g In American roulette, the wheel has the 38 numbers, 00, 0, 1, 2, …, 34, 35, and 36, marked on equally spaced slots. If a player bets $1

Question

g In American roulette, the wheel has the 38 numbers, 00, 0, 1, 2, …, 34, 35, and 36, marked on equally spaced slots. If a player bets $1 on a number and wins, then the player keeps the dollar and receives an additional $35. Otherwise, the dollar is lost. Calculate the expected value for the player to play one time. Round to the nearest cent.

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Serenity 1 week 2021-10-08T05:32:33+00:00 1 Answer 0

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    2021-10-08T05:34:17+00:00

    Answer:

    The expected value for the player to play one time is -$0.05.

    Step-by-step explanation:

    The expected value of a random variable X is given by the formula:

    E(X)=\sum x\cdot P(X)

    The American roulette wheel has the 38 numbers, {i = 00, 0, 1, 2, …, 34, 35, and 36}, marked on equally spaced slots.

    The probability that the ball stops on any of these 38 numbers is same, i.e.

    P (X = i) = \frac{1}{38}.

    It is provided that a a player bets $1 on a number.

    If the player wins, the player keeps the dollar and receives an additional $35.

    And if the player losses, the dollar is lost too.

    So, the probability distribution is as follows:

        X : $35      -$1

    P (X) :   \frac{1}{38}         \frac{37}{38}  

    Compute the expected value of the game as follows:

    E(X)=\sum x\cdot P(X)

             =[\$35\times \frac{1}{38}]+[-\$1\times \frac{37}{38}]\\\\=\frac{\$35-\$37}{38}\\\\=-\frac{\$2}{38}\\\\=-\frac{1}{19}\\\\=-0.052632\\\\\approx -\$0.05

    Thus, the expected value for the player to play one time is -$0.05.

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45:7+7-4:2-5:5*4+35:2 =? ( )