G.U.T. strategy in a Nutshell
This is one of the most talked about and cryptic roulette strategies and it is based on the law of the third – meaning that, on average, in 36 spins one third of the numbers hits once, one third hits more than once and one third doesn’t hit.
Method: Basically GUT revolves around one major bet. Smaller bets are also available.
In a nutshell, (the major bet) you have un-hit and hit once (or more) numbers.
Un-hit // Hit 36 0 (when you start playing)
Main bet
As the numbers come in, the unhit numbers will decrease and the hit numbers will increase.
(You have to keep track of the unhit ones (which ones they are)
Unhit // Hit (example)
15 15
or
16 15
Now, when the unhit numbers equal total hit numbers (preferably 18 numbers or less) it is time to bet all of the un-hit numbers one time.
Smaller bets
Of the hit numbers, you can wager when once hit, twice hit or more equal each other…
Example:
Once hit // Twice Hit 8 8
You would bet the 8 once hit numbers (one time)
Winkel wrote (in 2008):
This is one of my ultimate systems that win flat bet, no progression needed.
| With pink are the sleepers (cold numbers), with yellow the numbers that hit once, with Blue the numbers tat hit twice, with Magenda the numbers that hit 3 times and with green all the numbers that hit |
As you see in this graph there is one thing for sure: The Lines have to cross!
to play that is very simple.
You can start at any point, at any casino, at any table.
First the basics:
We know at the first spin there has to appear one number, so we count:
36 not appeared – 1 appeared
35 – 2
34 – 3 etc.
then we might come to this point:
19 – 18
now we bet the 19 numbers that not appeared because the statistic has to change to 18 – 19
during the spins we bet when the following combinations appear:
19 – 18
18 – 17
17 – 17
17 – 16
16 – 16
16 – 15
15 – 15
15 – 14
14 – 14
14 – 13
etc.
The difference between appeared and not appeared numbers has to be 0 or 1.
The second bet situation:
we have numbers that appeared once and numbers that appeared more than once:
every time there is a difference of 0 or 1, we bet the numbers that appeared once.
The third situation to bet:
We have numbers that appeared twice and numbers that appeared more than twice.
every time there is a difference of 0 or 1, we bet the numbers that appeared twice.
if we have to bet 19 or 18, we just bet once
if we have to bet less than 18, then we bet as often as there is a win or 0 possible:
17 to 13 numbers – we bet twice
12 to 10 – we bet three times
9 to 8 – we bet four times
One example
0x 1x >1x
19 13 5 bet 19numbers
19 12 6 loss -19
19 11 7
18 12 7
18 12 7
17 13 7
17 13 7
17 13 7
16 14 7
15 15 7 bet 15 numbers
14 16 7 win -15+36-19=+2
14 15 8
14 14 9 bet 14 numbers
14 14 9 loss -14 +2 = -12 and bet again
13 15 9 win -14 +36 -12 = +10
12 16 9
12 15 10
TCS wrote:
Here’s the way I’m seeing it.
a whole perm no. 0x 1x >1 18 36 1 0 15 35 2 0 33 34 3 0 36 33 4 0 13 32 5 0 At this point we have had no repeaters. We have counted down from 36 to 32 in the 0x column and have counted up to 5 in the 1x column. The >1 column is empty. Then we get our first repeater. 36 32 4 1 We stay at 32 in our 0x column and reduce our 1x column by one. We add one to our >1 column. 8 31 5 1 7 30 6 1 33 30 5 2 Here we get another repeater and the count changes. 24 29 6 2 22 28 7 2 16 27 8 2 2 26 9 2 8 26 8 3 again 36 26 8 3 1 25 9 3 5 24 10 3 7 24 9 4 again 11 23 10 4 22 23 9 5 again 4 22 10 5 13 22 9 6 again 27 21 10 6 16 21 9 7 again 30 20 10 7 10 19 11 7 play 19 numbers that not yet appear
What do we think?
Sam
Wildcard’s reply
Sam, from what I can gather, you don´t really need to add…
You are under the (false) impression that you need to sum the 1x and the >1 together. On the 1st example it happens to be true, however, the NOT APPEARED count keeps decreasing as the other two keep increasing, so there will be a point from which one on-wards you will never get that sum to be equal or lower to the NOT APPEARED value.
Forget about adding, concentrate on subtracting.
Here you have : (10) 19 11 7
19- (11 + 7) = 1 = betting opportunity …….. Now we are at school ? On this example the sum is fine because 19 numbers that did not appear minus the sum of the numbers that appeared once and those that appeared more than once is one (1).
But then you wish to do the same about (31) 13 12 12 … In this case, the sum would be worthless.
The idea here is relating 13 and 12…… betting the 13 (numbers) left un-hit.
WHY ? Because you have 13 minus 12 = 1 = betting opportunity, as per the rules.
Capice ?
Now, the question here is what to do about the 12 12 since I believe it might be considered a trigger also. Either that or I am also lost here. HELP winkel
Winkel’s reply
Now, the question here is what to do about the 12 12 since I believe it might be considered a trigger also.
That is a problem that sometimes appears.
It is a law that the numbers that are in the game go straight to Zero (see the diagram in first post “red line)
So we play the Non-appearers at first.
On the other hand:
If you would have decided to play the 12 once-appearers (numbers that appeared once) what would have happened?:
31 13 12 12
25 12 13 12 loss -12
27 12 13 12 loss -12
17 11 14 12 loss -12 and now the difference is “2” so stop playing and look for another “crossing-situation”
27 11 14 12
This is a build-in stop-loss
If there is no other crossing in sight, just start collecting new numbers starting with 37 – 0!
Same if we have continued playing the 13 N – 12 F
25 12 13 12 betting 13 once-app.
27 12 13 12 loss -13
17 11 14 12 loss -13 stop playing because we play 13 only twice!
Argument why G.U.T may not really work: The GUT and Law Of The Third Nonsense.
G.U.T. tracker
| Click the notes to zoom |