# A fair die is rolled 5 times resulting in outcomes $X_i$, with $i=1,\dots,5$. Find $P(X_2+X_3<X_3+X_4)$

I was looking at the solutions to this problem:

A fair die is rolled 5 times resulting in outcomes $X_i$, with $i=1,\dots,5$.

Find $P(X_2+X_3<X_3+X_4)$.

Why does that probability equate to $P(X_3<X_4)$? Shouldn't it be $P(X_2<X_4)$?

Note that if $i\not=j$ then $P(X_i<X_j)$ are all equal to $$\frac{5+4+3+2+1}{36} =\frac{15}{36}=\frac{5}{12},$$ or $$\frac{1-\frac{6}{36}}{2}=\frac{15}{36}=\frac{5}{12}.$$
$\newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{P}\pars{X_{2} + X_{3} < X_{3} + X_{4}} & = {1 \over 6}\sum_{x_{1} = 1}^{6}{1 \over 6}\sum_{x_{2} = 1}^{6} {1 \over 6}\sum_{x_{3} = 1}^{6}{1 \over 6}\sum_{x_{4} = 1}^{6} {1 \over 6}\sum_{x_{5} = 1}^{6}\bracks{x_{2} + x_{3} < x_{3} + x_{4}} \\[5mm] & = {1 \over 6^{2}}\sum_{x_{2} = 1}^{6}\sum_{x_{4} = 1}^{6}\bracks{x_{2} < x_{4}} = {1 \over 6^{2}}\sum_{x_{2} = 1}^{6}\sum_{x_{4} = x_{2} + 1}^{6}1 = {1 \over 6^{2}}\sum_{x_{2} = 1}^{6}\pars{6 - x_{2}} \\[5mm] & = {1 \over 6^{2}}\bracks{6 \times 6 - {6\pars{6 + 1} \over 2}} = \bbox[8px,border:1px groove navy]{5 \over 12} \end{align}