Register to reply 
Roulette probability 
Share this thread: 
#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: 4,513

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: 4,513

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: 4,513

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: 1,776

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 


Register to reply 
Related Discussions  
Roulette ball trajectory (after impact)  Introductory Physics Homework  2  
Here is the roulette cheat cookbook (Fake?)  General Physics  8  
Roulette F=mv^2/r  Classical Physics  0  
Answers to a Roulette problem  General Physics  1  
Man bets all on Las Vegas roulette spin  General Discussion  5 