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