HMMT 十一月 2013 · 团队赛 · 第 1 题

HMMT November 2013 — Team Round — Problem 1

[ 3 ] Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons, and regular octagons. Tim makes a random sequence F , F , F , F , . . . of figures as follows: 0 1 2 3 • F is a square. 0 • For every positive integer i , F is randomly chosen to be one of the 2 figures distinct from F i i − 1 1 (each chosen with equal probability ). 2 • Tim takes 4 seconds to make squares, 6 to make hexagons, and 8 to make octagons. He makes one figure after another, with no breaks in between. Suppose that exactly 17 seconds after he starts making F , Tim is making a figure with n sides. What 0 is the expected value of n ?

解析

[ 3 ] Tim the Beaver can make three different types of geometrical figures: squares, regular hexagons, and regular octagons. Tim makes a random sequence F , F , F , F , . . . of figures as follows: 0 1 2 3 • F is a square. 0 • For every positive integer i , F is randomly chosen to be one of the 2 figures distinct from F i i − 1 1 (each chosen with equal probability ). 2 • Tim takes 4 seconds to make squares, 6 to make hexagons, and 8 to make octagons. He makes one figure after another, with no breaks in between. Suppose that exactly 17 seconds after he starts making F , Tim is making a figure with n sides. What 0 is the expected value of n ? Answer: 7 We write F = n as shorthand for “the i th figure is an n -sided polygon.” i If F = 8, the F = 6 or F = 4. If F = 6, Tim is making a 6-gon at time 13 (probability contribution 1 2 2 2 1 / 4). If F = 4, F = 6 or F = 8 will take the time 13 mark (1 / 8 contribution each). 2 3 3 If F = 6, F = 8 or F = 4. If F = 8, it takes the 13 mark (1 / 4 contribution). If F = 4, F = 6 or 1 2 2 2 2 3 F = 8 will take the 13 mark (1 / 8 contribution each). 3 1 1 1 1 1 1 Thus, the expected value of the number of sides at time 13 is 0(4) + ( + + )(6) + ( + + )(8) = 7. 4 8 8 8 4 8