Distribution of the difference of two uniformly distributed variables subject to a condition! Let $A$, $B$ and $C$ be three independent uniformly distributed random variables on $(0,1)$. The variables are admissible if and only if $a<b<c$.
I want to find the distribution of $B-C$ given that the triple is admissible.
Thanks.
 A: My answer does not give you that distribution but reaches you the possibility to find it yourself. I presume that $A$, $B$ and $C$ are independent. This should be mentioned in your question. Note that $P\left[A<B<C\right]=\frac{1}{6}$. If I understand you correctly then you are searching for:$$P\left[B-C\leq x\mid A<B<C\right]=6P\left[B-C\leq x\wedge A<B<C\right]=$$
$$6\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left[b-c\leq x\wedge a<b<c\right]dadbdc$$ 
where $x$ is fixed and $\left[b-c\leq x\wedge a<b<c\right]$ denotes
the characteristic function on $\left(0,1\right)^{3}$ of set $\left\{ \left(a,b,c\right)\in\left(0,1\right)^{3}\mid b-c\leq x\wedge a<b<c\right\} $.
A: Given: random variables $A$, $B$ and $C$ have joint pdf $f(a,b,c)$:

(source: tri.org.au) 
Let $Z = B-C$. Then, the cdf of $Z$ is $P(Z<z) = P(B-C<z)$:

(source: tri.org.au) 
Take the derivative of the cdf wrt $z$ to yield the desired pdf, say $g(z)$:
$$g(z) = 3(1+z)^2   \qquad \text{ for } -1<z<0$$
All done.
Notes


*

*The Prob function used above is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors.

A: $A,B$ and $C$ are independent uniformly distributed variables. We want to find the distribution of $B-C$ given that $A<B<C$. Note that $P[A<B<C]=\frac{1}{6}$. Call $F_{B-C}$ the distribution function of $B-C$.
$F_{B-C}(z)=P[B-C<=z|A<B<C]=\frac{P[B-C\leq z \wedge A<B<C]}{Prob[A<B<C]}=6P[B-C\leq z \wedge A<B<C]$.
Now we look closer at the condition $B-C<=z \wedge A<B<C$. For a given triple $(a,b,c)$, these three conditions must be satisfied (1) $0<a<1+z$ (2) $a<b<1+z$ (3) $b-z<c<1$.
If follows $F_{B-C}(z)=6 \int_{0}^{1+z} \int_{a}^{1+z} \int_{b-z}^{1} dc db da =(1+z)^3$. This is only valid if $-1<z<0$.
The density function is therefore $\partial F_{B-C}(z)/\partial z= 3(1+z)^2$ for $-1<z<0$ and zero otherwise.
