Question

David and Carol play a game as follows: David throws a die, and Carol tosses a coin. If die falls "six", David wins. If the die does not fall "six" and the coin does fall heads, Carol wins. If neither the die falls "six" nor the coin falls heads, the foregoing is to be repeated as many times as necessary to determine a winner. Whant is the probability that David wins?

Answer is 2/7

Answer #1

p(getting 6 on die) = 1/6

P(not getting 6 on die) = 1 - 1/6 = 5/6

P(getting heads on coin toss) = P(not getting heads) = 1/2

Thus, P(david win) = P(win in 1st attempt) + P(win in 2nd attempt) and so on

Thus, P(david win) = 1/6 + 5/6*1/2*1/6 + 5/6*1/2*5/6*1/2*1/6 + 5/6*1/2*5/6*1/2*5/6*1/2*1/6 + ..

= 1/6 + 5/12*1/6 + (5/12)^2*1/6 + (5/12)^3*1/6 + ....

This is GP with a = 1/6, r = 5/12

Sum of GP = a/(1-r) when r < 1

Thus,

P(david win) = 1/6 / (1- (5/12))

= 1/6 * 12/7

= 2/7

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