Vaddis Holy Grail

[nottophammer]

Well close to the 185 spins.

[Proofreaders2000]

Good graph Notto, but Vaddi also said no progressions.

[nottophammer]

Hello Proof
I think from the few previous posts, that members/readers can see I'm not a fan of the Vaddi method.
I spent some time yesterday and the pairings still has to many numbers or pockets as aka: General will spout, against you.
But nothing ventured nothing gained.
Here just gained more proof for me to just use ROTT; or as Herbie says use your "gamblers intelligence"

Try to see how over 40 spins, the known to me, 1-3-5-7=16 repeats, you can make decisions on whether to bet for repeats or non-hits.

[Proofreaders2000]

Here is an idea
==============================
Procedure: Start by betting the last decision outcome and its pair.

Example: #30 (newest outcome)

Bet 30,31:    1.) 13(x)-2

Bet that last two numbers and the newest trigger together.

Bet 30, 31, 13, 14:    2.) 6(x)-4

Again bet the last four numbers and the newest trigger together.

Bet 30, 31, 13, 14, 6, 7:    3.) 0(x)-6

And the last bet of this series, the last six numbers and the new trigger.

Bet 30, 31, 13, 14, 6, 7, 0, 1:    4.) 19(x)-8

If none of those hit, start over with the last decision.

Bet 19, 20....

[Elite]

Guys Vaddi talking about a formula in their postings, regarding balance.

What do you think?

Formula he gave later , 12/4=3 and 24/8=3.   

[Elite]

12/4 =3 is base.,its knowable fact which no one can mess up,,here is true explabation
Divide 36 nbers in 9 blocks, and put 12  unique nbers in that block,u will get 3 blocks having 2 nbers each, 9 blocks will come if we have one block of 4 nber,

[d80]

12/4 =3 is base.,its knowable fact which no one can mess up,,here is true explabation
Divide 36 nbers in 9 blocks, and put 12  unique nbers in that block,u will get 3 blocks having 2 nbers each, 9 blocks will come if we have one block of 4 nber,

Hello elite..you mean 36numbers /9 blocks =4  ??  How you will have 3 blocks with 2 numbers?? can give me a example?

[Elite]

enter 12 nbers in 9 blocks, u can do this on paper

[Elite]

12  unique nbers entered in nine blocks of 4, result will  b 3 having two nbers, this is 12/4=3, and six having 1 number, this is extreme case,with 12 unique, and this rise to balance nmber  greater than 6 and less than 9,

[Elite]

Example, 9 nbers drop and we put 9 in 9 blocks , now what about reaming 3 nvers, it will go in blocks having already 1 nber, so 3 blocks will have 2  6 will have 1.  Now situation is , 6 blocks having 6 nners and 3 blocks having 6 nvers, double and single 50 50  '6+6 =12 nvers 

[Elite]

Law of third, 12 doubles , 12 singles in 36. So above case was with 12 ,, if we expand it to 2 times or 3 timesstill it will b same, two times case 24/8=3, , 48/16=3, 96/32=3

[alexlaf]

I had a gap playing 9 nr for 12 round and there must more higher gaps  .. How you will play that bet?

[Proofreaders2000]

May I try?

Bet last decision as Vaddi said in the beginning.

Example: #1 (newest)

Bet #1:  1.) 33(x)-1

Bet 1, 33:    2.) 4(x)-2

Bet 1, 33, 4...

If eight total numbers miss, just bet those same eight continuously
and change the last decision with each spin. (nine numbers in play)

If one of the eight uniques hits, bet the pair number and continue.

Bankroll suggestion: 180 units

[Proofreaders2000]

As far as law of third, Vaddi imo was focusing on unique outcomes. 

The challenge is to make it playable, profitable (and time friendly).

[Elite]

Bank roll,  12 *  is numbers to bet .e.g
8 nbers*12=96 . how that will process is ,12/4=3,,   e.g with, 8 , we loose 4 times, and it happens ,3 times, e.g, 8, loss, 8 loss,8 loss, 8 loss, repeat this process again, then again, 3 times,total