Don't gamble!

A gambler has 1 dollar in his pocket. He tosses an unbiased coin unless he has lost all his money or unless the coin is tossed for 5 times. For each head he wins a dollar, and for each tail he loses a dollar. Find the probability that the gambler will end up with $0.

If the answer is in the form a b \frac{a}{b} , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 27.

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1 solution

The gambler's winnings must go up or down 1 dollar after each coin toss unless they have lost all their money. The only possible sequences of dollars on hand that result in them losing all their money are

( 0 ) , ( 2 , 1 , 0 ) , ( 2 , 1 , 2 , 1 , 0 ) (0), (2,1,0), (2,1,2,1,0) and ( 2 , 3 , 2 , 1 , 0 ) (2,3,2,1,0) .

Bottoming out in a sequence of n n coin tosses occurs with probability 1 2 n \dfrac{1}{2^{n}} , so as there is 1 losing sequence involving 1 toss, 1 involving 3 tosses and 2 involving 5 tosses, the probability that the gambler will lose all their money is

1 2 + 1 2 3 + 2 × 1 2 5 = 1 2 + 1 8 + 2 32 = 8 + 2 + 1 16 = 11 16 \dfrac{1}{2} + \dfrac{1}{2^{3}} + 2 \times \dfrac{1}{2^{5}} = \dfrac{1}{2} + \dfrac{1}{8} + \dfrac{2}{32} = \dfrac{8 + 2 + 1}{16} = \dfrac{11}{16} , and thus a + b = 11 + 16 = 27 a + b = 11 + 16 = \boxed{27} .

Comment: The probabilities P ( k ) P(k) of possessing k k dollars after 5 tosses are

P ( 0 ) = 11 16 , P ( 1 ) = P ( 3 ) = P ( 5 ) = 0 , P ( 2 ) = 5 32 , P ( 4 ) = 1 8 , P ( 6 ) = 1 32 P(0) = \dfrac{11}{16}, P(1) = P(3) = P(5) = 0, P(2) = \dfrac{5}{32}, P(4) = \dfrac{1}{8}, P(6) = \dfrac{1}{32} .

Follow-up question: If this game were to continue for potentially an infinite number of tosses, what is the probability that the gambler will eventually lose all their money?

Could you give a hint to follow up question ?

Vishal Yadav - 4 years, 1 month ago

Follow-up question is a one-dimension random walk question. The gambler will lost all his money eventually, but the expectation of times of tosses does not exist.

Alstream Yang - 2 years, 3 months ago

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