System based purely in Maths

[winkel]

I can´t believe it.

br
winkel

[gingermolloy]

cant believe what?

[TOPCAT]

hi gingermolloy. i was kinda expecting a system on the even chances ( black n red )   il give this one some testing tho, cheers gingermolloy

[hermes]

Ginger, I am a little bit confused. You said that it must be a 10 unique numbers without repeat (we know already system strategy like that around) and than you take # 3 which is repeater into team. That's not kosher? I think from your explanation you should wait for one more unique number before bets. That means all 10 must be kosher, no gentile between.
Another thing, you started debate about completely different theme, this conclusion is a real surprise - not only for Winkel.
Thanks anyway and keep discovering you are on the right way.
Hermes

[gingermolloy]

This is the end point of my system, which I believe is rooted in maths and I believe I can prove that.

I returned to the forum to give my system over for testing.

I decided to jump to the end because people were being impatient and i thought it was what people wanted.

Lets see if the system works. If it does and people are interested in how I came to it then i will elaborate.

I was trying to take it step by step but that just caused problems.

I have run this system with massive success, can anyone confirm that it works!

I will then resume my step by step explanation.

cheers

ginger

[VLSroulette]

I for sure am glad you are back Ginger.

This system can be easily coded. I just want your confirmation you aren't making any human-based decisions at any point (i.e. 100% mechanical system approach).

Thank you.

[Marven]

Hi Ginger,

Assuming that my win/loss calculations for the progression below are correct, I think this system wouldn't work as a failure of the 10 steps progression would mean a -584 loss. I think the system can't recover for itself.

Step     Win     Loss

20p,     +5.2      -2
80p,     +18.8    -10
£1.20,   +21.5    -22
£2,       +30       -42
£3,       +36       -72
£4.40,   +42.4    -116
£6.40,   +50.4    -180
£9.20,   +59.2    -272
£13,      +66       -402
£18.20.  +71.2    -584

I have done a quick hand test for this and hit the wall twice, ending up with a -481.5 loss.

Your contribution is well appreciated though.

Regards,
Marven

[Kon-Fu-Sed]

Hi ginger, and all.


Warning: This is...

An analysis based purely in maths, of a system based purely in maths.


Suppose you find 10 numbers to bet in 100,000 sessions (to cut at least some decimals)


The probability for 10 numbers to hit at the first trial is 10/37 and so you will win the first stage (26u) in 27,028 sessions = 702,728u
You started 100,000 sessions, won and ended 27,028 so 72,972 sessions to bet for the 2nd stage.

The probability for 10 numbers to hit at the second trial is 10/37 and so you will win the second stage (96u) in 19,723 sessions = 1,893,408u = total 2,596,136u
You bet the 2nd stage in 72,972 sessions, won and ended 19,723 so 53,249 sessions to bet for the 3rd stage.

The probability for 10 numbers to hit at the 3rd trial is 10/37 and so you will win the 3rd stage (106u) in 14,392 sessions = 1,525,552u = total 4,121,688u
You bet the 3rd stage in 53,249 sessions, won and ended 14,392 so 38,857 sessions to bet for the 4th stage.

The probability for 10 numbers to hit at the 4th trial is 10/37 and so you will win the 4th stage (150u) in 10,502 sessions = 1,575,300u = total 5,696,988u
You bet the 4th stage in 38,857 sessions, won and ended 10,502 so 28,355 sessions to bet for the 5th stage.

The probability for 10 numbers to hit at the 5th trial is 10/37 and so you will win the 5th stage (180u) in 7,664 sessions = 1,379,520u = total 7,076,508u
You bet the 5th stage in 28,355 sessions, won and ended 7,664 so 20,691 sessions to bet for the 6th stage.

The probability for 10 numbers to hit at the 6th trial is 10/37 and so you will win the 6th stage (212u) in 5,593 sessions = 1,185,716u = total 8,262,224u
You bet the 6th stage in 20,691 sessions, won and ended 5,593 so 15,098 sessions to bet for the 7th stage.

The probability for 10 numbers to hit at the 7th trial is 10/37 and so you will win the 7th stage (252u) in 4,081 sessions = 1,028,412u = total 9,290,636u
You bet the 7th stage in 15,098 sessions, won and ended 4,081 so 11,017 sessions to bet for the 8th stage.

The probability for 10 numbers to hit at the 8th trial is 10/37 and so you will win the 8th stage (296u) in 2,978 sessions = 881,488u = total 10,172,124u
You bet the 8th stage in 11,017 sessions, won and ended 2,978 so 8,039 sessions to bet for the 9th stage.

The probability for 10 numbers to hit at the 9th trial is 10/37 and so you will win the 9th stage (330u) in 2,173 sessions = 717,090u = total 10,889,214u
You bet the 9th stage in 8,039 sessions, won and ended 2,173 so 5,866 sessions to bet for the 10th stage.

The probability for 10 numbers to hit at the 10th trial is 10/37 and so you will win the 10th stage (356u) in 1,586 sessions = 564,616u = total 11,453,830u
You bet the 10th stage in 5,866 sessions, won and ended 1,586 so 4,280 sessions were lost.

4,280 sessions were lost.


First: Is that according to math - is the figure reasonable? (Reality check, you know)

The probability for at least one hit betting ten numbers for up to ten spins (quit on hit) is 95.71825892718769% or in 100,000 sessions 95,719 should win.
(Figure found here nolinks://vlsroulette.com/grabb/)

That is 4,281 lost sessions.
The calculated figure above was 4,280 lost sessions. A difference of ONE session (decimal error).

So our figure of 4,280 lost sessions is probably correct.

SO! 4,280 sessions lost - and they lost 2,920u each so that's 12,497,600u. Lost.

The total won was 11,453,830u so...

The net, based purely in maths, is -1,043,770u...
For 100,000 sessions = -10.44u/session.


Sorry.
KFS

PS. The loss is like -2.6% but it's because I rounded all the winning sessions UP when cutting decimals.
Had I not, it would probably end on -2.7% as it's based purely in maths.

[Proofreaders2000]

Here's a hand test I did of your unique numbers system and I'm afraid it was not good.
(Riverbelle Live Wheel Casino: Date: January 10,2009: 8:00 am

(I converted the pence figure to American Dollars, starting at $1):: <$> means in the negative<---

1. (win) (Bet $1 on 10 unique numbers): $36-$9=$27
2. (x)--(1x)----<$10>
3. (x)--(4x)----<$40>
4. (x)--(6x)----<$60>
5. (x)--(10x)---<$100>
6. (x)--(15x)---<$150>
7. (x)--(22x)---<$220>
8. (x)--(32x)---<$320>
9. (x)--(46x)---<$460>
10.(x)--(65x)---<$650>
11.(x)--(91x)---<$910>

------------------------Total for the session: <$2920>+$27=<$2893>

[gingermolloy]

I for sure am glad you are back Ginger.

This system can be easily coded. I just want your confirmation you aren't making any human-based decisions at any point (i.e. 100% mechanical system approach).

Thank you.

The bit of this system that is 100% mechanical is the progression, which is derived from the Grand Martingale.

The human decision is the waiting for 10 consecutive numbers with no repeaters.

It would be nice to do a comparison of betting on 10 arbitrary numbers using the progression (100% mechanical) as apposed to waiting for the 10 numbers (which does have a human decision).

The decision to wait for the consecutive numbers is based around the theory of repeating numbers.

Yes you could say that this is in contradiction of what I said earlier, but the mathematics of random numbers does suggest that numbers are more statistically likely to repeat than not to.

for example if this system is to fail for just 1 progression, we must see 20 of 37 numbers in a row with no repeaters, this is sooooo unlikely.

Just look al Winkels GUT! Check out his graphs.

You hardly ever get a graph where you get to the 20th spin without a repeater.

ginger

[gingermolloy]

Here's a hand test I did of your unique numbers system and I'm afraid it was not good.
(Riverbelle Live Wheel Casino: Date: January 10,2009: 8:00 am

(I converted the pence figure to American Dollars, starting at $1):: <$> means in the negative<---

1. (win) (Bet $1 on 10 unique numbers): $36-$9=$27
2. (x)--(1x)----<$12>
3. (x)--(4x)----<$48>
4. (x)--(6x)----<$72>
5. (x)--(10x)---<$120>
6. (x)--(15x)---<$180>
7. (x)--(22x)---<$264>
8. (x)--(32x)---<$384>
9. (x)--(46x)---<$552>
10.(x)--(65x)---<$780>
11.(x)--(91x)---<$1092>

------------------------Total for the session: <$1092>+$27=<$1065>

sorry i have no idea what all this means can you clarify for me with no jargon

plain English

sorry

ginger

[gingermolloy]

Hi ginger, and all.


Warning: This is...

An analysis based purely in maths, of a system based purely in maths.


Suppose you find 10 numbers to bet in 100,000 sessions (to cut at least some decimals)


The probability for 10 numbers to hit at the first trial is 10/37 and so you will win the first stage (26u) in 27,028 sessions = 702,728u
You started 100,000 sessions, won and ended 27,028 so 72,972 sessions to bet for the 2nd stage.

The probability for 10 numbers to hit at the second trial is 10/37 and so you will win the second stage (96u) in 19,723 sessions = 1,893,408u = total 2,596,136u
You bet the 2nd stage in 72,972 sessions, won and ended 19,723 so 53,249 sessions to bet for the 3rd stage.

The probability for 10 numbers to hit at the 3rd trial is 10/37 and so you will win the 3rd stage (106u) in 14,392 sessions = 1,525,552u = total 4,121,688u
You bet the 3rd stage in 53,249 sessions, won and ended 14,392 so 38,857 sessions to bet for the 4th stage.

The probability for 10 numbers to hit at the 4th trial is 10/37 and so you will win the 4th stage (150u) in 10,502 sessions = 1,575,300u = total 5,696,988u
You bet the 4th stage in 38,857 sessions, won and ended 10,502 so 28,355 sessions to bet for the 5th stage.

The probability for 10 numbers to hit at the 5th trial is 10/37 and so you will win the 5th stage (180u) in 7,664 sessions = 1,379,520u = total 7,076,508u
You bet the 5th stage in 28,355 sessions, won and ended 7,664 so 20,691 sessions to bet for the 6th stage.

The probability for 10 numbers to hit at the 6th trial is 10/37 and so you will win the 6th stage (212u) in 5,593 sessions = 1,185,716u = total 8,262,224u
You bet the 6th stage in 20,691 sessions, won and ended 5,593 so 15,098 sessions to bet for the 7th stage.

The probability for 10 numbers to hit at the 7th trial is 10/37 and so you will win the 7th stage (252u) in 4,081 sessions = 1,028,412u = total 9,290,636u
You bet the 7th stage in 15,098 sessions, won and ended 4,081 so 11,017 sessions to bet for the 8th stage.

The probability for 10 numbers to hit at the 8th trial is 10/37 and so you will win the 8th stage (296u) in 2,978 sessions = 881,488u = total 10,172,124u
You bet the 8th stage in 11,017 sessions, won and ended 2,978 so 8,039 sessions to bet for the 9th stage.

The probability for 10 numbers to hit at the 9th trial is 10/37 and so you will win the 9th stage (330u) in 2,173 sessions = 717,090u = total 10,889,214u
You bet the 9th stage in 8,039 sessions, won and ended 2,173 so 5,866 sessions to bet for the 10th stage.

The probability for 10 numbers to hit at the 10th trial is 10/37 and so you will win the 10th stage (356u) in 1,586 sessions = 564,616u = total 11,453,830u
You bet the 10th stage in 5,866 sessions, won and ended 1,586 so 4,280 sessions were lost.

4,280 sessions were lost.


First: Is that according to math - is the figure reasonable? (Reality check, you know)

The probability for at least one hit betting ten numbers for up to ten spins (quit on hit) is 95.71825892718769% or in 100,000 sessions 95,719 should win.
(Figure found here nolinks://vlsroulette.com/grabb/)

That is 4,281 lost sessions.
The calculated figure above was 4,280 lost sessions. A difference of ONE session (decimal error).

So our figure of 4,280 lost sessions is probably correct.

SO! 4,280 sessions lost - and they lost 2,920u each so that's 12,497,600u. Lost.

The total won was 11,453,830u so...

The net, based purely in maths, is -1,043,770u...
For 100,000 sessions = -10.44u/session.


Sorry.
KFS

PS. The loss is like -2.6% but it's because I rounded all the winning sessions UP when cutting decimals.
Had I not, it would probably end on -2.7% as it's based purely in maths.

Sorry, I am struggling with your abbreviated explanation.

You are however correct about the %age chance of this system failing. It is 95.71825892718769%

that is the number i get.

[Proofreaders2000]

Here's a hand test I did of your unique numbers system and I'm afraid it was not good.
(Riverbelle Live Wheel Casino: Date: January 10,2009: 8:00 am

(I converted the pence figure to American Dollars, starting at $1):: <$> means in the negative<---

1. (win) (Bet $1 on 10 unique numbers): $36-$9=$27
2. (x)--(1x)----<$10>
3. (x)--(4x)----<$40>
4. (x)--(6x)----<$60>
5. (x)--(10x)---<$100>
6. (x)--(15x)---<$150>
7. (x)--(22x)---<$220>
8. (x)--(32x)---<$320>
9. (x)--(46x)---<$460>
10.(x)--(65x)---<$650>
11.(x)--(91x)---<$910>

------------------------Total for the session: <$2920>+$27=<$2893>

I see you started out with .20 pence on the first attempt, so I started the betting at $1.00 and used a progression of 1,4,6,10,15,22,32,46,65,91 to coincide with you progression in British Pounds. 

I betted each time there were 10 unique numbers that were hit only once.
Each time a number hit, I resetted the progression back to $1.

Out of eleven attempts, there was only one hit.  Attempts 2-11 were misses.

[Kon-Fu-Sed]

gingermolly,

I'm sorry that you have trouble understand what I showed above.

Let's do it piece by piece:

The probability for 10 numbers to hit at the first trial is 10/37

If you bet ten numbers on a 0-wheel, your probability to hit is 10/37 (27.027027%)

and so you will win the first stage (26u) in 27,028 sessions

If you start 100,000 sessions, you will hit on the very first trial in 27,027.027027 sessions - I rounded that to 27,028 sessions.
Each time you win at the first stage of your progression you will have a net of 26u.

= 702,728u

You have won 27,028 times 26u = 702,728u

You started 100,000 sessions, won and ended 27,028

Now, as you won 27,028 sessions on the first bet you quit those sessions...

so 72,972 sessions to bet for the 2nd stage.

...so you will bet for a second time in 72,972 sessions.

Each stage is the same.


Now:

SO! 4,280 sessions lost - and they lost 2,920u each so that's 12,497,600u. Lost.

If you lose 4,280 sessions and each loses 2920u, that's a total of 12,497,600u
And you agree to the 95.7% figure... 4,280 sessions is 4.28% of 100,000...

As the total units won by the progression was only 11,453,830u... well.


I hope it's clear now.
KFS

[gingermolloy]

Allow me to offer my mathematical analysis of the 100% mechanical part of this system, i.e the progression.

It is a Grand Martingale progression. That means that if we win at any point we win 1 unit for each spin.

Based the numbers I gave you i.e. in £
0.2
0.8
1.2
2
3
4.4
6.4
9.2
13
18.2

the unit is £7.

The actual profit at each stage is not exactly a multiple of 7 but the mathematics mean that the bets need to be approximated. (i.e one cannot bet £2.132, one would round this up to £2.20) we always round up so that the maths always give an under estimation if how good it is.

i.e.

If i win on the first spin the payout is £7.20 (£0.20 x 36) Therefore £7 profit. (£7 per spin)

if i win on the second spin the payout is £28.80 (£0.80 x 36) therefore £18.80 profit (£9.40 per spin)

if i win on the 3rd spin the payout is £43.20 (£1.20 x 36) therefore £21.20 profit (approx.£7.07 per spin)

if i win on the 4th spin the payout is £72 (£2.00 x 36) therefore £30 profit (£7.50 per spin)

if i win on the 5th spin the payout is £108 (£3 x 36) therefore £36 profit (£7.20 per spin)

if i win on the 6th spin the payout is £158.40 (£4.40 x 36) therefore £42.40 profit (approx.£7.07 per spin)

if i win on the 7th spin the payout is £230.40 (£6.4 x 36) therefore £50.40 profit (£7.2 per spin)

if i win on the 8th spin the payout is £331.20 (£9.2 x 36) therefore £59.20 profit (£7.4 per spin)

if i win on the 9th spin the payout is £468.00 (£13 x 36) therefore £66 profit (£7.3333 per spin)

if i win on the 10th spin the payout is £655.2 (£18.20 x 36) therefore £71.20 profit (£7.12 per spin)

So if from now on i say that we will win £7 a spin, that is an acceptable approximation because we will actually be winning more than that. (just by a little)

Now the way i see the mathematical  chance of losing is this:

(27/37) x (27/37) x (27/37) x (27/37) x (27/37) x (27/37) x (27/37) x (27/37) x (27/37) x (27/37) = 0.0428174107281231 (i.e. 4.28%) or 1 in 23.355.

So, the mathematical chance of this progression failing is 1 in 23 (again rounding on the conservative side.)

So, 1 out of every 24 progressions will fail, and we will lose our bankroll of £584.

But, the other 23 will win.

Now, the amount of money we win from a successful progression depends on how long it takes before we win. i.e. if we win after 1 spin we will win £7, after 5 spins we will win £36.

We could say that the average number of spins taken to win would be the average.

i.e. (1+2+3+4+5+6+7+8+9+10)/10 = 5.5 spins

The amount of money won in 5.5 spins (underestimated) is 5.5 x £7 = £38.5

So the average money won from the successful progression is £38.50.

and for every progression lost (£584) we will win 22 (22 x £38.5) = £847.

for every £584 lost we will win £847.

This combined with the waiting for 10 consecutive numbers results in a system that I personally have never seen fail, EVER!!!.

Comments???

ginger

[gingermolloy]

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