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Math Help - Expected Loss for American Roulette

  1. #1
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    Expected Loss for American Roulette

    Hey everyone! So I'm getting a bit stuck when it comes to calculating expected loss for American Roulette (38 spots: 18 are red (odd numbers between 1 and 35), 18 are black (even numbers between 1 and 36), and 2 spots are green (one is 0 and the other is 00)).

    For example, I understand that If I bet on red, my expected loss is (18/38 - 20/38) = -1/19 = -0.0526

    However, I am confused when it comes to betting on a colour as well as numbers. For example, if I bet on red and the numbers 1, 2, 3, and 4, how would I go about calculating the expected loss then?

    I figure it should be (18/38 - 20/38 + 20/38 - 18/38), but this would give 0, so this seems wrong.

    Any help would be very much appreciated!
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  2. #2
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    Re: Expected Loss for American Roulette

    Hey Shapeshift.

    Can you outline how roulette pays out (for people like me not familiar with the game)?
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  3. #3
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    Re: Expected Loss for American Roulette

    Hey sorry I forgot to include that!
    The payoff for red is 1 to 1, so a $1 bet would have a net gain of $1.
    The payoff for the block of four numbers is 8 to 1, so a $1 bet would have a net gain of $8.

    So if I place a $1 bet on the red, and a $1 bet on the block of four numbers, I end up getting

    18/38 + 8(4/38) - 20/38 - 34/38 = -0.105263 which I believe is right?
    Last edited by Shapeshift; February 8th 2013 at 10:47 AM.
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  4. #4
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    Re: Expected Loss for American Roulette

    That looks right.
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