Question about roulette probability

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Discussion Overview

The discussion revolves around calculating the probability of a sector of 6 numbers winning in roulette after a certain number of spins, specifically aiming for a probability of 0.99. The context includes theoretical probability calculations related to European and American roulette wheels.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant asks how to calculate the number of spins required for a sector of 6 numbers to win with a probability of 0.99.
  • Another participant suggests calculating the probability of not getting a number from the selected sector in the first n spins as a potentially easier approach.
  • A participant points out the need to specify whether the discussion is about a European or American roulette wheel, explaining the differences in the number of total outcomes and the corresponding probabilities for each type.
  • It is noted that for a European wheel, the probability of not hitting a number in the selected group on one spin is 27/33, while for an American wheel, it is 28/34.
  • The formula for the probability of n spins without hitting a number in the selected group is provided for both types of wheels, leading to equations that need to be solved for n.
  • A participant clarifies that they are referring to a European wheel and expresses uncertainty about the mathematical syntax used in the discussion.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the calculation method, and there are varying levels of familiarity with the mathematical concepts involved.

Contextual Notes

There is a lack of clarity regarding the mathematical syntax and notation used in the discussion, which may affect understanding for participants not familiar with such expressions.

ninko
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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 [tex](9/11)^n[/itex] on a European wheel, [itex](14/17)^n[/itex] on an American wheel. You need to solve [itex](9/11)^n= 0.01[/itex] or [itex](14/17)^n= 0.01[/itex].[/tex]
 
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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