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

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

Answered by Frix last updated on 30/Oct/22

$$\mathrm{the}\mathrm{probability}\mathrm{to}\mathrm{get}7\mathrm{with}2\mathrm{dices}\mathrm{is}\frac{1}{6}$$$$\Rightarrow$$$$P\left(1^{\mathrm{st}}\mathrm{player}\mathrm{has}\mathrm{no}7\right)=\frac{5}{6}$$$$P\left(2^{\mathrm{nd}}\mathrm{player}\mathrm{has}7\right)=\frac{1}{6}$$$$P\left(2^{\mathrm{nd}}\mathrm{player}\mathrm{wins}\mathrm{in}{1}^{\mathrm{st}}\mathrm{round}\right)=\frac{5}{36}$$$$$$$$P_1=P\left(1^{\mathrm{st}}\mathrm{player}\mathrm{wins}\mathrm{in}{n}^{\mathrm{th}}\mathrm{round}\right)=\frac{1}{6}{\left(\frac{5}{6}\right)}^{2n-2}$$$$P_2=P\left(2^{\mathrm{nd}}\mathrm{player}\mathrm{wins}\mathrm{in}{n}^{\mathrm{th}}\mathrm{round}\right)=\frac{1}{6}{\left(\frac{5}{6}\right)}^{2n-1}$$

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