Library / /44 of 62

, 2020

Combinatorics Difficulty 5.8 AIME, harder Prove it United States

Problem:
A fair coin is flipped eight times in a row. Let $p$ be the probability that there is exactly one pair of consecutive flips that are both heads and exactly one pair of consecutive flips that are both tails. If $p=\frac{a}{b}$ , where $a, b$ are relatively prime positive integers, compute $100 a+b$ .

Solution

Solution:
Separate the sequence of coin flips into alternating blocks of heads and tails. Of the blocks of heads, exactly one block has length $2$ , and all other blocks have length $1$ . The same statement applies to blocks of tails. Thus, if there are $k$ blocks in total, there are $k-2$ blocks of length $1$ and $2$ blocks of length $2$ , leading to $k+2$ coins in total. We conclude that $k=6$ , meaning that there are $3$ blocks of heads and $3$ blocks of tails.

The blocks of heads must have lengths $1,1,2$ in some order, and likewise for tails. There are $3^{2}=9$ ways to choose these two orders, and $2$ ways to assemble these blocks into a sequence, depending on whether the first coin flipped is heads or tails. Thus the final probability is $18 / 2^{8} = 9 / 128$ .

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.