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## Homework Statement

What is the expected number of flips of a biased coin with probability of heads 'p', until two consecutive flips are heads?

## Homework Equations

## 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.

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

[itex] = \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)[/itex]

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

[itex] = \frac{1+p}{p^2} [/itex]

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!