Question about roulette probability

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To calculate the number of spins needed for a sector of 6 numbers to win with a probability of 0.99 in European roulette, first determine the probability of not hitting any of those numbers in n spins. The probability of not hitting a number in one spin is 27/33, leading to the formula (27/33)^n = 0.01. Solving this equation will give the required number of spins. The discussion emphasizes understanding the calculations rather than just the results, indicating a need for clarity in mathematical expressions. This approach helps in grasping the underlying probability concepts in roulette.
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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:)
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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