Stopping Coin

Solution

Answer: 6

We will define: \[ E_0 = \text{expected remaining flips when no consecutive heads have just occurred} \] \[ E_1 = \text{expected remaining flips when the last flip was one head}. \] The game ends once two heads in a row occur. From state \(E_0\): You make one flip. With probability \(1/2\) you get a tail and remain in state \(E_0\). With probability \(1/2\) you get a head and move to state \(E_1\). \[ E_0 = 1 + \frac12 E_0 + \frac12 E_1. \] From state \(E_1\): You again make one flip. With probability \(1/2\) you get a head and the game ends. With probability \(1/2\) you get a tail and return to state \(E_0\). \[ E_1 = 1 + \frac12(0) + \frac12 E_0. \] So \[ E_1 = 1 + \frac12 E_0. \] Substitute this into the first equation: \[ E_0 = 1 + \frac12 E_0 + \frac12\left(1 + \frac12 E_0\right). \] \[ E_0 = \frac32 + \frac34 E_0. \] \[ \frac14 E_0 = \frac32. \] \[ E_0 = 6. \] Since the game begins with no previous head, the expected number of flips required is 6.