Dependence creates bias

[redhot]

“Each spin is independent” â€" This forms the basis for the argument that roulette can’t be beaten (along with unfair pay-outs).

On the other side of the argument, several members here have stated that it can be beaten by creating dependence as - “dependence creates bias”.

I wanted to start this thread to discuss how dependence can be created and if so, does it create bias?

I’ll start out with my thoughts on this, the only way to create dependence is if spin B is determined in some way by the outcome of spin A.

I think bet stitching can help here, for example if we stitch together an EC (High/Low) with the dozens, we now have 6 possible outcomes over 2 spins:

H1
H2
H3
L1
L2
L3

The outcome is dependent on both spins. If spin 1 is High, this affects the outcome of spin 2 as it can no longer be L1, L2, or L3. So out of the 6 equal possibilities, a bias has been created towards H1, H2 and H3.

Is this bias useful? Can it be used to gain an edge?

[Firefox]

Not useful as it stands as the sequence can complete with any dozen equally. If you could find an example where the next spin was influenced to alter the chance of certain partitions occurring, then maybe.

All of  this kind of theory never constrains the partition forming the next result. When a cycle is due to complete, it can always complete with any partition, or when an arithmetic progression is due, any partition can complete it. So in fact there is no dependence on a past spin which offers a betting advantage.

Unfortunately, you will find on this forum, some baiters,  hoaxers, and liars who will try to waste your time saying they have found a mathematical advantage, betting to beat the house edge based on the results of previous spins.

When challenged, they won't provide examples, because there is no example.

[Person S]

Perhaps, more precisely, it will look like this:
H + 1 - 1
H + 2 - 2
H + 3 - 3
L + 1 - 4
L + 2 - 5
L + 3 - 6,
then we can create a parallel game using DS /
And, perhaps, when connected, this creates a certain bias. It remains to find out at which points to attack and bet - it’s unclear whether we need the L + 3 or H + 1 combination, we need to cover both a dozen and the EU - so we risk 2 chips to win 1, or lose or make a return . The remaining 4 combos give us a little more chance if they play together.

[ego]

I give it a try ...

Even money:

One single is one outcome
Series of two is one outcome
Series of three and higher is one outcome

As you can see we have 1/3 because a series of three and higher is regarded as the same.

Now if you look at the random bits using any results you want, from Casino or random.org - it does not matter - what you will find is the following:
During 300 trails samples you will see two elements or two outcomes strike from 23456789 ... 10 11 12 13 14 15 ... 20 21 22 23 24 25 ...
That is a true bias.

I can compare this with dozen 1 2 3 where two dozen can strike 23456789 ... 10 11 12 13 14 15 ... 20  21 22 23 24 25 times in a row.
Also the principal with three possibilities.

Problem is how to take advantage of such bias - dominance.
I have been trying different random walks and some are better than others.

To explain this further you will see two things happen.
Either you get a TRIPLET or a true BIAS

A triple is when all three elements/outcomes alternate and show once each.
For example one single and one series of two and one series of three or higher in any combination.

True bias is when two elements/outcomes show three times or more in a row.
There can only be a true bias or triplet.

The bias combines the following combinations.

Singles and series of two
Singles and series of three or higher
Series of two and series of three or higher

This is a sample of today's random bits from random org and the number to the right at the end indicate how many elements/outcomes the bias has when it strikes as a sequence 3456789 and so on ... a zero "0" indicate a triplet with no present bias.

This is known as Sputnik's March and I am the creator.
The largest strike/bias was 13 in a row with this 300 trail sample.

Here can you see the even money with a small bias, a medium bias, a large bias.

Code Select Expand



2
1
1
2
2
1
2
2
1
2
1
2 9
---
1
1
1
2
2
2
2
2
2
1
2
2
2
1
1
1
2
2
2
2
2
1
1
1 7
---
2
2
1
2
2
1
1 4
---
2
2
2
1
1
2
2 3
---
1
2
2
1
2 4
---
1
1
1
2
1
2
2
2
1 5
---
2
2
1
2
1 4
---
2
2
2
2
1
1
1
1
1
2
2 3
---
1
2
2
2
1 3
---
2
2
1
1
2
2
2
2
1
1
1
2
2
2
2
2
2
2
2
1
1
1 6
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2
1
2
2
2
2
1
1
1
2
1
1
1
1
2 7
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1
1
2
2
2
1
1 3
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2
1
2
1
2
1
2
1
2
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2
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1
2 11
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1
1
2
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2
1
1
1
1 3
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2
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1
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1
2 7
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1
1
1
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1
1
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1
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2
1
2 8
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1
1
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1
1
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1
2
1 7
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2
2
2
1
2
2 0
---
1
1
1
2
2
2
1
1
1
2
1
2
2
2 6
---
1
1
2
2
2
1
1
1
1 3
---
2
1
1
2
2
2 0
---
1
2
2
2
1 3
---
2
2
1
1
2
2
1
1
2
2
1
1
1
1
1
2
2 7
---
1
2
2
2
1 3
---
2
2
1
2 3
---
1
1
1
2
2
1 0
---
2
2
2
1
2
1
1
1
1
1 4
---
2
2
1
2
1
2
1
1
2
1
1
2
1
1
2
1
2 13
---
1
1
1
2
1
2 4





[Blueprint]

This is not an opinion or belief like Firefox's view but a FACT.   Keep going, Redhot.

[Firefox]

If my view is not a fact, then disprove my view by providing an example of how the next spin can be influenced by previous ones and how this can be exploited to produce a bet with positive expectation.

[ego]

Redhot my example is among the best-known description of true bias using even money bets.
No one else has such a clear, simple and clever solution.

Even if the outcomes are independent so is there no other way to explore bias using even money.
Someone can claim they have a better solution, but that would be empty words as no one has shown any better solution then Sputnik's March.

Cheers

[Herby]

Hi Ego, how do you decide that ...

2
1
1
2
2
1
2
2
1
2
1
2   
---   here is the end of this sequence ?
Tnx

[Hairyballs]

The outcome is dependent on both spins. If spin 1 is High, this affects the outcome of spin 2 as it can no longer be L1, L2, or L3. So out of the 6 equal possibilities, a bias has been created towards H1, H2 and H3.

Where is the bias? There will be a bias only if you are able to say even before the first spin. 

[Hairyballs]

So in fact there is no dependence on a past spin which offers a betting advantage.

FACT

[Herby]

Tnx, got it.

[Blueprint]

If my view is not a fact, then disprove my view by providing an example of how the next spin can be influenced by previous ones and how this can be exploited to produce a bet with positive expectation.

When you ask the right question (and do the work I asked you to in the other thread) then I’ll consider answering you.   

[redhot]

Perhaps, more precisely, it will look like this:
H + 1 - 1
H + 2 - 2
H + 3 - 3
L + 1 - 4
L + 2 - 5
L + 3 - 6,
then we can create a parallel game using DS /
And, perhaps, when connected, this creates a certain bias. It remains to find out at which points to attack and bet - it’s unclear whether we need the L + 3 or H + 1 combination, we need to cover both a dozen and the EU - so we risk 2 chips to win 1, or lose or make a return . The remaining 4 combos give us a little more chance if they play together.

This is close to where I was going next...

We have 6 possible outcomes:

H1 - 1
H2 - 2
H3 - 3
L1 - 4
L2 - 5
L3 - 6

We've essentially created our own "double streets".

Any stats which apply to double streets will also apply to our game, for example - "92% of repeats in a double street cycle will come from the last 3 double streets". This is something I proved in another thread and holds true over 1 million cycles. A repeat in a DS cycle therefore has a bias towards the last 3 DS.

So now we have another bias to work with, can we use them together?

19
6  (H1)
27 (L3)
20 (H2)

Here the last 3 "double streets" were H1, L3 and H2. Betting for the repeat using standard double streets we would need to bet all 3 however by combining the first bias we can eliminate some options.

The last spin was H so we have a bias towards H1, H2 and H3. Therefore our only options for a repeat are H1 and H2.

Does eliminating some options give us an advantage here?

[Person S]

Statistics promises us to repeat in the first half.
But it is difficult to understand what to bet on.
Well, we excluded the L + 3 combination.
There are 2 options left: H + 1, H + 2.
If this is an external rate, then we must put EC + Doz, i.e. high and two dozen.
It turns out we cover the whole field.

Here's how victory will affect:
dozen 1 win - breakeven
a dozen 2 wins - 1 part (13-18) brings breakeven, 2 parts (19-24) win + 1 chip from H. Total + 1
dozen 3 win - lose - 1.
Hmmm not so cool

[Person S]

I apologize, a dozen 2 (19-24) brings + 2, from the bet 3.