Bet selection that won't have huge losing streaks?

[Jordsie669]

Hey, for the HP Johnson progression (in money management)
I'll need a betting selection that doesn't have to win more than lose,
But that just doesn't have large losing streaks!
Any ideas? (The selection should be for R/B)

[Jordsie669]

Maybe by betting uncommon patterns (While not looking to the spun numbers!)
like: RRRR BBBB RRRR BBBB
Would this have large losing streaks?

Because the only way that this pattern will have large losing streaks is when this happens:
BBBB RRRR BBBB RRRR BBBB (The exact opposite of the bet selection)

(I think that this is really uncommon so???)
Cheers, Jordsie.

[superman]

Any ideas?

What have you tried so far?

[ego]

I am not sure about that progression, but you could try this.

Clustering outcomes into patterns of two.

RB
RR
RR
BR
BB
RB
RR
BR

Then one complete cycle would be four events.
So you could bet against one side to show four times in a row.

But you don't have to start at the first level, you could wait and play at second level.
This mean when three loses chop you have one trigger to play opposite.

Lets assume you have RB as your first clustering pattern.
Then you following the left column and play against red.

You continue to play against the last colour that show in the left column clustering patterns into two.
When you have three loses you watch what formation the first clustering pattern has with that cycle.

RB
BR W
RR W
RB L
BB W
BB L
RR W
RB L
RB L
RR L

This is the end with three loses and as you can see RR is the first clustering pattern with that cycle.
Then you play against R in the right column.

So you play against 4xRR.

RR W
RB L W
RB L
RR L

Now lest assume one side would show for four times in a row, one complete cycel, then you would have three loses.

  RR
RR L L
RR L L
RR L L

Then you would wait for the next clustering pattern by two and repeat the formula all over again.

Now if you can play four levels then you can start from the first level and gain some long strings of winnings.
That would be 12 step progression.
If you start from secound level that would be 9 step progression.

[Le_Chiffre]

Jordsie. have you tried the HP method on double dozens?  Bet selection bet on the dozen that won last plus the one next to it.  If it was the middle one then bet that plus the one nearest to which number came up in the middle dozen.

Same thing, write down a string 2 2 2 2 2 2. bet the first and last number combined on EACH dozen, then cross off on wins, on losers distribute the losses as before but this time x2, so loss on the 1st go the string would now be 2 2 4 4 4 4.

[Jordsie669]

Ego, I don't really  understand what you are doing there 
What do you mean by clusters and columns,
I only see one column, can you explain it a bit more clear please?

Cheers.

[TwoCatSam]

Jordsie

First, what you are trying to do is reduce volatility.  Good luck!  He who does that will be wealthy!

Second, referring to your second post, you want to reduce volatility in your life time.

Sam

[Jordsie669]

But can someone explain ego's bet selection?
'Cause it's not very clear to me.

Cheers, Jordsie.

[TwoCatSam]

Ego's writings are not understandable by the common man.  I looked at it.  I think the refers to columns as the left notations are one column and the right are another.

RR............The first R is in the right column; the second R is in the second column.

Anyway, this will not reduce volatility which is what you're wanting to do.

Sam

[ego]

You see red bet black.

R
R L
B W

Now you have black bet red.

R
R L
B W
R W

Just at all time bet opposite to the last hit outcome until three loses.

R
R L
B W
R W
R L
R L
R L

Now you have four red in a row, next outcome is black and you put it next to the top red as one full cycle is four clustering patterns of two.

R
R
B
R B
R
R
R

Now you would play against black at the secound column, that means you would play against the clustering pattern RB to hit four times in a row.

RB
RB
RB
RB

But if the next outcome after three loses would been red, then it would look like this.

R
R
B
R R
R
R
R

Then you would play black and aginst that the clustering patterns of two reds, RR would hit four times in a row, one complete cycle.

There exist four clustering patterns of two.

1. RR
2. RB
3. BB
4. BR

Conlusion, you bet against the last hit colour until three loses, first level.
Then you wait for one outcome to see what kind of clustering pattern show for that cycel.
Then you bet against it as i show above.

There exist four complete cycels with eight outcomes with clustering patterns of two hittting four times in a row.

RR
RR
RR
RR

RB
RB
RB
RB

BB
BB
BB
BB

BR
BR
BR
BR

So if you use all four levels you have 12 attempts or a 12 step progression.
The negative expectation is for two cycles to hit four times in a row each.

If you use one trigger of three loses you have 9 attempts or a 9 step progression.
The negative expecation is the same as i mention above.

[Jordsie669]

I still don't really get it sorry 
Why are you clustering the numbers in parts of 2?

And is:
RR
RR
RR
RR
6 reds in a row?

And is RBBRRRBB the same as:
RB
BR
RR
BB

When when is it a full cycle? It's all really vague to me.

Cheers, and thanks for all your help.

[ego]

Well if you have a dice with four sides, then each side has a number and are uniq.

1 2 3 4

Now if you throw this dice you expect to see a new side show each time or at least most of the times.

Is the same if you have a regular dice with six sides.

1 2 3 4 5 6

Now if you throw this dice you expect to see a new side show each time or almost.

Now if the dice has four sides, then one cycle would be that each side show one time each during four trails with out any repeats, then we would say that the dice alternate, one complete cycel.

Now if the dice has six sides, then one cycle would be that each side show one time each during six trails with out any repeats, then we would say that the dice alternate, one complete cycel.

So if you have a dice with four sides and one side show four times in a row during four trails, then no other side show, then its a complete cycel.

So if you have a dice with six sides and one side show six times in a row during six trails, then no other side show, then its a complete cycel.

[Jordsie669]

Now, I now what a cycle is (4 times same color in a row or RBRB, am I right?)
Now what does this mean?

RR
RR
RR
RR

RB
RB
RB
RB

BB
BB
BB
BB

BR
BR
BR
BR

And what/when do I have to bet after three losses, And do I have to spin without betting after 4 losses? 

[ego]

I find it more easy to apply the same math and probability for even money bets as with a dice.
Then at least you have some degree of something to base your selection upon.
And to do that you have to clustering 1 in 2 into patterns.

We also have the same for sequential distribution using the Ecart/STD.
Where we use the law of series and divide the distribution into different opposite events based upon    that.

[ego]

I apologies, for me its clear as water and don't know how to describe it more clear then what i mention above.
If you try to PM Albertjonas, then he might help you to explain this.
He understand all my writings, easy as complex.

Cheers