2026-04-23

Problem — 2026-04-23

Medium

Mediumprobability

A fair die is rolled until a $6$ appears. What is the probability that the number of rolls is even?

Express as a fraction

Show solution

Let

p

be the probability that the first

6

appears on an even roll. The first

6

on roll

2k

requires

2k-1

non-sixes then a six: sum of

(5/6)^{2k-1}(1/6)

for

k \ge 1

. This equals

\frac{1}{6} \cdot \frac{5/6}{1 - 25/36} = \frac{1}{6} \cdot \frac{5/6}{11/36} = \frac{1}{6} \cdot \frac{30}{11} = \frac{5}{11}