Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:

A fat coin is one which, when tossed, has a 25\frac{2}{5} probability of being heads, 25\frac{2}{5} of being tails, and 15\frac{1}{5} of landing on its edge. Mr. Fat starts at 00 on the real line. Every minute, he tosses a fat coin. If it's heads, he moves left, decreasing his coordinate by 11; if it's tails, he moves right, increasing his coordinate by 11. If the coin lands on its edge, he moves back to 00. If Mr. Fat does this ad infinitum, what fraction of his time will he spend at 00?

Solution

Solution:

For nZn \in \mathbb{Z}, let ana_n be the fraction of the time Mr. Fat spends at nn. By symmetry, an=ana_n = a_{-n} for all nn.

For n>0n > 0, we have
an=25an1+25an+1, a_n = \frac{2}{5} a_{n-1} + \frac{2}{5} a_{n+1},
or
an+1=52anan1. a_{n+1} = \frac{5}{2} a_n - a_{n-1}.
This Fibonacci-like recurrence can be solved explicitly to obtain
an=α2n+β2n a_n = \alpha \cdot 2^{|n|} + \beta \cdot 2^{-|n|}
for all nZn \in \mathbb{Z}. Now we also have
nZan=1, \sum_{n \in \mathbb{Z}} a_n = 1,
so we better have α=0\alpha = 0, so that a0=βa_0 = \beta and a±1=β2a_{\pm 1} = \frac{\beta}{2}.

Now we also have
a0=25a1+25a1+15, a_0 = \frac{2}{5} a_{-1} + \frac{2}{5} a_1 + \frac{1}{5},
so β=13\beta = \frac{1}{3}.

This matches perfectly with nZan=1\sum_{n \in \mathbb{Z}} a_n = 1.

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