Brainstorming LOTT

[nottophammer]

bets total 1212 the 34 wins return 1224 so minus bets from return +12 units for flat bet every spin from spin3 to spin 108

[6th-sense]

If you work it out you could play 8 spin mini games as 1+2+3+4+5+6+7+8=36

How many times would you get a repeat in 8 spin time frame ..with 35 unit bank roll?  8 spins to 1 loss?

Each 8 spin seperate games?

[6th-sense]

Not hammer I was looking at the 24 spin example

[6th-sense]

ANd in those mini games take the loss and virtual play the rest of the spins reducing the 35 unit loss minus your wins...would be interesting to know the stats on this

[nottophammer]

Not hammer I was looking at the 24 spin example

Ok  lets use a touchscreen, you have a BR of 35 $/£, you insert and the machine shows 35 units. Now the green chart shows spin3 theres 35 non-hit, so you place the 35units and BR shows zero, you win now the meter shows 36, you are +1 unit. So we know its 35+1, so if you take all wins as 36 units, 17*36= 612 and the 23 bets which total 585, take one from the other and theres the profit 27

[Bigbroben]

If you work it out you could play 8 spin mini games as 1+2+3+4+5+6+7+8=36

How many times would you get a repeat in 8 spin time frame ..with 35 unit bank roll?  8 spins to 1 loss?

Each 8 spin seperate games?

46% repeaters after 8 spins.

[6th-sense]

DId a quick test on nothammers result in the 100 spin list didn’t do so good ..more repeats and the odd zero in an 8 spin time frame than I would imagine

[Roulettebeater]

Dollar for Dollar there is no better investment in the world than a winning system.

[Bigbroben]

Big Bro
Let me see if i can get this across to some members. you talk of flat betting. Well heres the green graph.

Hi,
let's say I'd play this system.  Betting on unhits and Alembert when a loss.  I'd have a br of 1000 units.  What are the odds of reaching +100u without busting 1000u?
If the odds are lower than 10%, it would make sense to play 10 games with a goal of 100, then going Fibonacci 1,1,2,3,5,8,... so:

If br of 1000-3000: games at 1u base.
3000-5000: 2u base ;
5000-8000: 3u base...
...
10 succesful games every step.

So, what are the odds of losing 1000u br before reaching 100u?  What about 200u?

[nottophammer]

Bigbroben
Look in Turbo if you know the average for Non-hits, then betting for them is not hard, nor is it hard to see when repeats should come, told you use countback to aid your decisions

[Bigbroben]

I was just reading The Law of the Third that is available online, just for inspiration.

One of the main point isn't really wether to bet on repeaters or unhits, 2x's, 3x's...  seems we've been arguing about this for quite a while and overseeing one point:  when to bet.

Look at the Pattern breaker tread: they get hit rates pretty high but must wait many spins before attacking.

Pb: what to bet on: opposite of last pattern;
       when to bet: when only one pattern left;
       how long: 3 spins.
       How much: subjective amounts, perhaps 1,2,4.
That is clear.

What about LOTT, so far?  Ok, some think unhits are good, others rep 1x, 2x.  How long: depends.
When to bet? Not a word (that I know of...).

So, why don't we check for higher accuracy by defining when to bet, when not to...

Put some extirps from the doc just for some thoughts...

[Bigbroben]

not the deviation, but this:
see attached picture

title line read: after x coups there were at least

first table line: no repeater in these spins
second line: minimum 1 repeater
third line: minimum 2 repeaters
and so on
repeater is any repeater hit 2times or more

Graphed it: interesting.  See attached.

[Herby]

Hi Bigbroben,
could you provide us your data in computer file form for further calculations ?
TNX

[Bigbroben]

Hi,
here!
I basically copied the nrs given by a member in an Excel sheet and created a graph.

What I want to be confirmed, though, is if my calculations are ok.  Pssstt... I think I got something...
I think I understand why the Hurried Split Test exposed in the LOTT document waits for 6 different splits to hit before betting.

Stats and maths!

Can anyone confirm it is about so, for 1st repeaters arriving after spin 11:

% 1st repeater coming at spin:
12: 6.32%
13: 4.68%
14: 3.51%
15: 2.50%
16: 1.62%
17: 1.04%
18: 0.57%
19+: 0.45%.

I'll explain further when someone checks and confirms.  The nrs I got are from 65000 cycles.  I think my %age are precise, but if someone can verify for more accuracy?

[Herby]

Hi Bigbroben,
tnx again for  the Excel sheet.

Can anyone confirm it is about so, for 1st repeaters arriving after spin 11:

% 1st repeater coming at spin:

Above are 2 different questiones.
- Repeat at spin 11 means: repeat exactly at  spin 11
- Repeat after spin 11 means:  spin12 or spin 13 or spin 14 ...

Here at list with repeats at exact positions:
example: {5, 9.14329} : 1. repeat at spin nr. 5 probability: 9.14 %

{2, 2.7027}, {3, 5.25931}, {4, 7.46254}, {5, 9.14329},
{6, 10.1935}, {7, 10.5792}, {8, 10.341}, {9, 9.58236},
{10, 8.44931}, {11, 7.10453}, {12, 5.70282}, {13, 4.3717},
{14, 3.2}, {15, 2.23535}, {16, 1.48879}, {17, 0.944243},
{18, 0.569417}, {19, 0.325898}, {20, 0.176651}, {21, 0.0904612},
{22, 0.0436414}, {23, 0.0197706}, {24, 0.00837944}, {25, 0.00330845}, {26, 0.00121086},
{27, 0.000408421}, {28, 0.000126093}, {29, 0.0000353412}, {30, 8.90354*10^-6},
{31, 1.99147*10^-6}, {32, 3.89324*10^-7}, {33, 6.51701*10^-8}, {34, 9.08199*10^-9},
{35, 1.01159*10^-9}, {36, 8.44331*10^-11}, {37, 4.69435*10^-12}}

Above are theoretical= calculated values.