Back where he started after nine random moves

Question

A footballer makes $9$ moves, each independently either $2$ steps forward or $1$ step back with probability $\tfrac12$ each. What is the probability that after the $9$ moves he is back at his starting point?

Short answer

$\tfrac{21}{128}$ . Returning needs $2f=b$ with $f+b=9$ , so exactly 3 forward moves and 6 backward moves. That happens with probability $\binom93/2^9=\tfrac{84}{512}=\tfrac{21}{128}$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Counting and Combinatorics
Sample Spaces, Events, and the Probability Axioms
Conditional Probability, Total Probability, and Bayes' Rule

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