Calculating Roulette Probability for Multiple Occurrences in Limited Spins

[TalonD]

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 occurring more than once in a fixed number of spins.
T.D.

 

[Phrak]

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)

 

[2RIP]

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?

 

[Phrak]

Yes, you're right, Rip.

 

[TalonD]

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.

 

[Phrak]

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

 

[TalonD]

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)

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 )

 

[jambaugh]

BTW,
This is the classic binomial probability distribution:
http://en.wikipedia.org/wiki/Binomial_distribution"

 

[TalonD]

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

 

[r30]

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

 

[mfarooq52003]

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