A fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) is repeatedly…
2024
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?
- A.
2
- B.
4
- C.
6
- D.
8
Attempted by 10 students.
Show answer & explanation
Correct answer: C
We want the expected number of throws until two consecutive even results (even faces are 2, 4, 6). Let p = 1/2 be the probability of an even on any throw.
Use two states for the process:
State "no previous even": call its expected remaining throws E0.
State "previous throw was even": call its expected remaining throws E1. If the next throw is even we stop; if it is odd we return to the no-previous-even state.
Write equations for the expectations.
From no previous even: on the next throw we always use 1 throw; with probability p we move to the previous-even state, with probability 1-p we stay in no-previous-even. So E0 = 1 + p·E1 + (1-p)·E0, which rearranges to E0 = 1/p + E1.
From previous-even: on the next throw we use 1 throw; with probability p we finish (no more throws), with probability 1-p we go back to no-previous-even. So E1 = 1 + (1-p)·E0.
Solve the system by substituting E0 into E1:
E1 = 1 + (1-p)(1/p + E1) = 1 + (1-p)/p + (1-p)E1, so p·E1 = 1 + (1-p)/p = 1/p. Thus E1 = 1/p^2.
Then E0 = 1/p + E1 = 1/p + 1/p^2 = (p+1)/p^2.
Substitute p = 1/2: E1 = 1/(1/2)^2 = 4 and E0 = 1/(1/2) + 4 = 2 + 4 = 6. Therefore the expected number of throws from the start until two consecutive evens is 6.
Answer: 6