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VERY Basic Expected Value Question

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data

    If you throw exactly 1 head in three tosses of a coin you win $27. If not you pay $21.


    2. Relevant equations
    I know, E(X) = sum[x subscript i * P(X = x subscript i)]


    3. The attempt at a solution
    Sorry y'all I don't really know how to use all the fancy letters and signs on here, but I'm tryin.
    OK so, I get stuck at the very beginning of the problem trying to find P(X = x)
    Here is what I have:

    Possible Outcomes:
    x = # of heads: P(X=x)
    0 1/8 = 0.125
    1 3/8 = 0.375
    2 ?
    3 ?

    My problem is that I don't really understand why the P(X=x) for 0 heads is 1/8 I know I should, I just can't seem to remember and I'm getting frustrated which is making everything harder. I do understand that there are 8 possible outcomes b/c 2 sides to a coin and 3 tosses so 2^3 = 8
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 5, 2010 #2

    Doc Al

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    Staff: Mentor

    Do you further understand that each of those 8 possible outcomes is equally likely, thus the probability of each is 1/8? And that '0 heads' corresponds to just one of those outcomes, namely TTT?
     
  4. Feb 5, 2010 #3
    Yes I do, I also understand that all possible outcomes are: (HHH,THH,HTH,HHT,TTT,HTT,THT,TTH)
     
  5. Feb 5, 2010 #4

    Doc Al

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    Staff: Mentor

    Good. So can you rephrase your question? (Or did you already figure it out?)
     
  6. Feb 5, 2010 #5
    Am I right in my thinking that with 1 head the probability is 3/8 because head only occurs once in (HTT,THT,TTH) so,(1/8)+(1/8)+(1/8) = 3/8
     
  7. Feb 5, 2010 #6
    I figured it out, ugh... I think I just psyched myself out over it... do you mind if I work the rest of it out so you can tell me if I am on the right track?
     
  8. Feb 5, 2010 #7

    Doc Al

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    Staff: Mentor

    Exactly.

    So what's the probability of you not getting exactly 1 head in three tosses? (No need to enumerate all the possibilities.)
     
  9. Feb 5, 2010 #8
    E(X) = (0.125+0.375+0.375+0.125)/4
    = 1/4
    = 0.25

    ok, why is it called an expected value? I think I don't get it, because I don't really understand the purpose of it... what does the 0.25 actually mean?
     
  10. Feb 5, 2010 #9
    The probability of not getting exactly 1 head is 7/8.... should I have used the 1-P(loose) formula?
     
  11. Feb 5, 2010 #10

    Doc Al

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    I'm not sure what you are doing here, but that's not the expected value. Your first post had the correct definition of E(x)--assuming I understand your notation. Use that.
    For this game, you can think of the expected value as how much you would 'expect' to win (or lose) per game--on average--if you played many, many games.
     
  12. Feb 5, 2010 #11

    Doc Al

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    Staff: Mentor

    Nope.
    Yep.

    You have two situations, A and not-A. A = getting exactly 1 head in three tosses. The probability of A plus the probability of not-A must add to 1.
     
  13. Feb 5, 2010 #12
    ok so I will try again

    P(win) = 1/8
    P(loose) = 7/8
    ?
    ugh, confused again... sorry is this at all right?
     
  14. Feb 5, 2010 #13

    Doc Al

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    No. You already figured out that the probability of winning was 3/8 in post #5. Make use of that.
     
  15. Feb 5, 2010 #14
    Deep Breath and trying again... so, the probability that I win (by getting 1 head in 3 tosses) is 3/8; so the probability that I loose (by getting anything else) is 5/8... am i on the right track
     
  16. Feb 5, 2010 #15

    Doc Al

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    Exactly! Keep going.
     
  17. Feb 5, 2010 #16
    so...
    E(X) = 27.00 *(3/8) - 21.00 * (5/8)
    = 10.125 - 13.125
    = - $3.00

    So the player should expect to loose $3 per game?
     
  18. Feb 5, 2010 #17

    Doc Al

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    Staff: Mentor

    Perfect!

    Yes, on average the player will lose $3 per game.
     
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