A B take turns in throwing two dice The first to throw a sum 1o,being awarded Show that - Maths - Probability

Question

A & B take turns in throwing two dice The first to throw a
sum 1o,being awarded Show that if A has the first throw, their
chances of winning is at the ratio of 12:11

Chetna Khanna · Accepted Answer

Let E = person A gets a sum of 10 F = Person B gets a sum of 10 Sum 10 is obtained in the throws (4, 6), (5, 5), (6, 4), $∴ P (E) = \frac{3}{36} = \frac{1}{12} P (\bar{E}) = 1 - \frac{1}{12} = \frac{11}{12}$ Similary, $P (F) = \frac{1}{12} and P (\bar{F}) = \frac{11}{12}$ Now, A will win in the 1st or 3rd or 5th or throws So, probability of winning of A $= P [E \cup (\bar{E} \cap \bar{F} \cap E) \cup (\bar{E} \cap \bar{F} \cap \bar{E} \cap \bar{F} \cap E) \cup] = P (E) + P (\bar{E} \cap \bar{F} \cap E) + (\bar{E} \cap \bar{F} \cap \bar{E} \cap \bar{F} \cap E) + = \frac{1}{12} + (\frac{11}{12})^{2} \cdot \frac{1}{12} + (\frac{11}{12})^{4} \cdot \frac{1}{12} + = \frac{1}{12} [1 + (\frac{11}{12})^{2} + (\frac{11}{12})^{4} +] = \frac{1}{12} \cdot [\frac{1}{1 - (\frac{11}{12})^{2}}] = \frac{1}{12} \cdot (\frac{144}{144 - 121}) = \frac{1}{12} \times \frac{144}{23} = \frac{12}{23}$ Probability of winning of B = 1 â€“ Probability of winning of A $= 1 - \frac{12}{23} = \frac{23 - 12}{23} = \frac{11}{23}$ P (Winning of A) : P (Winning of B) $= \frac{12}{13} : \frac{11}{23} = 12 : 11$

A & B take turns in throwing two dice.The first to throw a sum 1o,being awarded.Show that if A has the first throw, their chances of winning is at the ratio of 12:11.