Probability that the largest of $X_1,X_2,X_3\sim \operatorname{Uniform}(0,1)$ is greater than the sum of the other two. We have $X_1,X_2,X_3 \sim \operatorname{Uniform}(0,1)$ and we want to find the probability that the largest realized value among them is greater than the sum of the other two. I know that we can make a symmetry argument, define limits, and integrate, but I want to know if there is a way to do it using order statistics so that
$$
3P(X_1>X_2+X_3)=P(X_{(1)}+X_{(2)}< X_{(3)}) =P(X_{(1)}< X_{(3)}-X_{(2)}) 
$$
I tried starting from $$P(X_{(1)}<X_{(3)}-X_{(2)})=F_{X_{(1)}}(x_{(3)}-x_{(2)})=\sum_{i=1}^{3}\binom{3}{i}F(x_{(3)}-x_{(2)})^i[1-F(x_{(3)}-x_{(2)})]^{3-i}$$
I haven't worked through all the algebra of the resulting expression, but it looks to me like I'm going to get something that is in terms of the difference $x_{(3)}-x_{(2)}$ which is good generally, but doesn't give me the straight numerical answer I'm looking for (which I know through direct calculation is 1/2).
Is my above approach using order statistics valid and, if so, why isn't it giving me the answer I want? Thanks!
 A: Suppose the largest of the three values is $X_{(3)}=m$. That means that $X_{(1)}$ and $X_{(2)}$ are both smaller than $m$, which happens with probability $m^2$.
Now we can calculate the probability that the sum $X_{(1)}+X_{(2)}\leq m$ using integrals:
$$\Pr[X_{(1)} + X_{(2)} \leq m] = \int^m_0\int^{m-y}_01\,dx\,dy=\int^m_0(m-y)\,dy=\frac{m^2}{2}.$$
However, this is the probability when you already know what the maximum is, so we need to correct for the probability that $m$ is in fact the maximum:
$$\Pr[X_{(1)} + X_{(2)} \leq m \mid X_{(3)} = m]=\frac{1}{\Pr[X_{(3)} = m]}\cdot \Pr[X_{(1)} + X_{(2)} \leq m]=\frac{1}{m^2}\cdot\frac{m^2}{2}=\frac{1}{2}.$$
That means the probability that the smaller two values do not exceed the maximum is $\frac{1}{2}$, so the requested probability is also $\frac{1}{2}$.
A: I don't understand your equality
$$P(X_{(1)}<X_{(3)}-X_{(2)})=F_{X_{(1)}}(x_{(3)}-x_{(2)})$$
You seem to be plugging the CDF , but that does make much sense to me.
If (say) you have two rv $X,Y$ and you want to calculate $P(X\le Y)$, you can't just write $P(X\le Y) = F_X(y)$ because $y$ means nothing. And you can neither write $P_X(Y)$ because that would be a random variable (and the probability should be a number). What you can write is $P(X\le Y \mid Y=y)=F_X(y)$, but that would hardly lead you anywhere, I think.
