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Math Help - Probability

  1. #1
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    Probability

    Help i've been trying to figure his out all week?

    In the game of roulet a player can place a $5 bet on the number 17 and have a 1/38 probability of winning ifthe metal ball lands on 17 the player wins $175, otherwise the casino takes the players $5. What is the expected value of the game to the player? If you played the game 1000 times how much would you expect to lose?
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  2. #2
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    Hello, kandela573!

    Are you familiar with the Expected Value formula?


    In the game of roulette, a player can place a $5 bet on the number 17.
    He has a 1/38 probability of winning $175.
    Otherwise, the casino takes the player's $5.

    (a) What is the expected value of the game to the player?
    (b) If you played the game 1000 times how much would you expect to lose?
    It's really straight-forward . . .

    . . \begin{array}{ccc}P(\text{win \$175}) &=& \dfrac{1}{38} \\ \\[-3mm]<br />
P(\text{lose \$5}) &=& \dfrac{37}{38} \end{array}


    E \;=\;\left(\frac{1}{38}\right)(+175) + \left(\frac{37}{38}\right)(-5) \;=\;-\frac{10}{38} \;=\;-0.263157895


    He can expect to lose an average of about 26.3 cents per game. (a)



    In 1000 games: . E \;=\;1000 \times (-0.263157895) \;= \;-263.157895

    . . We can expect to lose about $263.16 (b)

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  3. #3
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    actuallly

    You got it wrong you forgot to take out the $5 dollars that he placed as his original bet so the answer actually was

    170*1/38+(-5)*37/38=-.395 for his expected loses for each role
    for 1000 games youthen expected to lose $395
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  4. #4
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    Hello, kandela573!

    Sorry, you're way off-base . . .


    You got it wrong.
    You forgot to take out the $5 dollars that he placed as his original bet.
    No, I didn't . . . Did you look at my reply?


    So the answer actually was:

    170(1/38) + (-5)(37/38) = -0.395 for his expected losses for each roll.
    . ?
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