Run of three ones in five die rolls

Question

A fair die is thrown five times and the sequence of results noted. What is the probability that somewhere in the sequence three or more $1$ s appear in a row?

Short answer

$\tfrac{1}{81}$ . Classify by the longest run of ones. A run of exactly three starting at roll 1, 2 or 3 gives $30+25+30=85$ sequences (a non-one must border the run); runs of exactly four give $5+5=10$ ; all five ones gives $1$ . That is $96$ of $6^5=7776$ sequences, i.e. $\tfrac1{81}$ .

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

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Lessons behind this question

Sample Spaces, Events, and the Probability Axioms
Counting and Combinatorics
Expectation, Mean, and Variance

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