Have I figured something new out about roulette?

[wyldegibson]

Hi everyone. I'm new to the forum but have paid attention to this site a lot as a guest and I think there truly are many great minds here. I do not pretend to be philosophically or mathematically so I wanted to run this by everyone to see if I'm on to something.

Many systems have been made where you bet black and column 3 or red and column 2 as to offset some losses by hitting more of the other color based on the fact that column 2 has 8 black numbers and only 4 red numbers and column 3 has 8 red numbers and only 4 black numbers. I feel like the thinking is wrong here.

Since we can theoretically assume black or red hits almost 49% of the time am I the only one that thinks its crazy to be betting column 2 as well if your betting black and column 3 if your betting red? Since if we choose red or black right we will also have an 8/18 chance to hit our blacks in column 2 or reds in column 3? That to my eye would seem like a mathematical edge to the game to me. We're getting paid 2/1 on the column that supports the most of the color we're betting on.

The tests of about 9600 live spins so far is supporting my feelings on this but as I am saying now they are just "feelings." I don't know the mathematics behind the edge if there is any. But so far I'm showing a nice steady income playing this way and only having to flat bet because in my eyes as long as I choose the right color I'm also hitting on the 8/18 chance on the color in that column. To further help me out with if I happen to hit one of the 4 blacks or reds in that column I win as well. 

I'm hoping someone with knowledge can help me understand why this is mathematically wrong because right now I can't understand where I'm wrong. To me I see a clear edge here. But how could no one of stumbled on to this before myself? Thats where I think I must be wrong.

Anyways any and all of your help would be appreciative.  I never join forums but I really respect this one and thought if anyone would know or deserved to know some knowledge it would be you guys.

Thanks, Wylde

[ati]

I'm not a math expert, so I'm probably doing something wrong, but here's what I think:
If you bet on red and column 3, you are winning on 12 numbers, and the win average is 2.5 units. With the same bet, you are losing on 15 numbers, average loss is 2 units. The rest is neutral.
Out of 37 spins, you can expect to win 12x2.5=30 units, and you can expect to lose 15x2=30 units. This would be 0 house edge, I hope someone can help me out how to calculate this. The 2.7% house edge has to be always there somewhere.

[Asxetos]

Wylde both ways have been tested over the so many years and none of them produced any advantage.
I hope I helped.

[wyldegibson]

Wylde both ways have been tested over the so many years and none of them produced any advantage.
I hope I helped.

Unfortunately that isn't of any help to me. Just saying something without any support of why it is doesn't really help me out. I think ATI is much more closer to the mark but I think it may be even better than a 0% house edge. The way I look at it I'm actually covering all 18 red numbers if betting red and than also getting paid 2 to 1 on column 3 where 44% of the reds lie in. Plus if the wheel hits 6, 15, 24 or 33 I also get paid 2 to 1. Basically in this bet I cover 22 numbers total. I get paid even money if it hits red and than I get paid 2 to 1 on 44% of where the red numbers lie plus 4 black numbers. I'm not saying its always going to hit column 3 I'm just saying the odds would seem much more likely that if red comes up it will be column 3 and if black comes up it will be column 2.

Am I crazy to think that there is a mathematical edge here than? Like I said so far in nearly 10, 000 spins on a live wheel I am showing a clear advantage for the player that seems higher than even the 0% house edge ATI was referring to.  So far it has looked to be more like an 8% edge for the player!

[GLC]

6 Blacks in col 1 costs -1 ea. = -6
6 Reds in col 1 costs -3 ea. = -18
12 Reds in col 2 & 3 costs 0 ea. = +/- 0
12 Blacks in col 2 & 3 wins  2 ea. = +24
0  costs -3  This is the house edge

We can never get away from it.

By the way, welcome wyldegibson

GLC

[wyldegibson]

6 Blacks in col 1 costs -1 ea. = -6
6 Reds in col 1 costs -3 ea. = -18
12 Reds in col 2 & 3 costs 0 ea. = +/- 0
12 Blacks in col 2 & 3 wins  2 ea. = +24
0  costs -3  This is the house edge

We can never get away from it.

By the way, welcome wyldegibson

GLC


Thanks for the reply GLC. I've read many posts from you and know you to be one of the best at money nanagement here.

My question has more to do with if we are betting Red than wouldn't it make sense to also bet column 3 because 44% of Reds are in column 3? We are being paid 2 to 1 in this column like any other and yet the only fair payout column would be column 1 where we have 6 each of black and red. If all 3 columns were built like that I would undoubtedly understand there being no player edge. But all 3 columns are not like this, thus if you are betting red it would seem to have a mathematical edge betting in column 3 and if you were betting black there would also seem to be an edge betting column 2. That is because when Red comes up there is an 8 in 18 chance of it hitting the 3rd column. If black comes up there is an 8 in 18 chance it will hit in column 2. The math on this is absolute barring a biased wheel, etc.

Therefore I'm trying to understand how we don't have a mathematical edge here. Basically if I think Red is coming up and bet only column 3 and red did indeed come up I would have a 44% chance it was in column 3 which is only 1 number (9/18) away from being even money and yet we're getting paid 2 to 1 here! Thus it would seem we have a very good advantage here!

I'm wanting someone to tell me how the math in my head is incorrect here. At first I wanted to simply disregard the idea and just assume the edge couldn't be gained this easily. And maybe it can't although no one has yet to explain how there is no advantage. And in nearly 10, 000 spins I do see an advantage for us. One could say its a small sample size and I wouldn't disagree with that but I'd like to know the logistics as to why its not an advantage. If I do another 100, 000 spins on a live wheel I have a hard time believing I will go all the way from an 8% advantage to a 2.7% disadvantage. Put in other words...the variance would be "extremely" shocking.

I'm really hoping a math guy on here gets ahold of this one.

Thanks GLC, Wylde

[wyldegibson]

Sorry my post got mixed in with my response to GLC. Here it is again.

Thanks for the reply GLC. I've read many posts from you and know you to be one of the best at money nanagement here. My question has more to do with if we are betting Red than wouldn't it make sense to also bet column 3 because 44% of Reds are in column 3? We are being paid 2 to 1 in this column like any other and yet the only fair payout column would be column 1 where we have 6 each of black and red. If all 3 columns were built like that I would undoubtedly understand there being no player edge. But all 3 columns are not like this, thus if you are betting red it would seem to have a mathematical edge betting in column 3 and if you were betting black there would also seem to be an edge betting column 2. That is because when Red comes up there is an 8 in 18 chance of it hitting the 3rd column. If black comes up there is an 8 in 18 chance it will hit in column 2. The math on this is absolute barring a biased wheel, etc. Therefore I'm trying to understand how we don't have a mathematical edge here. Basically if I think Red is coming up and bet only column 3 and red did indeed come up I would have a 44% chance it was in column 3 which is only 1 number (9/18) away from being even money and yet we're getting paid 2 to 1 here! Thus it would seem we have a very good advantage here! I'm wanting someone to tell me how the math in my head is incorrect here. At first I wanted to simply disregard the idea and just assume the edge couldn't be gained this easily. And maybe it can't although no one has yet to explain how there is no advantage. And in nearly 10, 000 spins I do see an advantage for us. One could say its a small sample size and I wouldn't disagree with that but I'd like to know the logistics as to why its not an advantage. If I do another 100, 000 spins on a live wheel I have a hard time believing I will go all the way from an 8% advantage to a 2.7% disadvantage. Put in other words...the variance would be "extremely" shocking.I'm really hoping a math guy on here gets ahold of this one.Thanks GLC, Wylde

[Azim]

Sorry my post got mixed in with my response to GLC. Here it is again.

Thanks for the reply GLC. I've read many posts from you and know you to be one of the best at money nanagement here. My question has more to do with if we are betting Red than wouldn't it make sense to also bet column 3 because 44% of Reds are in column 3? We are being paid 2 to 1 in this column like any other and yet the only fair payout column would be column 1 where we have 6 each of black and red. If all 3 columns were built like that I would undoubtedly understand there being no player edge. But all 3 columns are not like this, thus if you are betting red it would seem to have a mathematical edge betting in column 3 and if you were betting black there would also seem to be an edge betting column 2. That is because when Red comes up there is an 8 in 18 chance of it hitting the 3rd column. If black comes up there is an 8 in 18 chance it will hit in column 2. The math on this is absolute barring a biased wheel, etc. Therefore I'm trying to understand how we don't have a mathematical edge here. Basically if I think Red is coming up and bet only column 3 and red did indeed come up I would have a 44% chance it was in column 3 which is only 1 number (9/18) away from being even money and yet we're getting paid 2 to 1 here! Thus it would seem we have a very good advantage here! I'm wanting someone to tell me how the math in my head is incorrect here. At first I wanted to simply disregard the idea and just assume the edge couldn't be gained this easily. And maybe it can't although no one has yet to explain how there is no advantage. And in nearly 10, 000 spins I do see an advantage for us. One could say its a small sample size and I wouldn't disagree with that but I'd like to know the logistics as to why its not an advantage. If I do another 100, 000 spins on a live wheel I have a hard time believing I will go all the way from an 8% advantage to a 2.7% disadvantage. Put in other words...the variance would be "extremely" shocking.I'm really hoping a math guy on here gets ahold of this one.Thanks GLC, Wylde

You have forgotten to double count the REDS...


18 Reds / 18 Blacks + Zero  =  Even Chance Payout.

Column 3 Reds are : 8 / 37 
Column 3 Blacks are : 4 /37

you are betting 18 / 37 + 12 / 37 = 30 / 64   This is your chance to hit on both the Even Bet and the Column Bet.

[ausguy]

Wylde  - I think you are off target here with your "NEW/OLD" roulette idea. This is because the EC bet & the Column bet are 2 seperate events. Your bet plan, or methods close to it, have been around for nearly as long as roulette has & that goes back at least a couple of 100 years.

If it worked consistantly than everybody would be on to it & winning which would either send casinos broke or force them to change their rules & so ban that type of bet. That just AIN"T the case as casinos are still alive & kicking everywhere.

On a single zero wheel we all all know that the EC odds are 50% - 2.7% = 47.3%. On the column it's 33.3% - 2 .7% = 30.6%.

To make it easier lets delete the zero from the calcs. here. If you put $10 on RED & $10 on Col. 3. If 9R drops you make $30.

If 19R drops you break even $10W - $10L = $0. If 33B drops you make $10, @ $20W - $10L. For column 1 & 2 any Red you break even & any black you lose $20.

The Red bet is always the same chance 50/50 & the column bet is 66.7% against & only 33.3% for it. On column 3 with the 8 reds you have a 44.44% chance to win & a 55.56% chance to lose. With the blacks you have a 22.22% chance to win & a 77.78% chance to lose.

I've got some live dealer spins here so lets do a 20 line mini test to see what's what here. I'm sitting on Red & Column 3 for the bets & flat betting $10 on each position..

1)26B LL -20, 2) 18R WW +30 - 20 = +10, 3)29B LL - 20 = - 10, 4) 15B LW +10 = 00, 5) 27R WW = +30, 6) 3R WW +30 = +60, 7)28B LL - 20 = +40,
8)26B LL -20 = +20, 9) 18R WW +20 = +40, 10) 29B LL -20 = +20, 11) 19R WL +/- 0 = +20, 12) 34R WL +/- 0 = +20, 13) 29B LL -20 = 00,
14) 19R WL +/- 0 = 00, 15) 22B LL -20 = -20, 16) 9R WW +30 = +10, 17) 11B LL -20 = -10, 18) 11bB LL -20 = - 30, 19) 5R WL +/- 0 = -30,
20) 2B LL - 20 = -50.

I realise it's only a short sample & winning & losing streaks can come at any time.

I suggest you run some longer tests as that is the best way I know to get your head around most ideas. You may also do well playing a progression if you aren't already doing so ? Variables make the betting equation more complex.

If you mostly win at testing then you could expect a reasonable chance of success with real money play.

[wyldegibson]

Yes Azim I agree that no matter which column I bet on it is still 12/37 chance. The difference is being that if I'm betting on red and column 3 and red hits it IS more likely that column 3 will hit more often than the other two columns. The same would be said if I'm betting on black and black comes up it IS more likely that column 2 will hit over the other two columns. There is no doubt in my mind about this. Column 3 is twice as likely to hit red over column 2 and 25% more likely over column 1. Column 2 is twice as likely to hit black over column 3 and 25% more likely over column 1.

I'm not really sure how I can  simplify this anymore. If I bet red and column 3 and red hits I have a 44% chance that I will hit column 3 because I have an 8/18 chance of my red number hitting this column. Based on a payout of 2 to 1 I should only have a 33% chance of hitting this column on red (like in column 1) but that isn't the case.

Can anyone understand what I'm saying here?

[wyldegibson]

Wylde  - I think you are off target here with your "NEW/OLD" roulette idea. This is because the EC bet & the Column bet are 2 seperate events. Your bet plan, or methods close to it, have been around for nearly as long as roulette has & that goes back at least a couple of 100 years.

If it worked consistantly than everybody would be on to it & winning which would either send casinos broke or force them to change their rules & so ban that type of bet. That just AIN"T the case as casinos are still alive & kicking everywhere.

On a single zero wheel we all all know that the EC odds are 50% - 2.7% = 47.3%. On the column it's 33.3% - 2 .7% = 30.6%.

To make it easier lets delete the zero from the calcs. here. If you put $10 on RED & $10 on Col. 3. If 9R drops you make $30.

If 19R drops you break even $10W - $10L = $0. If 33B drops you make $10, @ $20W - $10L. For column 1 & 2 any Red you break even & any black you lose $20.

The Red bet is always the same chance 50/50 & the column bet is 66.7% against & only 33.3% for it. On column 3 with the 8 reds you have a 44.44% chance to win & a 55.56% chance to lose. With the blacks you have a 22.22% chance to win & a 77.78% chance to lose.

I've got some live dealer spins here so lets do a 20 line mini test to see what's what here. I'm sitting on Red & Column 3 for the bets & flat betting $10 on each position..

1)26B LL -20, 2) 18R WW +30 - 20 = +10, 3)29B LL - 20 = - 10, 4) 15B LW +10 = 00, 5) 27R WW = +30, 6) 3R WW +30 = +60, 7)28B LL - 20 = +40,
8)26B LL -20 = +20, 9) 18R WW +20 = +40, 10) 29B LL -20 = +20, 11) 19R WL +/- 0 = +20, 12) 34R WL +/- 0 = +20, 13) 29B LL -20 = 00,
14) 19R WL +/- 0 = 00, 15) 22B LL -20 = -20, 16) 9R WW +30 = +10, 17) 11B LL -20 = -10, 18) 11bB LL -20 = - 30, 19) 5R WL +/- 0 = -30,
20) 2B LL - 20 = -50.

I realise it's only a short sample & winning & losing streaks can come at any time.

I suggest you run some longer tests as that is the best way I know to get your head around most ideas. You may also do well playing a progression if you aren't already doing so ? Variables make the betting equation more complex.

If you mostly win at testing then you could expect a reasonable chance of success with real money play.

I think you may have involuntary supported my argument. Despite only winning 9/20 of bets note that 5 out of the 9 winning red numbers came out of column 3  which is 55% (higher than the 44% expectancy). If there was 10 reds and 10 blacks in these 20 spins you would of came out ahead despite only winning 50% of your bets. This is flatbetting here.

This only confirms more to me that this idea has legs....

[ausguy]

Wylde - I just remembered a bet with some elements of what you talk about except that it involved H or L & a single dozen bet to cover 30 numbers on the table.

It went like this : 3 units on Low & 2 units on 3rd Doz. or 3 on High & 2 on 1st doz. At loss risk is zero & 6 numbers in 2nd doz.

Any win gives a +1unit gain. Any loss is - 5 units. To get ahead 1 unit needs 6 wins for every 1 loss.

Thinking about your EC & 2 : 1 bets I'm now uncertain if the base bets would work better with 3 units on Red & 2 units on Col.3 ?

A loss gives you -5. An EC only win gives you +1. A Col. win only gives you +1 & a dbl. win gives you +7. Any win is at least +1 with nil break evens.

As always it's risk VS reward.

[wyldegibson]

Wylde - I just remembered a bet with some elements of what you talk about except that it involved H or L & a single dozen bet to cover 30 numbers on the table.

It went like this : 3 units on Low & 2 units on 3rd Doz. or 3 on High & 2 on 1st doz. At loss risk is zero & 6 numbers in 2nd doz.

Any win gives a +1unit gain. Any loss is - 5 units. To get ahead 1 unit needs 6 wins for every 1 loss.

Thinking about your EC & 2 : 1 bets I'm now uncertain if the base bets would work better with 3 units on Red & 2 units on Col.3 ?

A loss gives you -5. An EC only win gives you +1. A Col. win only gives you +1 & a dbl. win gives you +7. Any win is at least +1 with nil break evens.

As always it's risk VS reward.

Yes the way I've been playing is either 5 on red and 10 on column 3 or 5 on black and 10 on column 2. In this way if I hit my red or black  but miss my column i am only out 5 bucks but make a sizeable profit if my column hits. Based on what I feel is a strong success rate if I am able to choose the right color I feel like this works very well. After about 10,000 real spins I already came to that conclusion. But the empowerment of having the math behind what I'm saying is what would really put me at ease playing this "long term."

Maybe I'm crazy but while I understand its a crapshoot in choosing red or black there is still almost a 50/50 chance we choose correctly. So in doing so wouldn't we want to have the power that if we choose the right color we have a 44% chance we chose the right column as well by simply just playing the column that has the most red or blacks in them? Now instead of having only a 33% chance in hitting our 2 to 1 we have a 44% chance!

Thanks for looking more into this Aus as I feel its certainly something worth discussing.

[ausguy]

Wylde - I've been trying to get my head tuned into this & I think I have it calc'd ? I was always thinking no matter how you tweak things you can't alter the odds the casino presets, especially relevant here are the 2 : 1 bets. As in Column 3 = 33.33%. So I thought we/I are missing something here & with a bit of number scribbling I got something sensible (well at least in my twisted mind) that adds up if you ask yourself the right kind of questions.

As always we're putting zero off to the side.  Now Red has a 50% chance of a hit out of 36. Then Red 8 reds in column 3 has a 44.44% chance out of 18 of a hit. (8/18).

BUT it's 44.44% of 50% = 22.22% the overall odds of a Red hit (8/36). Now not forgetting the 4 blacks.  It's 4/18 =  22.22% chance of being in column 3 (notice black % is 1/2 of red which is how it must be) & again, BUT it's 22.22% of 50% = 11.11% the overall odds of a black hit (4/36).

Now when you add the the red & black column overall % we get 22.22 + 11.11 = 33.33 %.

Another way to look at it is, when the ball is spinning the odds of Red hitting is 50%. The odds of a Red being in Column 3 is 22% The odds of black hitting is 50% & the odds of Black being in Column 3 is 11%.

It's only after the ball lands that the chances are relevant as 50% of the equation is eliminated as soon as the ball stops. So if Red spins up the chances of it being in column 3 is 44% & if Black comes in then it's 22%. Halving the 36 automatically doubles the remaining %'s.

I hope that makes some sense ?

For numbers covered you have 10 Red 1 to 34 & 12 all up in Col.3 = 22 in total & 14 blacks not covered. This is layered with the Col. overlap of 8 Reds & 4 Blacks. To add further to the degree of difficulty here (calc'ing) you have stepped value bets. Do you just static bet of use a progression ?

For testing I'd suggest some simulated sessions of how you would actually play the game, timewise if you follow what I mean. 2 or 3  x2 hour sessions over a week should show you if it's long term viable ? If your looking to play this live dealer many on line casinos plays are 1 spin/min. so 60 p/hr.
I'm only saying Live dealer as I don't play RNG or Autowheel because I think they cheat.

Following the wheel might work ? That is if last spin was Red then back Red & Col 3 if Black then B & Col 2. Perhaps if it's Col 1 you could stop Column bets in case of a C1 run & resume betting when C2 or C3 hits. Probably no need to stop R/B bets ?

[RouletteKnight]

There is no advantage whatsoever in betting red and 3rd Columns (which have more reds then black).
To investigate this phenomenon, we try betting 1 unit each on red and 3rd column and 1 unit each on red and 2nd column
to see if there is any advantage from the first style of betting...

Bet 1 unit on Red and 3rd column each in European Roulette

Chance of losing 2 units: is 15/37 (Blacks in 1-2 column and Zero)
Chance of no win or loss: is  10/37 (Reds in 1-2 column)
Chance of winning 1 unit: is 4/37 (Blacks in 3rd column)
Chance of winning 3 units: 8/37 (Reds in 3rd column)

Out of 37 spins you are expected to have a net gain/loss of:
(-2*15)+(10*0)+(1*4)+(3*8 )
=-2

bet 1 unit each on Red and 2nd Column

Chance of losing 2 units: 11/37 (Blacks and zeros in columns 1 and 2)
Chance of no win/loss: 14/37 (Reds in 1st and 3rd column)
Chance of winning 1 unit: 8/37 (Blacks in 2nd column)
Chance of winning 3 units: 4/37 (Reds in 2nd column)

Out of 37 spins you are expected to have a net gain/loss of:
(-2*11)+(14*0)+(1*8 )+(3*4)
=-2

So you can see the end result is the same of -2 units.

The reason for this is simply because although betting red and 3rd columns seems logical seems you gain
more profit more times when the 8 red numbers in 3rd column hit, but you forget to include the fact that you cover "less numbers". While betting red and 2nd column seems illogical, more numbers are covered thus there's lesser chance that you win 3 units, but there's also a lesser chance you loss both your bets. Thus both of this discrepancies negate each other.