Results 1 to 4 of 4

Math Help - Probability Help expected value

  1. #1
    Newbie
    Joined
    Jan 2007
    Posts
    10

    Smile Probability Help expected value

    Will someone help me step by step on this problem. Im sure its not as hard as I make it out to be.

    When you bet $6.00 in a common gamble, there is a probability of 133 that you will win $6.00 and a probablilty of 138 that you will lose 271
    your $6.00. 271
    What is the expected value?

    Thanks!!!!!!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by air2slb View Post
    Will someone help me step by step on this problem. Im sure its not as hard as I make it out to be.

    When you bet $6.00 in a common gamble, there is a probability of 133 that you will win $6.00 and a probablilty of 138 that you will lose 271
    your $6.00. 271
    What is the expected value?

    Thanks!!!!!!!!
    There is something about this problem that I don't understand. Probabilities
    are always between 0 and 1, and the probability of a favourable outcome
    and that of all other outcomes sum to 1.

    Here we have probabilities of 133 and 138, which in terms of probabilities
    make no sense. Also when you lose what is the 271 you lose?

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2007
    Posts
    10

    Your right, it posted funny.

    The problem posted funny. Its 133 over 271 and 138 over 271 133/271 and 138/271. Does that make more sense??? Thanks!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    Posts
    136
    ok so the expectation simply represents the average amount one "expects" as the outcome of the random trial when identical odds are repeated many times.

    so if we have a random variable R that has values -> x1, x2
    with probabilities -> p1,p2

    The expected value of R

    E(R) = p1x1 + p2x2

    __________________________________________________ _____________

    So here we have two probabilities
    p1 = 133/271 (win)
    and
    p2 = 138/271 (loose)

    The values of x which go with these are

    x1 = 6 (amount if u win)
    x2 = -6 (amount if u loose)

    thereore the expected value = (133/271)*6 + (138/271)*-6 = -0.110701107

    Which means on average you loose 11 cents for every 6 dollar bet.

    __________________________________________________ _______________
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability & Expected Value
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 2nd 2010, 03:06 PM
  2. Expected Probability
    Posted in the Statistics Forum
    Replies: 8
    Last Post: June 28th 2009, 06:37 AM
  3. Probability and Expected Value problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 29th 2008, 05:32 AM
  4. expected value - probability
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: October 4th 2007, 11:14 AM
  5. Expected Value (Probability) Help Pls
    Posted in the Statistics Forum
    Replies: 6
    Last Post: October 24th 2006, 09:31 PM

Search Tags


/mathhelpforum @mathhelpforum