Find probability of $X_1 < X_2 < X_3 $ where $X_i$ is a random variable I have encountered this problem in the book Probability and Random Processes by Grimmett and Stirzaker.  The problem is: 
Let $X_1, X_2, X_3$ be independent (discrete) random variables taking values in the positive integers and having mass functions given by $\mathbb{P}(X_i=x)=(1-p_i)p_i^{x-1}$ for $x=1, 2, ....$ and $i=1,2,3$.
Show that  $\mathbb{P}(X_1<X_2<X_3)=\dfrac{(1-p_1)(1-p_2)p_2p_3^3}{(1-p_2p_3)(1-p_1p_2p_3)}$
I actually find this kind of problems too vague, because I know how to calculate $\mathbb{P}(X\leq x)$ but I can't calculate the probability of (in)equality of different random variables $X, Y$ like $\mathbb{P}(X<Y)$. Is there a method for that?
What I tried so far is to write it down as $\mathbb{P}(X_1<X_2<X_3)=\mathbb{P}(X_1<X_2, X_2<X_3)=\mathbb{P}(X_1<X_2)\mathbb{P}(X_2<X_3)$. But then I don't know how to go further...
I apperciate your help and hints. Thanks!
 A: We use this law https://en.wikipedia.org/wiki/Law_of_total_probability, and the fact that the variables are mutually independent
$P(X_1<X_2<X_3)=\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{P(X_1<X_2<X_3|X_2=j,X_1=i)}P(X_2=j,X_1=i)}$
$P(X_1<X_2<X_3)=\sum_{i=1}^{\infty}{\sum_{j=i+1}^{\infty}{P(j<X_3|X_2=j,X_1=i)}P(X_2=j)P(X_1=i)}$
By independence, $P(j<X_3|X_2=j,X_1=i)=P(X_3>j)$, and also by the law of total probability,
$P(X_3>j)=\sum_{k=j+1}^{\infty}{P(X_3=k)}$
So we have 
$P(X_1<X_2<X_3)=\sum_{i=1}^{\infty}{\sum_{j=i+1}^{\infty}{\sum_{k=j+1}^{\infty}{P(X_3=k)}}P(X_2=j)P(X_1=i)}$
$=(1-p_1)(1-p_2)(1-p_3)\sum_{i=1}^{\infty}p_{1}^{}{\sum_{j=i+1}^{\infty}p_{2}^{j-1}{\sum_{k=j+1}^{\infty}{p_{3}^{k-1}}}}$
Using the fact that $\sum_{j=i+1}^{\infty}{u^{j-1}}=\frac{u^{i}}{1-u}$, you can conclude
A: For two random variables the answer is:
\begin{eqnarray*}
\mathbb{P}(X_2<X_3)&=&\sum_{1\leq j<k} (1-p_2)(1-p_3)p^{j-1}_2 p_3^{k-1}\\[8pt]
&=&\sum_{1\leq j}(1-p_2)(p_2p_3)^{j-1}\\[8pt]
&=&{1-p_2\over 1-p_2p_3}.
\end{eqnarray*}
I will leave the rest to you.
