probability 2 numbers are larger than a third iid drawn from uniform I draw three numbers $ a $, $ b $, and $ c $ independently from a uniform distribution $[0, \theta] $. I am told which one is the smallest. What is the probability that the average of the two larger numbers is $ x $ times larger than the smallest?  
 A: To get you underway: 
Without loss of generality, scale everything down and assume the first number is the smaller.
Let $A,B,C$ be iid $\mathcal {U}[0;1]$ and be given $A$ is the least order statistic.  Find: $$\mathsf P(B+C\leqslant 2x~A\mid A<B, A<C) \\=\\ \dfrac{\mathsf P(A<B<1\,, A<C<1\,, 2A<B+C<\min(2,2xA))}{\mathsf P(A<B, A<C)}$$
Can you proceed now?
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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I'll assume, of course, that $\ds{\theta, x > 0}$. Hereafter, I'use Iverson Brackets $\ds{\bracks{\cdots}}$ which are convenient to handle restrictions.

In general I can assume $\ds{a < b\,,\ a < c}$ and evaluate, $\ds{\ \underline{for\ example},\ }$ the 'case' $\ds{{b + c \over 2} > xa}$,
with $\ds{x > 0}$, since the original post $\underline{is\ not\ quite\ clear}$ about it. Namely,
\begin{align}
&\mbox{Note that}\quad
{1 \over \theta}\int_{0}^{\theta}{1 \over \theta}
\int_{0}^{\theta}{1 \over \theta}\int_{0}^{\theta}\bracks{a < b}\bracks{a < c}
\bracks{{b + c \over 2} > xa}\,\dd a\,\dd b\,\dd c
\\[5mm] & \stackrel{a/\theta\ \mapsto\ a\,,b/\theta\ \mapsto\ b\,,c/\theta\ \mapsto\ c}{=}\,\,\,
\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\bracks{a < b}\bracks{a < c}
\bracks{{b + c \over 2} > xa}\,\dd a\,\dd b\,\dd c
\end{align}
which shows that the result is $\ds{\theta}$-independent.

Then,
\begin{align}
&\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\bracks{a < b}\bracks{a < c}
\bracks{{b + c \over 2} > xa}\,\dd a\,\dd b\,\dd c
\\[5mm] = &\
\int_{0}^{1}\int_{a}^{1}\int_{a}^{1}\bracks{b + c > 2xa}
\,\dd a\,\dd b\,\dd c
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1 - a}\int_{0}^{1 - a}
\bracks{b + c > 2\pars{x - 1}a}\,\dd a\,\dd b\,\dd c
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\bracks{b + c > 2\pars{x - 1}\,{a \over 1 - a}}\pars{1 - a}^{2}
\,\dd a\,\dd b\,\dd c
\\[5mm] \stackrel{a/\pars{1 - a}\ \mapsto\ a}{=}\,\,\,&\
\int_{0}^{1}\int_{0}^{1}\int_{0}^{\infty}
{\bracks{b + c > 2\pars{x - 1}a} \over \pars{a + 1}^{4}}\,\dd a\,\dd b\,\dd c
\\[5mm] = &\
\bracks{x < 1}\int_{0}^{\infty}{\dd a \over \pars{a + 1}^{4}} +
\bracks{x > 1}\int_{0}^{1}\int_{0}^{1}
\int_{0}^{\pars{b + c}/\bracks{2\pars{x - 1}}}{\dd a \over \pars{a + 1}^{4}}
\,\dd b\,\dd c
\\[5mm] = &\
{1 \over 3}\bracks{x < 1} +
{1 \over 3}\bracks{x > 1}\ \underbrace{\int_{0}^{1}\int_{0}^{1}\,
{\dd b\,\dd c \over \braces{1 + \pars{b + c}/\bracks{2\pars{x - 1}}}^{3}}}
_{\ds{3x - 2 \over x\pars{2x - 1}}}
\\[5mm] = &\
\bbox[20px,#ffe,border:1px dotted navy]{\left\{\begin{array}{lcrcl}
\ds{1 \over 3} & \mbox{if} & \ds{x} & \ds{\leq} & \ds{1}
\\[2mm]
\ds{{1 \over 3}\,{3x - 2 \over x\pars{2x - 1}}} & \mbox{if} & \ds{x} & \ds{>}
& \ds{1}
\end{array}\right.}
\end{align}

