What is $P(X_1>X_2+X_3)$ if $X_i$s are i.i.d uniform $(0,1)$ variables?

Question

Let $X_1,X_2$ and $X_3$ be iid $U(0,1)$ random variables. Then $P(X_1>X_2+X_3)$ equals?

What I think may be correct $$P(X_1>X_2+X_3) =\int\int \int _{x_1>x_2+x_3} 1.1.1\,dx_1\,dx_2\,dx_3=\int^1_0 \int^x_0 \,dx \,dx_1\,,$$ where $x_2+x_3=x=\frac{1}{2}$ .

Answer 1

We may directly compute the density of $Y:=X_2+X_3$ by convolution: $$ (f_{X_2}\star f_{X_3})(t) = \begin{cases} \int_0^t \ \mathsf dt = t,& 0\leqslant t<1\\ \int_{t-1}^1 \ \mathsf dt = 2-t,& 1\leqslant t\leqslant 2. \end{cases} $$ We can then compute \begin{align} \mathbb P(X_1>Y) &= \mathbb P(X_1>Y,0\leqslant Y <1) + \mathbb P(X_1>Y,1\leqslant Y <2)\\ &= \mathbb P(X_1>Y,0\leqslant Y <1)\\ &= \int_0^1 \int_t^1 t\ \mathsf ds\ \mathsf dt\\ &= \frac16. \end{align}

Answer 2

The correct value of the triple integral is $\int_0^{1}\int_0^{1-x_3} \int_{x_2+x_3} ^{1} dx_1dx_2dx$ which evaluates to $\frac 1 6$.