Roulette Odds Help Requested

I'm new to this forum and was hoping to get help on a few odds questions pertaining to roulette.
The Givens:
6 lines (1-6), (7-12), (13-18), (19-24), (25-30), (31-36)
First 4 spin results, each number landing in a different line.
Examples: 3, 11, 14, 31 / 1, 8, 18, 25 / 12, 19, 30, 31
Questions:
1/ What are the odds on the 5th spin in all of the examples above,
that the winning number will be in any of the (4) lines that just
came out? (NOT in (1) of the (2) lines that didn't come out).
2/ Lets assume that the 5th winning number came out in (1) of those
(2) unhit lines. In other words, first 5 winning numbers were in
(5) different lines, i.e. 3, 11, 14, 31, 26. What are the odds in
this case that the next winning spin will be any number in any of
the (5) lines that just came out? (not in the only line that did not come out).

And I have read about a strategy that sounds good. Place equal bets on the
first 5 lines above and the 2nd and 3rd columns. Same bet, win or lose,
no progression. Thoughts on this one both short and long term?

I am trying to come up with a strategy utilizing both of the above.
For example, playing every spin, start with the 7 bets above on the
lines and 2nd and 3rd columns. Then when 4 numbers come out in
4 different lines, switch the bet to those 4 lines, perhaps doubling
the size.

Any help will be much appreciated from this newbie.

Retroplaces

Past results are no indication of future events.



If I understand your question correctly, for your first question, you want to know the odds of the ball falling on any of 6 numbers. The odds are 6:38. (or 6:37 for a single zero wheel).

For your second question, I think you want to know the odds of the ball falling on any of 30 numbers. The odds are 30:38. (or 30:37).


The odds of ANY particular number hitting are ALWAYS 1:38 (or 1:37).

And past results are no indication of future events.

Whoa, look out! I think you're about to get lambasted.

The important thing to remember is that the wheel has no memory. What happened before has no effect on what will happen in the future. So your basic question of "Assume A happened, what is the probability that B will happen?" just boils down to "What is the probability that B will happen?"

So the answer to 1/ is 24/38 = 63.15% on an American roulette wheel. And the answer to 2/ is 30/38 = 78.95%.

You can develop any strategy you want for how to place your bets on the layout. Different strategies will result in many different possibilities of wins and losses. But you can always expect to lose 2/38 = 5.26% of every dollar you place on the board, on average.

Thanks for your answers.
I didn't explain myself clearly.
My first question is the odds of the 5th winning number
being in any of the lines that just came out. In other words,
the 5th number would be any of the 24 numbers in the 4 lines.

Same for my next question,assuming the first
5 numbers came out in different lines. Now what are the odds
that the 6th number will be any number among those 30 numbers
in the 5 lines?

Example: 1, 7, 18, 22, 27 came out.
These are in 5 different lines.
What are the odds that the 6th number will
be any number from 1-30 (the 5 lines that just came out).
That 31-36 will NOT come out.

Let me know if this makes more sense.

Thanks

Thanks, a bell just went off in my head.
That's why I'm still a newbie at this (o:
I just have to remember the no memory factor
and do the math like you did.

That's a tremendous help for me.
Winning possibilities over 38 on American
and over 37 on European, regardless of
previous spins.

Many thanks.

There are 38 pockets on a 00 Roulette wheel. A "fair" payout would be 37:1, which means if you bet $1 on a number and it hits, you should get paid $37 and keep your original $1. A total of $38 returned. If you were to bet all 38 numbers (1 => 36 + 0/00) you would lay out $38 and, no matter what number comes up, you would get $38 back. That is the essence of a "fair" game.

But, the game isn't fair. You only get paid 35:1. The fastest and most surest way to lose $2 in a casino is to bet all of the 38 numbers on a 00 Roulette wheel. You lay $38. No matter what number comes up, you get $36 back. This is the essence of the house advantage (HA), which is computed as follows:

HA = 1 - ((total return) / (total bet))

Roulette HA = 1 - (36 / 38) = 5.26%

That's pretty much all you need to know about the game.

The only way to "beat" the game is to reduce the probability that the number(s) you are betting have a 1 in 38 chance in coming up. This is what bias wheel and VB (Visual Ballistics) play is all about. In prime bias & VB play, the numbers that are bet have a 32:1 or better chance of coming up. But, you still get paid 35:1. Sweet.

That's pretty much all you need to know about beating the game.

Thanks, makes sense.
I'm sure that finding bias wheels and VB takes a lot of skill and a lot of time learning how to, right?
No easy way is my guess.
If they were easy to spot, everyone would be spotting them, right?
Thanks for your help.

I know absolutely nothing about odds. But what I do know about roulette is that the layout on the table is designed to mislead you. Since you seem to want to play the 'inside' you should look and see that the numbers as they are grouped on the layout are strategically placed so they are not next to each other on the wheel.

Good luck with your strategies!

Thanks, I'm picking up a lot from those on this forum including yourself.
Thanks again.