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Probability - Roulette

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    This particular roulette wheel has 37 slots - 0,1,2...36. A gambler can bet on different combinations of numbers. Louise loves to bet on a block of 4 numbers, called a corner. The payout on a corner is 8 to 1.

    (a) Let X be a gambler's winning from a $1 bet on a corner. What is the probablity distribution for X? (Hint X can be negative)

    (b) What is the expected value of a $1 bet on a corner.


    2. The attempt at a solution

    (a) Okay i figured that the chance of a corner would be.

    Pr(corner) = (1/37)X4 = 4/37.

    However, i dont think this counts as the probablity distribution?

    Not quite sure how to approach this.

    (b) Excepted value for a $1 bet. Need to take account both winning and losing?

    Pr (corner) = 4/37
    Pay out is 8 to 1.
    therefore is excepted value is 4/37 X $8 = 32/37

    Pr (no corner) = 33/37 ((1-(4/37))
    Expected value is 33/37 X -$8 = -264/37

    Excepted value of 1 dollar bet = (-264/37) + (33/37) = -232/37 = -6.27

    You would expect to lose -6.27??

    Thanks
     
  2. jcsd
  3. Aug 25, 2009 #2
    A probability distribution is a listing of all outcomes and their respective probabilities. Obviously one can only win or lose at Roulette, so there are only two outcomes to worry about here. How much would one win and how much would one lose in a single bet? Find the probability of each occurring.

    As for expected value - you have the right formula, but the incorrect values. Think about how much you're losing when you lose.

    The answer should be negative (or casinos wouldn't exist), but not -6.27.

    --Elucidus
     
  4. Aug 25, 2009 #3
    oh yes. i think i got it. if you put in 1 dollar you can only lose it. you can you more than you put in...

    Pr(no corner) = 33/37
    Excepted value = -1 X 33/37 = -33/37

    Excepted value of one dollar bet = -33/37 +32/37 = -0.02702 = -2.7%

    that shows slightly wrong though?
     
    Last edited: Aug 25, 2009
  5. Aug 25, 2009 #4
    Can anyone help me through the probability distribution?
    X is in the case refers to Winning a corner. You can only will or lose a corner..
    If Pr (Corner) = 4/37, Pr(Corner) = 33/37
     
  6. Aug 25, 2009 #5

    sylas

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    Looks correct to me.

    You've got the distributions. The variable X is the winnings.

    The value of X when you lose is -1. The value of X when you win is +8.

    The probability distribution is the probability for each value of X. You can give it as a table.

    Cheers -- sylas
     
  7. Aug 26, 2009 #6
    Cheers Sylas,

    I was thinking about the expected value of a $1 bet further.
    Shouldnt the EXPECTED VALUE of the $1 bet be $1 minus $0.02 = 98 cents?

    Got it from http://en.wikipedia.org/wiki/Expected_Value#Examples

    Not very sure because the definition is E(X)= xP(x) well from my book...
     
  8. Aug 26, 2009 #7

    HallsofIvy

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    I suggest you look at your book again. The definition of expected value is the sum of xP(x) where x ranges over all possible values. Here, the two possible values of x are 8 (if you win) and -1 (if you lose). The expected value is 8Pr(win)- 1Pr(lose)= [math]8\frac{1}{37}- \frac{33}{37}[/math]. You don't think you are going to have a positive expected value gambling do you?
     
  9. Aug 26, 2009 #8

    HallsofIvy

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    You mean "Pr (Corner) = 4/37, Pr(NOT Corner) = 33/37", of course.
     
  10. Aug 26, 2009 #9
    Thanks for the prompt reply. Oh my book is correct. I mistyped that.

    It is the sum of xP(X), therefore it is (32/37) + (-33/37) = -1/37

    Therefore, the expected value of a $1 bet is $(-1/37), which is approx -2 cents.

    It was wikipedia confusing me.
     
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