Roulette probability

  1. I'm not a math whiz but I would like to know how to calculate this and similar things. if you have a roulette wheel with 37 spots where the ball can land. If you have a limited number of spins, say 10 spins. what is the probability that the same number will appear two times, three times, etc. ? I know one time would be 1/37. but I don't know how to calculate the probabilty of the same number occuring more than once in a fixed number of spins.
    T.D.
     
  2. jcsd
  3. The wheel doesn't have double aughts?

    That's a probability of 1 in 37 for one spin, where the number is given in advance.

    Assumed is that each outcome of a spin is independent of other outcomes. To get a prepicked number exactly 3 times in 5 spins, independent of order, means that in 2 spins the number does not turn-up.

    P = (1/37)(1/37)(1/37)(36/37)(36/37)
     
  4. Don't you also have to consider the possible order of outcomes?

    YYYNN
    YNYNY
    YNNYY
    ... etc

    so 5C3 = 10. Multiple that probability by a factor of 10?
     
  5. Yes, you're right, Rip.
     
  6. European wheel only has one zero.
    Thanks for the lesson. So if order doesn't matter then this is correct for 3 occurences out of five ?

    (1/37)(1/37)(1/37)(36/37)(36/37)

    I know every bet in roulette has a negative expectation, but there is something called the andruchi system (see google) that has some interesting looking probability analysis and I was wondering how to do the actual calculations to properly analyze it.

    thanks for the tutorial
    T.D.
     
    Last edited: Feb 7, 2009
  7. Your welcome, but Rip was right, its P=10(1/37)(1/37)(1/37)(36/37)(36/37).

    See 'math combinations' to get C(5,3)=10
     
    Last edited: Feb 8, 2009
  8. Tell me if I'm understanding this right. to get the same number 3 times in 5 spins when you have 37 possible numbers, it would be ?

    P = ( (1/37)(1/37)(1/37)(36/37)(36/37) * 5 )
     
  9. jambaugh

    jambaugh 1,801
    Science Advisor
    Gold Member

  10. binomial distribution, that looks like what I was after
    thanks everyone
    T.D.
     
  11. r30

    r30 1

    Hello TalonD and all,

    According to the
    P=10(1/37)(1/37)(1/37)(36/37)(36/37)
    posted above, the probability of a specific number appearing at least one time in 37 spins would be:

    P= 37 (1/37) (37/37) (37/37)......(37/37) = 37(1/37)(37/37)^36=37(1/37)=1
    which of course is incorrect.
    What did I do wrong in interpreting the initial equation?

    I would calculate the probability of a specific number appearing at least one time in 37 spins as:
    P= 1-(36/37)^37

    ps:the andruchi system is no better than any other system inho
     
  12. Hi There
    I wonder if someone can help me with this roulette wheel command. I have to solve a problem in matlab using a set of algorithms. If I have two algorithms then I use rand() command to choose one of them randomly but if I have more than two algorithms I need to use this roulette wheel command but I am not a programmer so having problem how to use this roulette wheel command. Suppose I have set of 4 algorithms and I have to use one of them to solve the problem choosing one of these four algorithms. I'll be grateful for your help. I can upload the M-file if anyone wants to see the whole directory and set of algorithms.

    Thank you all in advance.

    Farooq
     
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