I found this:
According to the theorem there must be a repeat of the same chance at least three times in nine outcomes with the same distance between them and that is a fact.
Example : LLHHLLHLL
See also here: Van De Waerden theorem (https://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem)
Nick is expert in this method.
I just copy from his .pdf Van de Waerden Theorem
In a set of 9 spins there will ALWAYS be one
of the following 16 Arithmetic Progressions (AP)
An Arithmetic Progession is formed when the
the differencel between numbers is equal (constant).
1-2-3 the constant is 1. 2-4-6 it is 2. 1-4-7 is 3.
The 16 Arithmetic Progressions (AP) are:
1-2-3
2-3-4
1-3-5
3-4-5
4-5-6
2-4-6
5-6-7
1-4-7
3-5-7
6-7-8
2-5-8
4-6-8
7-8-9
1-5-9
3-6-9
5-7-9
RRR is an AP of 1-2-3
RBRBR is an AP of 1-3-5
BBRBRBR is an AP of 3-5-7
etc.
Nick you are welcome to post your .pdf here as well.
1-2-3
1-3-5
1-4-7
1-5-9
2-3-4
2-4-6
2-5-8
3-4-5
3-5-7
3-6-9
4-5-6
4-6-8
5-6-7
5-7-9
6-7-8
7-8-9
5 32 34 32 14 36 25 13 22
L H H H L H H L H
So in this case LOW does not meet the theorem [1 5 8] but its opposite HIGH does with 2 3 4?
So our uncertainty will always be which one of the chances are going to meet the theorem?
I guess there is a technique where we play both theorems at once?
There appear to be "built in" triggers where one chance eliminates itself such as:
LHHHH <=== LOW can no longer provide a trigger
But the problem here is that HIGH has already produced one: 1 2 3