Three ants on a triangle avoiding a shared corner

Question

An equilateral triangle has one ant on each side. Each ant independently walks to one of the two ends of its side, choosing each with probability $\tfrac12$ . What is the probability that no corner ends up with two or more ants?

Short answer

$1/4$ . The ants occupy distinct corners only if they all go the same way around the triangle, all clockwise or all anticlockwise: $2\cdot\left(\tfrac12\right)^3=\tfrac14$ .

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Lessons behind this question

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

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