Maths Olympiad Prep

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HMMT November, 2021

Combinatorics Difficulty 4.6 AIME Find the answer United States

Problem:
A counter begins at 00. Then, every second, the counter either increases by 11 or resets back to 00 with equal probability. The expected value of the counter after ten seconds can be written as mn\frac{m}{n}, where m,nm, n are positive integers and gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1. Find 100m+n100m+n.

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

Solution

Solution:
The probability that the counter is equal to kk corresponds to the last kk seconds all being increases by 11 and the second before that being a reset to 00, which happens with probability 2−k−12^{-k-1}. The only contradiction to this is when k=10k=10 and the counter gets there by only counting 11's. Therefore, the expected value is simply the sum of probabilities times the counter, which is
10210+∑k=19k2k+1=(1210+∑k=1912k+1)+(1210+∑k=2912k+1)+…+1210=12+14+…+1210=10231024 \frac{10}{2^{10}} + \sum_{k=1}^{9} \frac{k}{2^{k+1}} = \left(\frac{1}{2^{10}} + \sum_{k=1}^{9} \frac{1}{2^{k+1}}\right) + \left(\frac{1}{2^{10}} + \sum_{k=2}^{9} \frac{1}{2^{k+1}}\right) + \ldots + \frac{1}{2^{10}} = \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^{10}} = \frac{1023}{1024}

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