Probability: Conditional expectation

[IniquiTrance]

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.

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

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

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

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

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!

 

[mighty2000]

Yes, it is acceptable to condition on the second flip in this way. This approach is known as the law of total expectation or the tower property, which states that the expected value of a random variable can be calculated by taking the weighted average of its conditional expected values, where the weights are the probabilities of the corresponding events. In this case, you are conditioning on the outcome of the second flip, which is a valid approach for calculating the expected number of flips until two consecutive heads.