Just another die game

Probability Level 3

Ari and Brian play a sequence of independent games. Ari throws a die first and wins on a " six ." If he fails, Brian throws and wins on a " five " or " six ." If he fails, Ari throws and wins on a " four ," " five ," or " six ." And so on. The probability of Ari winning the sequence is a b \color{#3D99F6}{\dfrac{a}{b}} , where a \large\color{#3D99F6}{a} and b \large\color{#3D99F6}{b} are positive coprime integers. Calculate a + b \large\color{#3D99F6}{a+b} .


The answer is 493.

Tunk-Fey Ariawan by Tunk-Fey Ariawan , 35, Indonesia

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1 solution

Santosh Yadav
Aug 31, 2014

Probability of winning in first chance is=1÷6 And probability of winning in 3rd chance is =(5/6)(2/6)(3/6) Probability of winning in 5th chance is=(5/6)(4/6)(3/6)(2/6)(5/6) So probability of winning is sum of all 3 conditions=(169/324) a=169 b=324 And=493

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Shouldn't the probability of winning in the third chance be p(Ari fails first throw) times p(Brian fails second throw) times p(Ari wins third throw), or ( 5 / 6 ) ( 4 / 6 ) ( 3 / 6 ) (5/6)(4/6)(3/6) ? The second fraction differs from the above answer. The total answer that I found, multiplying together the following fractions, is ( 1 / 6 ) × ( 5 / 18 ) × ( 25 / 324 ) = 125 / 34992 (1/6)\times(5/18)\times(25/324)=125/34992 with a+b=35117

Clayton Wu - 6 years, 1 month ago

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