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Can one predict an outcome over (n) spins?

  1. Jun 29, 2005 #1
    I’m interested in following statistical question regarding the random chance of a Roulette Wheel:

    Suppose I were to place a single bet on the first column of numbers (1, 4, 7,10,13,16, 19, 22, 25, 28, 31, 34), which pays 2-1, what is the probability that I will hit at least one of these numbers in the column in any one of 3 consecutive spins of the wheel; 6 consecutive spins; 9 consecutive spins; 12 consecutive spins and 15 consecutive spins.

    Given the statistical laws of distribution, at what point (spin :rofl: # (n)) would one expect the first occurrence of one of these selected numbers appearing?

    Please show probability separately for each group of spins as to clearly show the mathematical differences and how they affect the outcome.
  2. jcsd
  3. Jun 29, 2005 #2


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    What have you done so far?
  4. Jun 29, 2005 #3
    Lacking the use of an actual Roulette wheel, I instead used a Die trial with numbers 1 & 2 to represent column 1, 3 & 4 as column2 and 4 & 5 as cloumn 3. After 200 throws found that the maximum numbers of throws (spins) before a 1 or 2 came out was 10. However, this does not take into consideration the affect of "0" and "00" has on this event.
  5. Jul 10, 2005 #4
    In this case, we have 38 outcomes, 2 of which are 0 or 00. Let p=12/38=6/19. Then the possibility of not hitting a number is 13/19. In three turns of the wheel we consider the Bernoulli trials: (p+q)^3. The number of no hits is q^3, so the probability of at least one hit is 1-q^3. Similarly the probabilities are 1-q^6, and 1-q^9. That is: 68%, 90%, 97% chance of at least one hit.
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