# Roulette probability

by TalonD
Tags: probability, roulette
 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.
 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 turn-up. P = (1/37)(1/37)(1/37)(36/37)(36/37)
 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?
 P: 4,512 Roulette probability Yes, you're right, Rip.
 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.
P: 4,512
 Quote by 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) 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
P: 177
 Quote by Phrak 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)
 Quote by Phrak 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 )
 Sci Advisor PF Gold P: 1,781 BTW, This is the classic binomial probability distribution: en.wikipedia.org/wiki/Binomial_distribution
 P: 177 binomial distribution, that looks like what I was after thanks everyone T.D.
 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
 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 M-file if anyone wants to see the whole directory and set of algorithms. Thank you all in advance. Farooq

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