Simple as 1,2,3

[RouletteGhost]

I've started double street threads about this before

It's very rare to have 6 unique lines in a row

So my idea was to wait for 3 unique lines (doesn't take long, happens fast) and bet them

If you lose now you have 4 unique lines bet those 4

If you lose now you have 5 unique lines bet those 5

99 percent of the time you will win

The problem is the progression

It's like 1 6 30 or something crazy like that

I haven't had the time to tweak it

[DoctorSudoku]

I've started double street threads about this before

It's very rare to have 6 unique lines in a row

So my idea was to wait for 3 unique lines (doesn't take long, happens fast) and bet them

If you lose now you have 4 unique lines bet those 4

If you lose now you have 5 unique lines bet those 5

99 percent of the time you will win

The problem is the progression

It's like 1 6 30 or something crazy like that

I haven't had the time to tweak it

Maybe a small tweak would be to wait for 4 unique double streets and then jump into the fray. Of course, the waiting time would increase, but that is a small price to pay if it makes your new resultant progression a little bit more tolerable.

[3Nine]

Or, you can play it as described in this thread.

[3Nine]

Here is another view including Splits and Streets:

[praline]

can this approach be useful?

[3Nine]

can this approach be useful?

Well, it looks nice!

[praline]

another observation...

[3Nine]

We were discussing a lot about repeaters and unique sequences without repeats etc. We also established that normally a repeat happens within half of the unique outcomes. Why not convert it into something that is playable in 3 simple steps. The key fact behind this system is "A repeat happens within the first half set of unique outcomes more than the second half". I tend to go with my favourite, the line.

Step 1
The first line that comes in, place 1 unit bet on that line. If a repeat happens at this stage, then the repeat has to be that line.
Win - Restart step 1. Lose - Proceed to step 2.

Step 2
Place 1 unit bet on the first line. If a repeat happens at this stage, then it is 50-50 between first line and second line. However, we are choosing first line. I dont know a very good reason why, so dont ask me why
Win - Restart step 1. Lose - Proceed to step 2.

Step 3
Place 1 unit bet each on the second and third line. I am taking the probability of repeat happening between first line and 2nd and 3rd lines. 2nd/3rd has twice the odds of repeating as compared to the first. So we go with 2nd and 3rd. Dont ask me why you did not compare 1st/2nd vs 3rd. No, I dont have an answer Win or lose with a repeat, go to step 1. Else, repeat Step 3 with last 3 uniques until a repeat happens.

Have loads of fun.

Thanks again for this, Priyanka.  Very easy on the mind.   

[3Nine]

I'm wondering if there's a way to integrate Dozens here...  so many ways to improve the game!

[maestro]

would we expect to hit in less spins if we aim for dozen,since there are 12 numbers

[Herby]

Else, repeat Step 3 with last 3 uniques until a repeat happens.

Maybe it's better not to continue:
The probabilities for any line repetition in % are:

{{1, 0}, {2, 16.6667 %}, {3, 27.7778 %}, {4, 27.7778 %}, {5, 18.5185 %},
{6, 7.71605 %}, {7, 1.54321 %}}

Alltogether (without zero) a zerosum, so everybody can easily calculate the break-even point.
But not me, because it's beyond midnight. 

[praline]

Maybe we can create lines from dozen, by stitching cycles and number of dozens before repeat happen?

[praline]

Playing Mortagon's spins from today, it gives -7

Are we sure?

[praline]

I'm wondering if there's a way to integrate Dozens here...  so many ways to improve the game!

I think there is a way, but it will have max 3 steps.

[3Nine]

hi Praline - you have mail!