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Roulette problem

  1. Mar 9, 2015 #1
    Hey guys long time since i've posted on here, got this random question a friend of mine and myself were wondering over.

    1. The problem statement, all variables and given/known data

    A player bets 100 dollars continuously on any number besides 0 (european roulette so 1/37 chance to win). He plays until he wins once and stops playing.

    a) Z is the amount of times he plays. What is the distribution of Z?
    b) What is the probability of the player winning at least 1000 dollars and the probability of the player losing 1000 dollars?


    2. The attempt at a solution
    I'm not so sure as how to go about this as i don't have any background in statistics and am currently studying calculus. But statistics is on my curriculum after the summer so it would be nice to have a little heads up on whats to come.
     
  2. jcsd
  3. Mar 9, 2015 #2

    LCKurtz

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    Start by figuring out the probability that he loses on the first ##n-1## plays and wins on the ##n##th play. That will give you the distribution.
     
  4. Mar 9, 2015 #3

    Ray Vickson

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    You need to tell us what is the payoff if the player wins; otherwise we cannot possibly tell if he/she can even win $1000 at all. Does this $1000 win mean the player quits and goes home $1000 richer than when he/she started (which could happen if he/she wins in game 2 and receives a $1200 payoff). Or, does it just mean that the payoff upon winning is $1000?
     
  5. Mar 9, 2015 #4
    The player pays 100 for a single spin and if he wins makes 200 back, so a 100 dollar profit, but each spin costs 100. and this is for any number besides zero so he has 1/37 gone already for the zero and another 1/37 probability for any other number.
     
  6. Mar 9, 2015 #5

    Ray Vickson

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    S
    So, ever winning $1000 is impossible, since he quits upon his first win.
     
  7. Mar 9, 2015 #6
    Aha thats true for b), didnt see that....as for losing 1000, he has to lose 10 times in a row before quitting, how likely is losing 10 times in a row?
     
  8. Mar 9, 2015 #7

    LCKurtz

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    Respond to post #2 and you will have your answer. I suppose you realize that, in the real world, your "roulette" game is completely absurd. You bet $100 with only 1/37 probability of winning, and in the unlikely event you are lucky, you just get $100 back??? So you lose $100 virtually every time you spin and win $100 once every blue moon. Some game.
     
  9. Mar 9, 2015 #8

    Ray Vickson

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    Your description does not match the actual way roulette is played. See, eg, http://en.wikipedia.org/wiki/Roulette .

    In particluar, the section entitled "Bet Odds Table" states that the payoff is ##\frac{1}{n}(36-n) = (36/n) - 1##, where ##n## is the number of squares the player is betting on. In your description you say the player bets on ONE square, so the payoff is 35 to 1; that is, they would receive a payoff of $3500 for their $100 investment, in the lucky event that they win. (Just to be clear, the $3500 is a net gain, so they paid $100 and got back $3600.) Betting on more than 1 square would be needed if they wanted to increase the probability of winning while lowering the possible payoff.

    BTW: If we assume they bet on single squares each time, the probability of losing 10 times in a row would be ##(36/37)^{10} \doteq 0.76034##.
     
    Last edited: Mar 10, 2015
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