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Expectanion value probabilities

  1. Apr 3, 2013 #1
    Flipping a coin 10 times. The currency brings crown with probability 0.57. If you bring a total of k heads they add 2 ^ k euros.

    What is the average profit in this game?

    i must find the expectanion value and i do this:

    ∑from k=1 to 10 --->((0.57)^k)*(2^k) =22.0445164

    i think that my solve is not correct any ideas?
     
  2. jcsd
  3. Apr 3, 2013 #2

    mfb

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    The probability of getting k heads in 10 flips is not 0.57^k.
     
  4. Apr 3, 2013 #3
    0.43?
     
  5. Apr 3, 2013 #4

    mfb

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    No, and guessing will not help.
    This is a basic problem in statistics, you should know which distribution you have to use.

    As a general remark: you should always check if your values, results, ... are plausible. Please do this before you ask questions.
    As an example, the probability to get k times heads in 10 flips is certainly different than the probability to get k times heads in 1 flip. Therefore, the value has to depend on the number of flips in some way, otherwise it cannot be right.
     
  6. Apr 3, 2013 #5
    we dont have learned distrinutions in school yet.....
     
  7. Apr 3, 2013 #6

    haruspex

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    in your sum, you need the probability of exactly k heads. So you have to take into account that 10-k are not heads. Also, you don't care which of the 10 are heads. how many different possibilities are there for which are?
     
  8. Apr 3, 2013 #7

    HallsofIvy

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    That makes this problem very difficult! I imagine you have learned enough to know that the probability of all 10 being crowns will be (.57)10 and the probability that all ten are not crowns will be (1- .57)10= (.43)10. But the probability that the first flip is a crown and the rest are not is (.57)(.43)(.43)(.43)(.43)(.43)(.43)(.43)(.43)(.43)= (.57)(.43)9 while the probability of "first not a crown, second a crown, then the rest not crowns" is (.43)(.57)(.43)(.43)(.43)(.43)(.43)(.43)(.43)(.43)= (.57)(.43)9 so that the probability of "one crown, nine non-crowns" is 10(.57)(.43)9.

    And things like "four crowns, 6 non-crowns" get much more complicated as the possiblities increase. Are you sure you have not learned things like "permutations and combinations", "binomial coefficents", and "binomial distributions"?
     
  9. Apr 3, 2013 #8
    how i can solve it with distributions way?
     
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