# Probability: Conditional expectation

1. Apr 19, 2012

### IniquiTrance

1. The problem statement, all variables and given/known data
What is the expected number of flips of a biased coin with probability of heads 'p', until two consecutive flips are heads?

2. Relevant equations

3. The attempt at a solution

Let T_1 = first flip is tails, H_1 = first flip is heads. and T_2, H_2 for second flip.

$\mathbb{E}[X] = \mathbb{E}[X|T_1]\mathbb{P}[T_1] + \mathbb{E}[X|H_1]\mathbb{P}[H_1]$

$= \mathbb{E}[X|T_1]\mathbb{P}[T_1] + \mathbb{P}[H_1]\left(\mathbb{E}[X|H_1H_2]\mathbb{P}[H_2]+\mathbb{E}[X|H_1T_2]\mathbb{P}[T_2]\right)$

$= (1-p)(1+ \mathbb{E}[X]) + p(2p+(2+\mathbb{E}[X])(1-p))$

$= \frac{1+p}{p^2}$

I know that the final answer is correct.

My question is whether I am allowed to condition on the 2nd flip the way I did.

Thanks!