Idea for new strategy based on sleepers/uniques

[ego]

Here is one example where the assumption from both is wrong.
For example lets take this string of formations

PPB PPP PBP PBB BPP BBP

Now the assumption is that if you get PP next you can not match with PPB or PPP as both alredy has hit.
But this is wrong.

If you look at the sequense again you also see that all P combinations has hit.
So even if we would get PB we could not match with PBB or PBP as both alredy hit.

Here is the sequense again.

PPB PPP PBP PBB BPP BBP

Now you bet Before you have a result and you bet P once with one single bet and you cover all previos four patterns to have a chance to repeat once, no matter if you get PP or PB.

PPB PPP PBP  PBB BPP BBP P

Easy and it takes a total of five bets to cover all possibilitys from this position.
Assume you got a B result, then you can see above that we have BPP and BBP.
This means that if you get BP next you would bet P and if you got BB you would bet P next.

PPB PPP PBP PBB BPP BBP BP
PPB PPP PBP PBB BPP BBP BB

But lest say you have a new situtaion similiar towards the PPB and PPP example where you can not bet both.
For example BPP and BPB where both begin with BP and you not match your next bet to cover the two possibilitys.

And again the solution is similiar as the first one.
You only have to wait for one B to show and bet P next to cath the begining of the formation BP.
One single bet.

PPB PPP PBP PBB BPP BPB BP
PPB PPP PBP PBB BPP BPB BP

This work as a cancelation march where you then force for a non repating formation to show.
So we cover both situation with one single bet each.

Now to the final stage where you have seven formation and only one missing pattern to complete the hole cycle to alternate with no present repeats.

For example with following sequense:

PPB PPP PBP PBB BPP BBP BBB
PPB PPP PBP PBB BPP BBP BPB

Then you have to bet up to Three times to cover all possibilitys.
Total five bets with the othe two bets.

And again we are using the cancelation term.
As all four formations with P patterns has a show, then the only way to cover all possibilitys to Catch one repat among the them is to bet P once.
If a loss you force the beginning of the last B pattern to show, which is missing or will repeat.

PPB PPP PBP PBB BPP BBP BBB P
PPB PPP PBP PBB BPP BBP BPB B

Now assume we don't get a repeat of the previos P patterns and we get a B result.
Then we again face a situation where we have to patterns that begin with the same formation and end with two differen results, BPP BPB.
So our next bet after B show is for P once to cover the BP formation and cover both patterns BPP BPB to repeat once.

Now to the last final cancelation bet and we have lost four bets with no present repeat.

PPB PPP PBP PBB BPP BBP BBB BB

Here you simply bet for BBB to repat once and if not then you have 8 uniq patterns show with no present repeat.

This might not improve the original or change any thing, but i just want to show have you cover all possibilitys to repeat once with placing a total of five bets.
As the autors of both Writings argue you can not cover a situation where two similiar patterns has a show, which is wrong.

Cheers

PPB PPP PBP PBB BPP BBP BBB BBB
PPB PPP PBP PBB BPP BBP BBB BBP

[ego]

I can reduce this attempts of betting down to eight or nine bets to cover all combinations - even less depending on how the random bits unfold.
Is like a sensetive march that follow the random bits unfold and follow the flow to Catch a repeat, so many times you just observe five or six no repeats unfold with out betting a single bet or just two bets.
That is the reducing part.

Intiuitiv approch - if some one would like to know how i can explain it tomorrow.

Cheers

[amk]

As always, it would be very interesting to hear your explanation ego.

Thanks for sharing high quality insights!