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Question Number 179542 by cortano1 last updated on 30/Oct/22

A and B play a game in which they alternately toss a pair of dice. The one who is first to get a total of 7 wins the game . Find probability that the one who tosses second will win the game

$$\:\:\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{play}\:\mathrm{a}\:\mathrm{game}\:\mathrm{in}\:\mathrm{which}\:\mathrm{they} \\ $$$$\:\:\mathrm{alternately}\:\mathrm{toss}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{dice}.\:\mathrm{The} \\ $$$$\:\:\mathrm{one}\:\mathrm{who}\:\mathrm{is}\:\mathrm{first}\:\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:\mathrm{7}\:\mathrm{wins} \\ $$$$\:\:\mathrm{the}\:\mathrm{game}\:.\:\mathrm{Find}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{one} \\ $$$$\:\:\mathrm{who}\:\mathrm{tosses}\:\mathrm{second}\:\mathrm{will}\:\mathrm{win}\:\mathrm{the}\:\mathrm{game} \\ $$$$ \\ $$

Answered by Frix last updated on 30/Oct/22

the probability to get 7 with 2 dices is (1/6) ⇒ P(1^(st) player has no 7)=(5/6) P(2^(nd) player has 7)=(1/6) P(2^(nd) player wins in 1^(st) round)=(5/(36)) P_1 =P(1^(st) player wins in n^(th) round)=(1/6)((5/6))^(2n−2) P_2 =P(2^(nd) player wins in n^(th) round)=(1/6)((5/6))^(2n−1)

$$\mathrm{the}\:\mathrm{probability}\:\mathrm{to}\:\mathrm{get}\:\mathrm{7}\:\mathrm{with}\:\mathrm{2}\:\mathrm{dices}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow \\ $$$${P}\left(\mathrm{1}^{\mathrm{st}} \:\mathrm{player}\:\mathrm{has}\:\mathrm{no}\:\mathrm{7}\right)=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${P}\left(\mathrm{2}^{\mathrm{nd}} \:\mathrm{player}\:\mathrm{has}\:\mathrm{7}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${P}\left(\mathrm{2}^{\mathrm{nd}} \:\mathrm{player}\:\mathrm{wins}\:\mathrm{in}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{round}\right)=\frac{\mathrm{5}}{\mathrm{36}} \\ $$$$ \\ $$$${P}_{\mathrm{1}} ={P}\left(\mathrm{1}^{\mathrm{st}} \:\mathrm{player}\:\mathrm{wins}\:\mathrm{in}\:{n}^{\mathrm{th}} \:\mathrm{round}\right)=\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}{n}−\mathrm{2}} \\ $$$${P}_{\mathrm{2}} ={P}\left(\mathrm{2}^{\mathrm{nd}} \:\mathrm{player}\:\mathrm{wins}\:\mathrm{in}\:{n}^{\mathrm{th}} \:\mathrm{round}\right)=\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}{n}−\mathrm{1}} \\ $$

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