Here's a matrix system that's so crazy random can't keep up with it to defeat it.
In the attachment you'll see a 3 by 3 matrix. It's built per normal. From left to right and from top to bottom. I numbered the points to show the order.
We are going to bet against 3 of the same dozen forming in a row.
There are 8 ways we're going to be betting against.
1. Points 1 2 3
2. Points 4 5 6
3. Points 7 8 9
4. Points 1 4 7
5. Points 2 5 8
6. Points 3 6 9
7. Points 1 5 9
8. Points 3 5 7
We're going to bet 1-1 on every bet. This will be addressed again later.
We don't bet on the 1st spin because it could be any of the 3 dozens. So no bet on spin 1, but once we have spin 1, now we can start betting that whatever the doz in spot 1 was it won't repeat in spot 2.
Please note that the number of arrow points pointing at a spot in the matrix represents the numbers of ways we can bet.
Spots 2, 3 and 4 only have 1 arrow point so we will just be betting against whichever dozen is on the spot the arrow is coming from.
So, at spot 2 we will be betting against the dozen in spot 1. If spot 2 loses, we will bet against the 2nd dozen forming in spot 3.
Note that spot 4's bet is against the dozen in spot 1 also.
Spots 6, 7 and 8 we will bet against 2 spots.
Spot 9 we will bet against 3 spots.
Spot 5 we will bet against 4 spots.
Here's an example: Let's say we got these 9 spins: dozens 2 2 1 3 2 1 1 3 2
Our matrix will look like this:
2 2 1
3 2 1
1 3 2
NO bet on spin 1. 2nd dozen spins and we have a 2 in spot 1
Now we can bet against the 2nd dozen showing for spot 2 which will be 1-1. Unfortunately the 2nd dozen hit so we lose. We now bet 1-1 that spot 3 won't be the 3rd 2nd dozen. Once again we bet 1-1 against the 2nd dozen and we win because dozen 1 hit.
On spot 4 we bet 1-1 against the dozen in spot 1 showing for the 2nd time on this vertical line. The 3rd dozen hits so we win.
Now we move to spot 5. Spot 5 bets 1-1 against the 2nd dozen from spot 1 hitting. Also against the 2nd dozen from spot 2 showing. Also against the 1st dozen from spot 3 showing and finally against the 3rd dozen from spot 4 showing. This is a little tricky because we bet 1-1 for each of the 4 bets and then we bet differentially. For spot 1 we bet 1-1 on dozens 1 and 3. For spot 2 we bet another 1-1 on dozens 1 and 3. For spot 3 we bet 1-1 on dozens 2 and 3 and for spot 4 we bet 1-1 on dozens 2 and 3. If you tally that up, we have 3 units bet on dozen 1; 2 units bet on dozen 2 and 3 units bet on dozen 3. Differentially that means we will bet 1-1 on dozens 1 and 3.
Spots 6, 7, 8 and 9 are a little different. If the 2 spots ahead of them were the same, we will be betting against the 3rd showing of that dozen. But if the 2 spots ahead of them were not the same, we will bet against the missing dozen hitting.
Let's take spots 1, 4, 7. Spot 1 was the 2nd dozen and spot 4 was the 3rd dozen. That means we will bet against dozen 1 in the 7 spot. In this case we will lost that bet. But look at spots 3, 5, 7. Spots 3 and 5 are dozens 1 and 2 so we bet against dozen 3 showing in the 7 spot and we win.
We will play a 9 spot matrix until we either get ahead by 1 unit or we don't get ahead by the 9th spin. As soon as we get ahead in the matrix, we immediately end that matrix and the next spin will be spot 1 of the next matrix.
If we don't bet ahead in the matrix, we will play the next matrix at the 2-2 level. This means that every bet blocking a dozen will be 2-2 instead of 1-1. All the bets in the 1st matrix were 1-1 and we stay at 1-1 as long as we reach a new high in the matrix. If we finish a matrix in the hole, we add 1 unit to our bet for the next matrix.
Sometimes we will only be betting a single dozen and on rare occasion we may have no bet because we're betting the same amount on all 3 dozens so differentially this would be a no bet.
This is a system in formation, so I'll write more details later.
If this doesn't belong in this section, will one of the mods please move it to the appropriate place?
Thanks,
GLC
Linkback: http://www.rouletteforum.cc/index.php?topic=16236.0
In the attachment you'll see a 3 by 3 matrix. It's built per normal. From left to right and from top to bottom. I numbered the points to show the order.
We are going to bet against 3 of the same dozen forming in a row.
There are 8 ways we're going to be betting against.
1. Points 1 2 3
2. Points 4 5 6
3. Points 7 8 9
4. Points 1 4 7
5. Points 2 5 8
6. Points 3 6 9
7. Points 1 5 9
8. Points 3 5 7
We're going to bet 1-1 on every bet. This will be addressed again later.
We don't bet on the 1st spin because it could be any of the 3 dozens. So no bet on spin 1, but once we have spin 1, now we can start betting that whatever the doz in spot 1 was it won't repeat in spot 2.
Please note that the number of arrow points pointing at a spot in the matrix represents the numbers of ways we can bet.
Spots 2, 3 and 4 only have 1 arrow point so we will just be betting against whichever dozen is on the spot the arrow is coming from.
So, at spot 2 we will be betting against the dozen in spot 1. If spot 2 loses, we will bet against the 2nd dozen forming in spot 3.
Note that spot 4's bet is against the dozen in spot 1 also.
Spots 6, 7 and 8 we will bet against 2 spots.
Spot 9 we will bet against 3 spots.
Spot 5 we will bet against 4 spots.
Here's an example: Let's say we got these 9 spins: dozens 2 2 1 3 2 1 1 3 2
Our matrix will look like this:
2 2 1
3 2 1
1 3 2
NO bet on spin 1. 2nd dozen spins and we have a 2 in spot 1
Now we can bet against the 2nd dozen showing for spot 2 which will be 1-1. Unfortunately the 2nd dozen hit so we lose. We now bet 1-1 that spot 3 won't be the 3rd 2nd dozen. Once again we bet 1-1 against the 2nd dozen and we win because dozen 1 hit.
On spot 4 we bet 1-1 against the dozen in spot 1 showing for the 2nd time on this vertical line. The 3rd dozen hits so we win.
Now we move to spot 5. Spot 5 bets 1-1 against the 2nd dozen from spot 1 hitting. Also against the 2nd dozen from spot 2 showing. Also against the 1st dozen from spot 3 showing and finally against the 3rd dozen from spot 4 showing. This is a little tricky because we bet 1-1 for each of the 4 bets and then we bet differentially. For spot 1 we bet 1-1 on dozens 1 and 3. For spot 2 we bet another 1-1 on dozens 1 and 3. For spot 3 we bet 1-1 on dozens 2 and 3 and for spot 4 we bet 1-1 on dozens 2 and 3. If you tally that up, we have 3 units bet on dozen 1; 2 units bet on dozen 2 and 3 units bet on dozen 3. Differentially that means we will bet 1-1 on dozens 1 and 3.
Spots 6, 7, 8 and 9 are a little different. If the 2 spots ahead of them were the same, we will be betting against the 3rd showing of that dozen. But if the 2 spots ahead of them were not the same, we will bet against the missing dozen hitting.
Let's take spots 1, 4, 7. Spot 1 was the 2nd dozen and spot 4 was the 3rd dozen. That means we will bet against dozen 1 in the 7 spot. In this case we will lost that bet. But look at spots 3, 5, 7. Spots 3 and 5 are dozens 1 and 2 so we bet against dozen 3 showing in the 7 spot and we win.
We will play a 9 spot matrix until we either get ahead by 1 unit or we don't get ahead by the 9th spin. As soon as we get ahead in the matrix, we immediately end that matrix and the next spin will be spot 1 of the next matrix.
If we don't bet ahead in the matrix, we will play the next matrix at the 2-2 level. This means that every bet blocking a dozen will be 2-2 instead of 1-1. All the bets in the 1st matrix were 1-1 and we stay at 1-1 as long as we reach a new high in the matrix. If we finish a matrix in the hole, we add 1 unit to our bet for the next matrix.
Sometimes we will only be betting a single dozen and on rare occasion we may have no bet because we're betting the same amount on all 3 dozens so differentially this would be a no bet.
This is a system in formation, so I'll write more details later.
If this doesn't belong in this section, will one of the mods please move it to the appropriate place?
Thanks,
GLC
Linkback: http://www.rouletteforum.cc/index.php?topic=16236.0