Two persons each make a single throw with a pair of dice. The probability that their scores are equal is

Required Probability $$\begin{gathered} \text{ There are }36\text{ outcomes for one roll of a pair of dice, so the total number of } \\ \text{ outcomes for the two persons making one throw each is }36\times 36=1296\text{. } \\ \text{ Now let }{\mathrm{a}}_{\mathrm{i}}\text{, with }2\le \mathrm{i}\le 12\text{, be the number of ways to get a sum of }\mathrm{i}\text{ showing on } \\ \text{ the pair of dice when they are rolled. } \\ \text{ Then }{\mathrm{a}}_2={\mathrm{a}}_{12}=1,{\mathrm{a}}_3={\mathrm{a}}_{11}=2,{\mathrm{a}}_4={\mathrm{a}}_{10}=3,{\mathrm{a}}_5={\mathrm{a}}_9=4,{\mathrm{a}}_6={\mathrm{a}}_8=5,{\mathrm{a}}_7=6 \\ \text{ Each player can throw an }\mathrm{i}\text{ in }{\mathrm{a}}_{\mathrm{i}}\text{ ways, so both of them will throw an }\mathrm{i}\text{ is }{\mathrm{a}}_{\mathrm{i}}^2 \mathrm{ways} \\ \text{ Summing over all values of }\mathrm{i}\text{, we see that the number of ways the throws of the } \\ \text{ two persons will be equal is } \\ {\mathrm{a}}_2^2+{\mathrm{a}}_3^2+\dots +{\mathrm{a}}_{12}^2=2\left({\mathrm{a}}_2^2+{\mathrm{a}}_3^2+{\mathrm{a}}_4^2+{\mathrm{a}}_5^2+{\mathrm{a}}_6^2\right)+{\mathrm{a}}_7^2 \\ \mathrm{required} \mathrm{probability} =\frac{1^2+2^2+3^2+4^2+5^2+6^2+5^2+4^2+3^2+2^2+1^2}{{36}^2} \end{gathered}$$