Seven-slot spinner: two fives before a four or six

Question

A spinner lands on one of the numbers 1 to 7, each equally likely, on every spin. You keep spinning and win if 5 comes up for the second time before either 4 or 6 has appeared at all. What is your probability of winning?

Short answer

$1/9$ . Only spins showing 4, 5 or 6 matter, and among those a 5 has probability $\tfrac13$ . You need the first two relevant spins both to be 5, giving $\left(\tfrac13\right)^2$ .

Adapted from OpenQuant; question and answer rewritten by SelfMadeQuant.com.

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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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