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Math Help - probability and expected gain

  1. #1
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    probability and expected gain

    i dunno how to do this question. could anyone possibly help? thx in advance.

    A banker throws three fair, ordinary dice. If the result is a total of 11 points or above, or three "1", or three "2", or three "3", he wins $900. Otherwise he loses $900.
    What is his expected gain?

    i wonder if there are other methods other than listing out all events.
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  2. #2
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    No. You need to know each three dice permutation that results in 11 or more.

    The "three 1's, 2's or 3's" one is easy.
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  3. #3
    Super Member Anonymous1's Avatar
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    Some useful facts:

    P(all 1's)+P(all 2's)+P(all 3's) = 3(\frac{1}{6})^3

    P(X_1+X_2+X_3 >11) = 1 - P(X_1+X_2+X_3<11)
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  4. #4
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    Hello, littleprince327!

    A banker throws three fair, ordinary dice.
    If the result is a total of 11 or above, or three 1's, or three 2's, or three 3's, he wins $900.
    Otherwise he loses $900.
    What is his expected gain?

    With three dice, there are 6^3 \,=\,216 possible outcomes.


    The totals range from 3 to 18.

    \text{Note that: }\;\underbrace{3,4,5,6,7,8,9,10}_{\text{108 ways}}\, , \, \underbrace{11,12,13,14,15,16,17,18}_{\text{108 ways}}

    Then: . P(\text{sum }\geq 11) \:=\:\frac{108}{216}

    Also: . P(\text{three 1's or three 2's or three 3's}) \:=\:\frac{3}{216}

    Hence: . P(\text{win}) \:=\:\frac{108}{216} + \frac{3}{216} \:=\:\frac{111}{216} \:=\:\frac{37}{72}

    . . and: . P(\text{lose}) \:=\:\frac{35}{72}


    Therefore: . E \;=\;\frac{37}{72}(+900) + \frac{35}{72}(-900) \;=\; +\,25

    He can expect to win an average of $25 per game.

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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Anonymous1 View Post
    Some useful facts:

    P(all 1's)+P(all 2's)+P(all 3's) = 3(\frac{1}{6})^3

    P(X_1+X_2+X_3 >11) = 1 - P(X_1+X_2+X_3<11)
    you left out the 11...

    P(X_1+X_2+X_3 \ge 11) = 1 - P(X_1+X_2+X_3<11)
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