Maths Olympiad Prep

Library / /55 of 71

Combinatorics Difficulty 5.4 AIME, harder Prove it United States

Problem:

Indecisive Andy starts out at the midpoint of the 1-unit-long segment HT\overline{H T}. He flips 2010 coins. On each flip, if the coin is heads, he moves halfway towards endpoint HH, and if the coin is tails, he moves halfway towards endpoint TT. After his 2010 moves, what is the expected distance between Andy and the midpoint of HT\overline{H T}?

Solution

Solution:

Let Andy's position be xx units from the HH end after 2009 flips. If Andy moves towards the HH end, he ends up at x2\frac{x}{2}, a distance of 1x2\frac{1-x}{2} from the midpoint. If Andy moves towards the TT end, he ends up at 1+x2\frac{1+x}{2}, a distance of x2\frac{x}{2} from the midpoint. His expected distance from the midpoint is then
1x2+x22=14. \frac{\frac{1-x}{2} + \frac{x}{2}}{2} = \frac{1}{4}.
Since this does not depend on xx, 14\frac{1}{4} is the answer.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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