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Roulette probability 
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#1
Feb709, 04:56 PM

P: 177

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


#2
Feb709, 05:31 PM

P: 4,512

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 turnup. P = (1/37)(1/37)(1/37)(36/37)(36/37) 


#3
Feb709, 05:58 PM

P: 62

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? 


#4
Feb709, 10:32 PM

P: 4,512

Roulette probability
Yes, you're right, Rip.



#5
Feb709, 11:23 PM

P: 177

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. 


#6
Feb809, 02:02 AM

P: 4,512

See 'math combinations' to get C(5,3)=10 


#7
Feb809, 07:00 AM

P: 177

P = ( (1/37)(1/37)(1/37)(36/37)(36/37) * 5 ) 


#8
Feb809, 07:11 AM

Sci Advisor
PF Gold
P: 1,781

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


#9
Feb809, 09:13 AM

P: 177

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


#10
Jun1910, 11:23 PM

P: 1

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
Jul2610, 05:00 AM

P: 22

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 Mfile if anyone wants to see the whole directory and set of algorithms. Thank you all in advance. Farooq 


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