Maths Olympiad Prep

Track / Stage 3 / 156 of 260 #636 of 2444

Problem 636

AMC 10/12, early questions
Algebra Difficulty 3.8 Find the answer China Mathematical Competition · China

Two persons roll two dice in turn. Whoever gets the sum number greater than 66 first will win the game. The probability for the person rolling first to win is ________.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Next problem →

Official solution

The probability for rolling two dice to get the sum number greater than 66 is 2136=712\frac{21}{36} = \frac{7}{12}. Therefore, the required probability is
712+(512)2712+(512)4712+=712×1125144=1217. \frac{7}{12} + \left(\frac{5}{12}\right)^2 \frac{7}{12} + \left(\frac{5}{12}\right)^4 \frac{7}{12} + \cdots = \frac{7}{12} \times \frac{1}{1 - \frac{25}{144}} = \frac{12}{17}.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.