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[split-monster]

Thanks Nick,

There are a few different ideas I have had regarding these intervals.

One area that looks ripe for attacking is to use parlays.

Quite often I see running down one column something like...

X
1
X
X
1
X
1
X
1
2

Keeping the interval to 1 makes 2 units profit or 3 units if an X follows straight after another X.

Assuming you started betting after seeing X, 1, X above, you would have the following.

X  +3
1  -1
X  +3
1  -1
X  +3
1  -1
2  -1

So 5 units profit. Tracking across all three e.c's may throw up some nice runs.

What progressions to use?

Star instantly comes to mind.

Another one would be a kind of modified 'foolproof' MM. That's the one by Stetson Bailey where you just keep raising a unit regardless of a win or loss. However there is no need to be so aggressive when betting parlays. So I thought about something like 1,1,1 2,2,2 3,3,3 etc or maybe even 1,1 2,2 3,3 and so on.

All just ideas for the melting pot really.

[split-monster]

Another idea regarding the 'foolproof' MM which may work out better than the above is just to concentrate on the missing pair.

Using a betting sequence like the following..

1/2
2/4
3/6
4/8
5/10
6/12
7/14
8/16 etc..

Example:

BB missing. So I am betting for RR.

B (bet 1 here) lost -1.
B (bet 2 here) lost -2. Total loss -3.

Now say RB is missing. So I bet BR.

R (bet 2 here) lost -2. Total loss -5.
R (bet 4 here) won +4. Total loss -1.

Now of course RB is still missing. So I bet for BR again.

R (bet 3 here) lost -3. Total loss -4.
R (bet 6 here) won +6. Total +2. Stop here and reset if you like or carry on.

I like this idea better than the two I gave above in the previous post. The missing pair is more likely to stay missing than getting a repeating X.

If you win the first bet in the pair, you don't need to bet for the second result. You could even then stay at the same level in the MM.

I will test some more with this one.

[split-monster]

Here is a longer example:

BB missing pair. Play RR.

B (bet 1. lose -1)
B (bet 2. lose -2) total loss -3.

RB missing pair. Play BR.

B (bet 2. lose -2) total loss -5.
R (bet 4. lose -4) total loss -9.

RR missing pair. Play BB.

R (bet 3. lose -3) total loss -12.
R (bet 6. lose -6) total loss -18.

BR missing pair. Play RB.

R (bet 4. win +4) total loss -14. don't play for second result. Stay at same level.

BR still missing pair. Play RB.

B (bet 4. lose -4) total loss -18.
B (bet 8. win +8) total loss -10.

BR still missing pair. Play RB.

B (bet 5. lose -5) total loss -15.
B (bet 10. win +10) total loss -5.

BR still missing pair. Play RB.

R (bet 6. win +6) total +1. Reset or carry on.

There were 8 losses and 4 wins here for a total of +1.

I always kind of liked the 'foolproof' method. This modified version for the missing pairs doesn't look too shabby. Time will tell I suppose.

[split-monster]

Now here is the whole idea of the Intervals transposed over to the dozens.

Yes, I know what some of you are thinking. [smiley]cps/suicide.gif[/smiley]

Can you figure it out?

[split-monster]

Here it is a bit further on.

[split-monster]

Here is how it works with the dozens.

There are 9 pairs.

1-1 = 1.
1-2 = 2.
1-3 = 3.
2-1 = 4.
2-2 = 5.
2-3 = 6.
3-1 = 7.
3-2 = 8.
3-3 = 9.

Now going back to the e.c's...remember that if RR was the missing pair, that then leaves RB, BB and BR. So I would play BB knowing that one of the results of the next pair would be B as long as the RR stayed missing.

So how about the dozens?

If 1-1 is missing. It then has to appear as either (2-2) (3-2) (2-3) (3-3) if no 1 continues to be absent in both the first and second result of the pair.

Here is the full list.

1-1. (2-2) (3-2) (2-3) (3-3)
1-2. (2-1) (3-1) (2-3) (3-3)
1-3. (2-1) (3-1) (2-2) (3-2)

2-1. (1-2) (3-2) (1-3) (3-3)
2-2. (1-1) (3-1) (1-3) (3-3)
2-3. (1-1) (3-1) (1-2) (3-2)

3-1. (1-2) (2-2) (1-3) (2-3)
3-2. (1-1) (2-1) (1-3) (2-3)
3-3. (1-1) (2-1) (1-2) (2-2)

Here is one type of bet you can make.

Looking at line 5 in the file I attatched in the above post. You can see that dozens 1,2,3 have all come out as the first result in the pairs. However only dozens 1,2 have come out as the second result of the pair. So this just leaves dozen 3 as the missing dozen. So you would bet here dozens 1,2 to appear as the second result of the pair. There were 5 wins of 1 unit before dozen 3 appeared as the second result of the pair. Then dozen 2 became the missing dozen on the first result of the pairs. So you would now bet dozens 1,3 to appear as the first result of the pair. Another 2 wins followed here.