Two players, P _ 1 and P _ 2 , play a game against each other. In...

Given,

$P$(draw in $1$ round)$= \dfrac{6}{36} = \dfrac{1}{6}$

$P$(win in $1$ round) $= \dfrac{1}{2}\left(1 - \dfrac{1}{6}\right) = \dfrac{5}{12}$

$P$(loss in $1$ round) $= \dfrac{5}{12}$

Now finding the probability of all we get,$P\left({X}_{2} > {Y}_{2}\right) = P\left(10, 0\right) + P\left(7, 2\right) = \dfrac{5}{12} \times \dfrac{5}{12} + \dfrac{5}{12} \times \dfrac{1}{6} \times 2 = \dfrac{45}{144} = \dfrac{5}{16} \to R$

$P\left({X}_{2} = {Y}_{2}\right) = P\left(5, 5\right) + P\left(4, 4\right) = \dfrac{5}{12} \times \dfrac{5}{12} \times 2 + \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{25 + 2}{72} = \dfrac{3}{8}$

$P\left({X}_{2} \ge {Y}_{2}\right) = \dfrac{5}{16} + \dfrac{3}{8} = \dfrac{11}{16} \to Q$

$P\left({X}_{3} = {Y}_{3}\right) = P\left(6, 6\right) + P\left(7, 7\right) = \dfrac{1}{6 \times 6 \times 6} + \dfrac{5}{12} \times \dfrac{1}{6} \times \dfrac{5}{12} \times 6 = \dfrac{2}{432} + \dfrac{75}{432} = \dfrac{77}{432} \to T$

$P\left({X}_{3} > {Y}_{3}\right) = \dfrac{1}{2}\left(1 - \dfrac{77}{432}\right) = \dfrac{355}{864} \to S$