Results 1 to 5 of 5

Math Help - Coin probability problem

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    70

    Coin probability problem

    I'm having some trouble with the following problem:

    Suppose you toss a fair coin. Let
    X be the amount of tosses you need to make until you get HTT

    and
    Y be the amount of tosses you need to make until you get HTH.
    (a) What is expected value of
    X?
    (b) What is the expected value of
    Y ?
    (c) The game turns out to be fair even though the expected values of
    X and Y are different. How could this be?

    Thanks!

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    HTT

    state 0 : initial state
    state 1 : H
    state 2 : HT
    state 3 : HTT (final state)

    Initially you're in state 0. You will remain in state 0 until you flip a H (in which case you'll move to state 1).

    Once in state 1, you will remain there until you flip a T (in which case you'll will move to state 2).

    And once in step 2, you'll move to state 3 if you flip a T, and move back to state 1 if you flip a H.


    Now let  E(X_{i}) be the expected amount of flips in takes to reach state 3 starting from state i.

    You can set up three equations by conditioning on your first flip after you enter a state.

     E(X_{0}) = E(X_{0}|H)P(H) + E(X_{0}|T)P(T) = E(1+X_{1}) \frac{1}{2} + E(1+X_{0}) \frac {1}{2}  = 1+ \frac{1}{2}E(X_{0}) + \frac {1}{2}E(X_{1})

     E(X_{1}) = E(X_{1}|H)P(H) + E(X_{1}|T)P(T) = E(1+X_{1}) \frac{1}{2} + E(1+X_{2}) \frac {1}{2}  = 1 + \frac{1}{2}E(X_{1}) + \frac{1}{2}E(X_{2})

     E(X_{2}) = E(X_{2}|H)P(H) + E(X_{2}|T)P(T) = E(1+X_{1}) \frac{1}{2} + (1)\frac {1}{2} =1 + \frac{1}{2}E(X_{1})

    now solve for  E(X_{0})

    I get 8 flips.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    for HTH, the expected number of flips is 10

    What game are they referring to in part (c)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    70
    thanks a ton!
    part (c) basically means explain why P(H) =1/2 and P(T)=1/2 but explain why E(HTT) and E(HTH) are different
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by morganfor View Post
    thanks a ton!
    part (c) basically means explain why P(H) =1/2 and P(T)=1/2 but explain why E(HTT) and E(HTH) are different
    If you flip a H when in state 2 of HTT you move back to state 1. But if you flip a T when in state 2 of HTH, you move back to state 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coin probability...
    Posted in the Statistics Forum
    Replies: 1
    Last Post: February 17th 2011, 09:08 AM
  2. Coin probability problem!
    Posted in the Statistics Forum
    Replies: 3
    Last Post: March 18th 2010, 04:18 PM
  3. coin probability
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: February 21st 2010, 09:47 AM
  4. even and odd probability of coin
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 24th 2010, 03:33 PM
  5. [SOLVED] coin probability problem
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 4th 2008, 06:20 PM

Search Tags


/mathhelpforum @mathhelpforum