let $e_1, e_2, \ldots, e_5$ be the standard bases of $\mathbb{R}^5$ and $\Delta_p(\lambda)$ be the simplex 
$$\big\{\; q \in \mathbb{R}^5 : q_i \ge p_i, i = 1\ldots5\;\text{ and }\; \ell(q-p) \le \lambda \;\big\}$$
The polytope on the right of $(*1)$ can be constructed by
cutting 5 small simplices based at $e_i$ from the simplex $\Delta_0(2)$. More precisely,
up to a set of measure zero, we have:
$$\big\{\; p \in [0,1]^5 : \ell(p) \le 2 \;\big\}
= [0,1]^5 \cap \Delta_0(2)
\approx \Delta_0(2) \setminus \left( \bigcup_{i=1}^5 \Delta_{e_i}(1) \right)$$
This leads to$\color{blue}{^{[1]}}$
$$\mathcal{P}
= 1 - \frac{2^5}{5!} + 5 \left(\frac{1^5}{5!}\right)
= 1 - \frac{32 - 5}{120} = \frac{31}{40}$$
Notes


*

*$\color{blue}{[1]}$ In general, in any dimension, the volume of the simplex  $\Delta_0(\lambda)$ inside the unit hyper-cube is given by following formula:
$$\verb/Volume/\big( [0,1]^n \cap \Delta_0(\lambda) \big) = \sum_{k=0}^{\lfloor\lambda\rfloor} (-1)^k \binom{n}{k} \frac{(\lambda-k)^n}{n!}$$
One can prove this formula using inclusion exclusion principle. Look at my answer for a similar question in $\mathbb{R}^3$,
it has more explanation of the underlying ideas.

A: $$\int_0^1 da \int_0^1 db \int_0^1 dc \int_0^{{\rm min}(1,a+b+c)} dd \int_0^{{\rm min}(1,a+b+c-d)} de = \frac{31}{40}=0.775$$
