# VLS Roulette Forum

## Study Groups => Study Groups => Law of the Third => Topic started by: simon on May 03, 2009, 12:07:51 PM

Title: question
Post by: simon on May 03, 2009, 12:07:51 PM
Hello Mr. Six, can you please tell me how to apply the law of the third to a situation where there are ten possible outcomes?  I am working on finals systems and there are ten possible outcomes for the finals (0 - 9) (although 3 sets of finals 7-9 have three number each, 6 sets of finals 1-6 have four numbers each, and one set of finals the zeroes has 5 numbers on American wheel) but basically I want to know within ten spins (assuming say 4 numbers in each final), how many of these ten possible outcomes can we expect to show within ten spins and how many can we expect to repeat?  Thank you!
Title: Re: question
Post by: Number Six on May 04, 2009, 10:54:19 AM
Simon,

The amount of numbers contained in each grouping is something of an irrelevance. The wheel might as well only have ten numbers, 0-9. Considering the number of combinations/permutations, or how those groups can appear in 10 spins, the answer you are looking for is 3 to repeat. However, 10 spins is a relatively small number, meaning 3 to repeat is an unreliable expectation. All 10 groups/finals could easily appear in some order in the 10 spins. I really don't think the law of the third can be applied with reliability to such a small sample. As the number of spins in the interval diminishes, so does the law. It doesn't appear as strongly anywhere as it would over 37 spins.
Title: Re: question
Post by: simon on May 04, 2009, 11:59:52 AM
Thank you Number Six.