Roulette LAW OF THE 2ND THIRD

MEISEL’S LAW OF THE 2ND THIRD
Did this during the co-vid shut in.
I put my name on this because I believe I discovered this a few days ago. Please pardon the ego. I like AYK’S TRACKER a lot and have been using it for different evaluations. A couple of days I ago I was wondering about the 37 Spin count and adding 18 Bets until it becomes a 55 Spin count. His TRACKER allows an automatic 37 or 50 or 100 or 200 spins.
So I pressed his automatic 50 button and added 5 random of my own. I received the numbers 17-16-29-0-29-28-28-11-20-8-34-28-18-19-32-23-21-14-2-13-0-29-29-36-8-11-19-4-11-33-5-18-13-22-4-30-28 (37)
-18-22-5-26-5-26-35-6-26-0-29-8-14-8-18-27-31-9 (55)
I looked at the 55 Spin chart and saw that there were 8 unhits, 16 uniques, and 13 repeaters.
I undid the last 18 Bets and looked at the 37 Spin chart and saw there were 14 unhits, 14 uniques, and 9 repeaters. (By the way some people that use the Law Of The Third like 12-12-12, someone named Winkle likes 14/15-14/15-8/9, others like 15-14-8. I personally like 14-14-9[6 triples & 3 doubles] on the European Wheel and 15-14-9[6 triples & 3 doubles] on the American Wheel).
I looked at the difference between the 55 Spin and the 37 Spin charts and saw the difference was -6/+2/+4. I then calculated a whole lot of comparisons between the 2 and came up with an average of -5/-1.7/+7.
This is now MEISEL’S LAW OF THE 2ND THIRD: -5 unhits, -1.7 uniques, and +7 repeaters. WOW! I noticed that many times the uniques were near 0 and 1. That means after 37 Spins you just Bet on the UNHITS and the REPEATERS! as the uniques won’t change very much.
Now I prefer an 18 or 19 number bet. On an American Wheel a 19 number Bet is slightly better than an 18 or 20 number Bet. Proof:
20 #s W20 L18 52.6% 47.4% 5% chance of Winning over Losing 52.6% of Winning 16 = 8.42
W35 -19 = W16 L20 need 1.3 Wins
W20x16=320 L18x20=360 = -40
19 #s W19 L19 50% 50% 0% chance of Winning over Losing 50% of Winning 17 = 8.5
W35 -18 = W17 L19 need 1.1 Wins
W19x17=323 L19x19=361 = -38
18 #s W18 L20 47% 53% -6% chance of Winning over Losing 47% of Winning 18 = 8.46
W35 -17 = W18 L18 need 1 Wins
W18x18=324 L20x18=360 = -36
So all my 18 Bets are given below. After every winning Bet, you delete that number from your Bet. After every Bet you can mark what happened, The choices are (1) unhit now a unique (2) unique now a single (3) single now a double, a repeater (4) double now a triple (5) triple now a quad, etc.
When your Bets go below 20 or 19, ADD A UNIQUE.
You will see in this example I Won 11 and Lost 7.
11 Wins = 17-17-17-18-19-18-18-18-18-18-18 = 196
7 Losses = 20-20-19-20-18-18-18 = 133 Total = 63
That is $63 in 18 Bets. Now I Bet $5 per unit so that is $315 in 18 Bets.
Here is my example. You might have to download it and straighten out the columns.
Before I post any examples can any of you Roulette mathematical geniuses determine if this is practical? I personally don’t play this, it is too tedious. I play my modification on an old Silverstone (yikes!) system.

 

I thought you had something there, when you said "just bet UNHITS AND REPEATERS."
But although the number of uniques doesn't change very much, those numbers are still getting hit. They just change to become repeaters. And some unhits are getting hit, and becoming uniques.
Do you see what I mean?

 

That's why I added:
After every Bet you can mark what happened, The choices are (1) unhit now a unique (2) unique now a single (3) single now a double, a repeater (4) double now a triple (5) triple now a quad, etc.
When your Bets go below 20 or 19, ADD A UNIQUE.
I didn't know how to check this out except by trial and error. Do you have an algorithm or program to check this system out?

 

It is possible for me to write a BASIC program to check a system, either with random numbers or inputted numbers.
But not sure what you mean by... (2) unique now a single. Surely a unique is a single?
And you prefer a 19 number bet. So what numbers do you choose, some of the unhits and repeaters? There are 23 in your example at spin 37. There are still 21 at spin 55.
So no need to add a unique. And I still cannot see the advantage in this.
Need a bit of clarification!

 

Do we have only three states [0 , 1 ,>=2] ?

 

Interesting, i'll have to test it thxs