Roulette 3rd's Strategy Odds Questions
I have a few questions and I was hoping someone could help me understand the odds? I like to play the 3rd's as I see it as an advantage on both American and European Roulette. For American I play 1st and 3rd sections same bet and for European I play 2nd and 3rd section. I think the odds are greater then 66% due to the large groupings of numbers for those wheels but, I don't know how to calculate that?
1) What are the odds when playing European Roulette Middle and Top 1/3's?
2) What are the odds of the ball landing in the 1st 3rd of European Roulette 6 times in a row. I know the ball does not have a memory but, it seems like this would be a rare occurrence.
3) How about American Roulette what are the odds of you place same bet on 1st and 3rd sections?
Thank You!
2. 12/37^3= 3.41%
3. 24/38ths chance of winning. 2 units bet, 3 units won when it hits. 3(24/38) - 2 = -.105 units/spin expected loss
Quote: BillionDollarFor American I play 1st and 3rd sections same bet and for European I play 2nd and 3rd section. I think the odds are greater then 66% due to the large groupings of numbers for those wheels but, I don't know how to calculate that?
If you're playing 24 out of 37 or 24 out of 38 numbers , which equates to 64.9% or 63.2% , why would you feel your chances of winning are greater than 66% ? Where your numbers are on that actual wheel doesn't matter
seconded. You are deluding yourself if you think that the way the numbers are arranged on the wheel is of any consequence.Quote: michael99000If you're playing 24 out of 37 or 24 out of 38 numbers , which equates to 64.9% or 63.2% , why would you feel your chances of winning are greater than 66% ? Where your numbers are on that actual wheel doesn't matter
Think... The probability of landing in an area of the wheel is equal to the ratio of that area to the total area. Breaking the area up or having it continuous makes no difference. If you bet all three columns at the same time, then that is one or two massive continuous areas. Doesn't help you miss the 0 and 00.
Quote: OP2) What are the odds of the ball landing in the 1st 3rd of European Roulette 6 times in a row. I know the ball does not have a memory but, it seems like this would be a rare occurrence.
Quote: rdw4potus2. 12/37^3= 3.41%
Probability of landing there once = (12/37)^1 = 32.432432432%
Probability of landing there 6 times = (12/37)^6 = 0.116379672% or 1 time in sets 829 sets of 6 rolls ( 1 time in 835 rolls )
The answer is no - no such strategy can succeed, this has been known for centuries. Move on.
Quote: michael99000If you're playing 24 out of 37 or 24 out of 38 numbers , which equates to 64.9% or 63.2% , why would you feel your chances of winning are greater than 66% ? Where your numbers are on that actual wheel doesn't matter
michael99000 thanks for your reply.
On both wheels the colors alternate red and black giving you an even chance of winning minus the green slots. Agree? When playing dozens the slots do not alternate lower, middle, top make sense? OK, so for American Roulette there are 2 sections of the wheel where all the numbers are lower and top dozen. 27, 10, 25, 29, 12, 8 giving you 6 in a row. Then you have 28, 9, 26, 30, 11, 7 another 6 in a row. All numbers in between are pretty much Low, Middle, Top plus green giving you a 6/7 chance of hitting if you bet Low and Top dozen at the same time. So you have a 100% chance of win if it hits in either of the two outlined sections and 6/7 chance on everything else.
If the dozens were equally distributed I would agree with you but, since there are large groupings I feel like that changes the odds? If you bet 2 columns your odds do not change you are covering 24 out of 38 numbers since the wheel does not follow the pattern of the columns. Same for inside numbers people bet 17 lets say and it hits 18 those numbers are no where near each other on the wheel only on the table so odds never change.
Quote: BillionDollarmichael99000 thanks for your reply.
On both wheels the colors alternate red and black giving you an even chance of winning minus the green slots. Agree? When playing dozens the slots do not alternate lower, middle, top make sense? OK, so for American Roulette there are 2 sections of the wheel where all the numbers are lower and top dozen. 27, 10, 25, 29, 12, 8 giving you 6 in a row. Then you have 28, 9, 26, 30, 11, 7 another 6 in a row. All numbers in between are pretty much Low, Middle, Top plus green giving you a 6/7 chance of hitting if you bet Low and Top dozen at the same time. So you have a 100% chance of win if it hits in either of the two outlined sections and 6/7 chance on everything else.
If the dozens were equally distributed I would agree with you but, since there are large groupings I feel like that changes the odds? If you bet 2 columns your odds do not change you are covering 24 out of 38 numbers since the wheel does not follow the pattern of the columns. Same for inside numbers people bet 17 lets say and it hits 18 those numbers are no where near each other on the wheel only on the table so odds never change.
I'm struggling to understand the corellation between your numbers being grouped together on the wheel and that fact giving you a better chance of winning than if they were spread out randomly. Unless you're telling me you found an unbalanced wheel or a dealer with the skill to hit your sections.
