Hello.

Can anybody please work out for me the probability of a particular number coming up on a roulette wheel and then not coming out in the next 129 spins. The wheel is of the single '0' variety. Would i be correct in assuming the calculation is:

(1/37) x (36/37)^129

I might have stumbled on a way to make lots and lots of money if the answer is what i want to hear... (although im not holding my breath)

Much obliged

Matt.

 

That seems reasonable to me, if you are seeking a specific number before the first spin, and that number must not come up again.

If you are watching/waiting for a number to come up, and then seeing whether or not it comes up again, or just taking whatever the first number is and seeing if that comes up, the probability does not require the first 1/37.

Many great minds have searched for a technnique to beat Roulette (other than characterising a table, which is considered cheating) and failed... stats is not an easy topic, please don't be too optimistic..

 

Fosters is telling me it's 0.085%

 

Ah, dont i know it! Probably needed a *raising eyebrows* smilie on the end of that first post.

It's just for a bit of recreational statistics really as i'm trying to prove a point to somebody. Im well aware you cant really beat it unless the Gods smile on you from cradle to grave. Still, gives me something else to think about i suppose!

Cheers!

 

=(1/37)*([36/37]*[36/37] {do this 129 times})*100 = 0.08%

Don't look like it'll happen

(first answer's not quite right only did it 126 times )