Problem:
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let m and n be relatively prime positive integers such that nm is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find m+n.
Solution:
The probability that Dave or Linda rolls a die k times to get the first six is the probability that there are k−1 rolls which are not six followed by one roll of six, which is pk=(65)k−1(61). The probability that Dave will need one roll and Linda will need one or two rolls is then p1(p1+p2). The probability that Dave will need k>1 rolls and Linda will need k−1,k, or k+1 rolls is then pk(pk−1+pk+pk+1). It follows that the desired probability is p1(p1+p2)+∑k=2∞pk(pk−1+pk+pk+1). This is
61⋅(61+65⋅61)+k=2∑∞(65)k−1(61)((65)k−2(61)+(65)k−1(61)+(65)k(61))=(61)⋅(366+365)+k=2∑∞(65)k−1(61)(65)k−2(61)(1+(65)+(65)2)=6311+6491⋅1−(65)265=6311+6391⋅115=63⋅11576=338.
Thus the final answer is 8+33=41.
The problems on this page are the property of the MAA's American Mathematics Competitions