Calculating Roulette Probability for Multiple Occurrences in Limited Spins

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Discussion Overview

The discussion revolves around calculating the probability of a specific number appearing multiple times in a limited number of spins on a roulette wheel, particularly focusing on a European wheel with 37 spots. Participants explore the application of binomial probability distribution and the implications of order in outcomes.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant seeks to understand how to calculate the probability of a number appearing multiple times in a fixed number of spins, starting with the basic probability of 1/37 for a single occurrence.
  • Another participant clarifies that the outcomes of spins are independent and suggests a formula for calculating the probability of a specific number appearing exactly three times in five spins.
  • There is a discussion about considering the order of outcomes, with one participant proposing the use of combinations to account for different arrangements of occurrences.
  • Some participants confirm the formula for calculating the probability when order does not matter, while others express uncertainty about the correctness of their interpretations.
  • A participant introduces the concept of the binomial probability distribution as relevant to the discussion.
  • Another participant questions their understanding of the probability calculation for a number appearing at least once in multiple spins, leading to a different interpretation of the initial equation.
  • A new participant seeks help with implementing a roulette wheel command in MATLAB for algorithm selection, indicating a practical application of the roulette concept.

Areas of Agreement / Disagreement

Participants generally agree on the use of binomial probability distribution for the calculations, but there are differing interpretations of the formulas and their applications. The discussion remains unresolved regarding the correct approach to calculating probabilities for specific scenarios.

Contextual Notes

Some participants express uncertainty about the assumptions underlying their calculations, particularly regarding the independence of spins and the implications of order in outcomes. There are also references to the limitations of certain betting systems without definitive conclusions on their effectiveness.

TalonD
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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 probability of the same number occurring more than once in a fixed number of spins.
T.D.
 
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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)
 
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?
 
Yes, you're right, Rip.
 
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:
TalonD said:
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)

thanks for the tutorial
T.D.

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:
Phrak said:
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)

Phrak said:
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

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 )
 
BTW,
This is the classic binomial probability distribution:
http://en.wikipedia.org/wiki/Binomial_distribution"
 
Last edited by a moderator:
binomial distribution, that looks like what I was after
thanks everyone
T.D.
 
  • #10
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
 
  • #11
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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