Parity of the running total until a six appears

Question

A fair die is thrown repeatedly, stopping right after the first $6$ (the $6$ is included in the total). What is the probability that the total of all values thrown is even?

Short answer

$\tfrac47$ . The final 6 does not affect parity. Condition on the first throw: a 6 makes the total even; a 2 or 4 leaves the parity question unchanged; an odd value flips it. So $p=\tfrac16+\tfrac26p+\tfrac36(1-p)$ , which solves to $p=\tfrac47$ .

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

See the hint and the full worked solution in the question bank (free account).

Open the worked solution

Lessons behind this question

Conditional Probability, Total Probability, and Bayes' Rule
Counting and Combinatorics
Expectation, Mean, and Variance

Start with a free lesson: Expectation, Variance and Covariance is part of a free topic. Create a free account, no card needed, to read it in full with worked solutions.

Read it free