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**SlipEternal** P(A wins|A rolls 10) = P(B and C roll lower than 10) + P(B and C both roll 10)*P(A wins next round vs all three) + P(B or C rolls a 10, the other rolls lower)*P(A wins next round vs either B or C)

There are six ways to roll a 10 or above: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), so there is a $\displaystyle \dfrac{30}{36} = \dfrac{5}{6}$ chance of rolling less than 10.

P(B and C both roll lower than 10) = $\displaystyle \dfrac{5}{6}\cdot \dfrac{5}{6} = \dfrac{25}{36}$

P(B and C both roll 10) = $\displaystyle \dfrac{3}{36}\cdot \dfrac{3}{36} = \dfrac{1}{144}$

In general, all three players start with equal chances of winning, so $\displaystyle P\left(A\text{ wins next round vs. both }B\text{ and }C\right) = \dfrac{1}{3}$

P(B or C rolls a 10 and the other rolls lower) = P(B rolls 10, C rolls lower)+P(C rolls 10, B rolls lower) = $\displaystyle 2\cdot \dfrac{1}{12}\cdot \dfrac{5}{6} = \dfrac{5}{36}$

In general, if there are only two players left, they have equal chances, as well, so $\displaystyle P\left(A\text{ wins next round vs. either }B\text{ or }C\right) = \dfrac{1}{2}$

Hence, $\displaystyle P\left(A\text{ wins}|A\text{ rolls }10\right) = \dfrac{25}{36} + \dfrac{1}{144}\cdot \dfrac{1}{3} + \dfrac{5}{36}\cdot \dfrac{1}{2} = \dfrac{331}{432}$