A fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) is repeatedly thrown independently.
What is the expected number of times the die is thrown until two consecutive throws of even numbers are seen?
Answer
(C)
Explanation
Each throw is even with probability \(\frac{1}{2}\). Let \(E_{0}\) be the expected number of further throws at the start or just after an odd throw, and \(E_{1}\) the same just after an even throw.
- \(E_{1} = 1 + \frac{1}{2}(0) + \frac{1}{2}E_{0}\): an even throw finishes, an odd one starts over.
- \(E_{0} = 1 + \frac{1}{2}E_{1} + \frac{1}{2}E_{0}\), which gives \(E_{0} = 2 + E_{1}\).
Substituting, \(E_{0} = 3 + \frac{1}{2}E_{0}\), so \(E_{0} = 6\).
Option B (4) is the expected wait for two even throws that need not be consecutive (\(2 + 2\)); insisting that they come back to back costs two more throws on average.
Answer: C