Question about roulette probability

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Can anybody help with this. How to calculate after how many spins the sector of 6 numbers will win, with probability of 0.99? What is the formula for calculating such problems?
 
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welcome to pf!

hi ninko! welcome to pf! :smile:

with questions like this, it's usually easiest to calculate the opposite …

what is the probability that none of the first n results will be in the sector? :wink:
 
Yes, maybe is easier to make that calc. but how? :)
Thanks anyway!
 
You will have to specify whether you are talking about a "European" or "American" roulette wheel. The standard European wheel has 32 numbers, red and black, together with "0" that is typically green. The standard American wheel has 32 numbers plus "0" and "00". That is the probability of a single turn coming up a specific number on a European wheel is 1/33, on an American wheel, 1/34. If you select 6 numbers, the probability of NOT getting a number in that group on one spin is 27/33= 9/11 on a European wheel, 28/34= 14/17 on an American wheel.

The probability of n spins without getting a number in your set of 6 is (9/11)^n[/itex] on a European wheel, (14/17)^n on an American wheel. You need to solve (9/11)^n= 0.01 or (14/17)^n= 0.01.
 
Roulette wheel has 36 numbers plus zero. I`m talking about European wheel.
Sorry, but I`m not very familiar with mathematics... could you please clarify the sintax tex...itex...
Hope is not very stupid question:)
 

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