Alice and Bob spin a roulette wheel alternately, each attempting to get one of s

Question

Alice and Bob spin a roulette wheel alternately, each attempting to get one of several numbers which they specify at the beginning of the game. Suppose that Alice spins first and has probability P1 of spinning one of her numbers any time she spins. Suppose that Bob spins second and has probability P2 of spinning one of his numbers any time he spins. The results of each spin are assumed to be independent. The first person who succeeds in spinning one of their numbers wins the game. (a) What is the probability that Alice wins the game? What is the probability that Bob wins? (b) If P1 < 1/2, what is a value of P2 that makes each player's probability of winning the game equal? Please be clear with your solutions.

Explanation / Answer

a) the probability that alice wins in first round = probability she gets the number she wants = P1 in her second chance = (1-P1)(1-P2)(P1) in her third chance = (1-P1)^2(1-P2)^2 (P1) so final probability that she wins = P1[(1+(1-P1)(1-P2) + ((1-P1)(1-P1))^2....] = P1 ( 1/1-(1-P1)(1-P2)) = P1/(1-1+P1+P2-P1P2) = P1/(P1+P2+P1P2) probability that Bob wins , for bob to win alice should not get any of her number, probability for that = 1-P1 probability for bob getting his number = P2 so probability that bob wins in first round= (1-P1)(P2) in second round = (1-P1)^2(1-P2)(P2) so probability that bob wins = (1-P1)(P2)[(1+ (1-P1)(1-P2) + (1-P1)^2(1-P2)^2....)] = (1-P1)(P2)/(1-(1-P1)(1-P2)) = (1-P1)(P2)/(P1+P2+P1P2) b)P1

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