Can one predict an outcome over (n) spins?

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The discussion focuses on calculating the probability of hitting at least one number from the first column of a Roulette wheel over various spins. The probabilities for hitting at least one number in 3, 6, 9, 12, and 15 spins are derived using the formula 1 - q^n, where q represents the probability of not hitting a number. The calculated probabilities are approximately 68% for 3 spins, 90% for 6 spins, and 97% for 9 spins. The analysis also highlights the impact of the "0" and "00" outcomes on the overall probabilities. Overall, the discussion emphasizes the mathematical approach to understanding outcomes in random chance scenarios like Roulette.
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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 anyone 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 :smile: # (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.
 
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What have you done so far?
 
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.
 
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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