Selection

[ego]

You cover high and dozen one and betting against line three.

The selection process:

One line has 50% probability to hit once within four attempts, that is equal to an even money bet.
That is based upon binomial probability calculation.

Charting and tracking:

You count each and every spin, each time you get four hits cycle with no hit on the line three you write across X.
If the line hit at attempt one within a cycle of four you write down 1 and start over counting to four attempts cycle.
Sometimes you will get several hits within a cycle of four attempts and sometimes you will get several crosses, both will have 50% probability to show.

When to bet:

Each time you get a hit within a cycle of four you bet that line three will not hit again and the same attempt.
For example, if you get attempt three hitting and start over counting and you reach three attempts then you bet against a repeat.

This is how it looks like:

34XX1X42X313312XXX4X13XXXX2X4X3XX3X
  W     W  W   W L WW    W WW      W W W  W

[poluvolo]

Hi could you please be more clarify?
one more example with a small sequence spin by spin
I am still a little confuse even it seems smart bet selection
thanks

[ego]

I will try to make one more explanation.

You see a live wheel and the first result is line 1 then you continue to watch the wheel and the next results are line 6 and next result line 6 again and result after that is line 4.
Now line 1664 has hit and are one cycle of four attempts, that has 50% probability to happen.

This means you write across X now line 3 has 50% probability to show at least once during four attempts.
So assume you get four more results with lines except line 3, then you have a new cycle with no hit and write across X.

Now you have XX and next, you get to line 2 and after that line 3, so line 3 hit at the second attempt and you write 2 standing for line 3 hitting at second attempts  XX2
Now you wait for next results, let's say line 5 hits and now you play that line 3 will not hit at the second attempt.
If it does you have a repeat and if line 3 is not hitting you have a winning bet.

Cheers

[Ross]

You see a live wheel and the first result is line 1 then you continue to watch the wheel and the next results are line 6 and next result line 6 again and result after that is line 4.
Now line 1664 has hit and are one cycle of four attempts, that has 50% probability to happen.

This means you write across X now line 3 has 50% probability to show at least once during four attempts.
So assume you get four more results with lines except line 3, then you have a new cycle with no hit and write across X.

Now you have XX and next, you get to line 2 and after that line 3, so line 3 hit at the second attempt and you write 2 standing for line 3 hitting at second attempts  XX2
Now you wait for next results, let's say line 5 hits and now you play that line 3 will not hit at the second attempt.
If it does you have a repeat and if line 3 is not hitting you have a winning bet.

Anyone figured this out?
I like the idea of the probability increasing
but can't get beyond the idea here that the
odds of a 1:3 bet becomes a 1:2 bet after three
spins.

[Thanatos]

I will try to make one more explanation.

You see a live wheel and the first result is line 1 then you continue to watch the wheel and the next results are line 6 and next result line 6 again and result after that is line 4.
Now line 1664 has hit and are one cycle of four attempts, that has 50% probability to happen.

This means you write across X now line 3 has 50% probability to show at least once during four attempts.
So assume you get four more results with lines except line 3, then you have a new cycle with no hit and write across X.

Now you have XX and next, you get to attempt 2 and after that line 3, so line 3 hit at the second attempt and you write 2 standing for line 3 hitting at second attempts  XX2
Now you wait for next results, let's say line 5 hits and now you play that line 3 will not hit at the second attempt.
If it does you have a repeat and if line 3 is not hitting you have a winning bet.

Cheers

Nice one, thanks Ego.

I made a small fix in the quote above 

I think what is confusing most people is that they cant figure out the win/loss example in post one as the forum changes the spacings so the 2 lines arnt the same length, so here i attempted to line em up (hope it works, or else copy into Noteblok/Wordpad, that should format the text equally):

34XX1X42X313312XXX4X13XXXX2X4X3XX3X
xWxxxWxWWxWWLxWWxxxWxWWxxxxWxWxWxxW

Line 2: W=win, L=Loss, x=no bet (wait for next circle/set of 4)

If i understand correct, you bet in sets(cycle) of 4 spins like:
1664 = X and you cant bet next set.
5622 = X and you cant bet next set.
4361 = 2 and next set you bet at 2nd spin only
1116 = X and you cant bet next set (but you won 2nd bet)
2113 = 4 and next set you bet at 4nd spin only
3166 = 1 and next set you bet at 1nd spin only (but you won 4nd bet)
3333 = Unknown (also you lost bet 1)

That brings me to the question: what do you bet if  you get multiple hits of Line 3 in the cycle/set of 4? Could happen 2 or 3 times (or worse) if really unlucky (rare).

Of cause the good thing about only betting once every cycle/set is you dont loose more than once so if line 3 gets super hot (that can happen) then you survive with just a single loss.

[Thanatos]

Wait! no! im wrong in above post it dosnt add up. The example i made is also wrong, as after a win/loss you appear to completely retrack 4 new spins (i think).

[ego]

34XX1X42X313312XXX4X13XXXX2X4X3XX3X

I will try to explain.

You count each hit if you get four hits without line 3 hitting once you have a cross.

X

Now assume you continue to spin the wheel and you get all kinds of lines hitting except line 3 for the next four attempts, then you have a cross again.

X X

Now you continue to count in cycles of four and spin the wheel, but now line 3 hits at second attempts, then you write a 2.
That indicates that line 3 has hit at the second attempt.

X X 2

Now you start over and count in four attempts cycles when you reach two attempts, then you bet that line 3 will not repeat at the same position and attempt during this new cycle you bet there will not be a repeat.
Let's say line 3 does not hit for the next four attempts and you get a new cross.
Then that bet won.

X X 2 X

Let's go back to another situation, let's say line 3 hits at first attempt and we write down a 1.

X X 2 1

Then we don't bet after the second attempt. Now we bet after the next first attempt.
Hope this is clear now.
And yes you bet once against one repeat and continue with next trigger.

Cheers

[Thanatos]

Thanks! I think i got it.

Also, waiting for a new trigger before each bet is definately also a lot more safe 

[Bigbroben]

I believe you do not have equal bets on high and doz1? 3u and 2u?

What do you estimate to be the w/l ratio?

[ego]

I attach system that was in my mind improving the selection method.
Having the staking plan in mind.

Cheers

[Thanatos]

Thanks for the detailed system description. Regression does make a lot of sense with that high a hitrate/coverage.

If anyone was wondering: yes the 3 & 2 bet on high+dozen (or low+dozen) is technical the same as betting 1 unit on 5 lines. However the 3 & 2 is usually mush faster to bet so is good to know if you are on a tiny timelimit.

[ego]

Don't forget there are online wheels with a low minimum and La Partage Rule - then if zero strikes you only lose half the staking on EC

[Ross]

There's a script for this idea on the Roulette Extreme web site.

Just for fun I wrote a programme to see how this idea performs.
Click on either of the Test buttons.  The maximum spins between
hits will shown - it's usually 8 or 9 so I haven't checked for what
other gaps there are.

Seems odd that the max gap happens usually late in the 1000 spin
test - but sometimes early. 

[ewarwoowar]

Hi Ego and thanks for your twist on an old favourite of mine.

I have a couple of questions if I may?

1. When tracking, do we just ignore 0 or count it as a no show of line 3? I.E. 2, 4, 1 and 0 = X, or 2, 4, 1, 0 and 5 =X?
2. Line 3 hits on 2nd spin on our cycle of 4. Do we spin out the other 2 spins and then begin our next cycle of 4, betting against the 2nd spin?
3. If the above is correct and line 3 hits on spins 1 and 3 of our cycle of 4, do we bet twice on the next cycle of our 4?

Hope this makes sense.
Thanks again,

Cheers.