1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability: Conditional expectation

  1. Apr 19, 2012 #1
    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.

    [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!
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Probability: Conditional expectation
Loading...