+ Reply to Thread
Results 1 to 3 of 3

Thread: 1 out 101

  1. #1
    Points: 34, Level: 1
    Level completed: 68%, Points required for next Level: 16

    Posts
    2
    Thanks
    0
    Thanked 0 Times in 0 Posts

    1 out 101




    Hi, I am hoping that someone can help me out with this?

    If I have 54 numbers on a roulette table, and there is a random generator picking the numbers, on average in 54 picks every number should in theory come out once. What if I had 6 of the 54 numbers, in theory my numbers should be picked 6 times out of 54.

    My situation is I had exactly 101 spins of this roulette table, I covered 6 of the 54 numbers, but it only dropped in 1 of my 6 numbers once, it should have dropped in nearly 12 times, what is the odds or probability of this happening?

    Really appreciate the help if any one can assist.

  2. #2
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: 1 out 101

    Assuming each spin is fair, then it is easy to see the probability that your number come out is \frac {6} {54} = \frac {1} {9}

    Assuming each spin is independent to each other. Now the number of times dropped in 101 spins follows \text{Binomial}\left(101, \frac {1} {9}\right).

    And the probability that it dropped exactly 1 time is

    \binom {101} {1} \left(\frac {1} {9}\right)^{1}\left(1 - \frac {1} {9}\right)^{101 - 1}

  3. #3
    Points: 34, Level: 1
    Level completed: 68%, Points required for next Level: 16

    Posts
    2
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Re: 1 out 101


    Thank you for your reply, so I take it that for this to happen then I would have to do this 101 times in theory for this to happen once?

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats