JEEOlympiad
A fair coin is flipped repeatedly until the pattern Heads-Tails-Heads…
Question
A fair coin is flipped repeatedly until the pattern Heads-Tails-Heads (HTH) first appears as three consecutive flips. Let E be the expected number of flips required. E is a rational number m/n in lowest terms; find m+n.
✓ Verified answer: 11checked by our engine — not a guess
Step-by-step solution
Use an absorbing Markov chain on the progress toward HTH.
States by the longest matching suffix: S0 = '' (no progress), S1 = 'H', S2 = 'HT', and absorbing 'HTH'.
Transitions (each flip H or T with prob 1/2):
- From S0: H -> S1, T -> S0.
- From S1: H -> S1 (the new H restarts a length-1 prefix), T -> S2.
- From S2 ('HT'): H -> absorb (HTH done!), T -> S0.
Let E0,E1,E2 be expected remaining flips from each state.
E0 = 1 + (1/2)E1 + (1/2)E0
E1 = 1 + (1/2)E1 + (1/2)E2
E2 = 1 + (1/2)(0) + (1/2)E0
From the first: (1/2)E0 = 1 + (1/2)E1 => E0 = 2 + E1.
From the second: (1/2)E1 = 1 + (1/2)E2 => E1 = 2 + E2.
Third: E2 = 1 + (1/2)E0.
Substitute: E1 = 2 + 1 + (1/2)E0 = 3 + (1/2)E0. And E0 = 2 + E1 = 5 + (1/2)E0 => (1/2)E0 = 5 => E0 = 10.
So E = 10 = 10/1, m+n = 10+1 = 11.
Final answer11
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Probability solutions
A Polya urn starts with 2 red and 3 blue balls. A ball is drawn…JEE · MathA permutation of the numbers 1,2,...,8 is chosen uniformly at random.…JEE · MathPlayer A repeatedly rolls a fair eight-sided die (faces 1-8) and…JEE · MathEight distinct people, three of whom are named A, B, and C, are…JEE · MathA gambler starts with 3 rupees. In each round he wins 1 rupee with…JEE · MathA fair six-sided die is rolled 5 times. Let p be the probability that…JEE · MathThe numbers 1,2,...,9 are arranged in a uniformly random permutation…JEE · MathTwo players A and B take turns rolling a single fair six-sided die,…JEE · Math