Author Topic: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!  (Read 14123 times)

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BlueAngel

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POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« on: December 19, 2015, 12:05:46 AM »

this is a thread which requires your full attention in order to fully comprehend the hidden opportunity.

In every 37 spins cycle there are 24 numbers which have appeared once or more and 13 sleepers
(numbers which haven't appeared within the last 37 spins)
Those are average numbers and deviations exist, for example I've seen up to 30 different numbers to show up within the 37 last spins and the least were 18 different numbers.
Those extremes are from my experience during gambling sessions and not from simulations.
In order to find the average we should determine the extremes,or in other words the limits, in that case 18 and 30 are the limits.

18 19 20 21 22 23 24 26 26 27 28 29 30 start by discarding the outer numbers from both sides:

And we conclude to 24 which is the average total of the appeared numbers within 37 last spins.
Another way to calculate the average is to add all the totals and then divide the total sum with the total of the averages,for example:

18 different numbers within the last 37 spins happened 1 time
19 different numbers within the last 37 spins happened 2 times
20 different numbers within the last 37 spins happened 3 times
21 different numbers within the last 37 spins happened 4 times
22 different numbers within the last 37 spins happened 5 times
23 different numbers within the last 37 spins happened 6 times
24 different numbers within the last 37 spins happened 7 times
25 different numbers within the last 37 spins happened 6 times
26 different numbers within the last 37 spins happened 5 times
27 different numbers within the last 37 spins happened 4 times
28 different numbers within the last 37 spins happened 3 times
29 different numbers within the last 37 spins happened 2 times
30 different numbers within the last 37 spins happened 1 times

[(1 x 18)+(2 x 19)+(3 x 20)+(4 x 21)+(5 x 22)+(6 x 23)+(7 x 24)+(6 x 25)+(5 x 26)+(4 x 27)+(3 x 28)+          (2 x 29)+(1 x 30) / 49] =>

[(18+38+60+84+110+138+168+150+130+108+84+58+30) / 49]=>

1176 / 49 = 24

This means that if someone was betting one number for 37 spins, 24 times he/she would have won and 13 times he/she would have lost.
To calculate the total sum of lost bets is easy:
13 times multiplied by 37 (1 x 37) equals minus 481 units.

In order to find the total amount of won bets, we should again calculate the averages as we did above,let's see:
We start from the middle numbers within a 37 spin cycle and add 2 more numbers from both sides till we have a total of 24.
18th and 19th spins
17th and 20th spins
16th and 21st spins
15th and 22nd spins
14th and 23rd spins
13th and 24th spins
12th and 25th spins
11th and 26th spins
10th and 27th spins
9th and 28th spins
8th and 29th spins
7th and 30th spins
So we have 24 wins from different spins, now let's calculate the total amount of net profit.

29+28+27+26+25+24+23+22+21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6 = 420

But since we would lose 481 - 420 = -61 loss, therefore the definition negative expectation.

Do we agree so far?
Is everything clear?

Let's see how we could turn the negative expectation into positive without changing a thing in terms of probability, in fact with exactly the same results which have lead us to the negative balance above.

The average expectation is to get twice the wins for every loss, with my method you only need approximately 1 positive cycle for every negative cycle, which means that the proportion of 2 to 1 changed to 1 to 1.

Let me explain how this is possible, the only thing which I'm not going to reveal you here is the criteria which I'm using to select the betting number.
You could pick a random number, or your "lucky" number or anything else you like.

I start flat betting 1 number with 1 unit for 36 spins.
When the betting number appears, no matter in which spin, I re-bet the same number plus its neighbour regarding the wheel layout (right or left doesn't matter)

So now I'm flat betting 2 numbers with 1 unit each for the next 18 spins.
When one of the two betting numbers hits within eighteen spins, then I re-bet those two numbers and adding the other wheel neighbour, thus in total three numbers for the next 12 spins, always flat bet 1 unit each.

When one of the three betting numbers comes, I add one more neighbour,this time from the table layout.
So far we have 4 numbers to flat bet with 1 unit each for the next 9 spins.
Let's see if you keep on winning what happens:

5 numbers flat bet with 1 unit each for the next 7 spins
6 numbers flat bet with 1 unit each for the next 6 spins
7 numbers flat bet with 1 unit each for the next 5 spins
8 numbers flat bet with 1 unit each for the next 4 spins
9 numbers flat bet with 1 unit each for the next 4 spins
10 numbers flat bet with 1 unit each for the next 3 spins
11 numbers flat bet with 1 unit each for the next 3 spins

Eleven is the maximum numbers you could bet, this may happen if your first number was 17 for example, because 17 has 2 neighbours on the wheel's layout (like every number) and 8 neighbours regarding the table layout 13,14,15,16,18,19,20,21 (yes,the diagonals too).

What if someone is lucky enough and after adding all neighbours one by one continues to win?
In this case you start adding 1 unit each time you win to only 1 of your betting numbers, you should start from your first selection and continue with the same order.
The minimum total of betting numbers are 6, for example 34 has 2 neighbours at the wheel's layout (17,6) and 3 neighbours at the table (31,32,35)
All together with 34 are six numbers.

We have seen how it might proceed if you are lucky and win frequently, this is not so rare because sometimes the croupiers are hitting specific sectors/numbers frequently (more than their probability)
Personally I've reached two times the 6 numbers bet selection after not many trials, of course this is not always the case.
Let's see what happens when we eventually lose, when you lose during 2 numbers bet selection or more, then you just pick another number and start from scratch, which means flat bet 1 number for the next 36 spins.
There is only one exception, if you lose during 1 number bet selection, in that case you pick another number but this time the betting wouldn't be flat, each and every time our number fails to hit we would add some interest by adding 1 unit.
So the bet on our number would be like this:
1st spin 1 unit
2nd spin 2 units
3rd spin 3 units
4th spin 4 units and so on till your number eventually appears or till you have lost for 37 spins in a row.

If you lose for second consecutive time your 37 bet cycle,this means that somewhere ahead are 4 more winning cycles/rounds but you would only need 2 of them in order to overcome the 2 negative ones.
You change again the number and continue with the dynamic progression till your winning cycles/rounds equals your losing ones.

Let's calculate the losing and winning totals by betting with the dynamic progression instead of the flat bet.
Remember that the average expectation is to lose 13 times within 37 cycles/rounds, so the total loss will be:
[13 x (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37)]=>
13 x 703 = -9139 units

Now let's calculate the total amount of profit:

[(7 x 35) - 21] + [(8 x 35) - 28] + [(9 x 35) - 36] +[(10 x 35) - 45] + [(11 x 35) - 55] + [(12 x 35) - 66] + [(13 x 35) - 78] + [(14 x 35) - 91] + [(15 x 35) - 105] + [(16 x 35) - 120] + [(17 x 35) - 136] +[(18 x 35) - 153] + [(19 x 35) - 171] + [(20 x 35) - 190] + [(21 x 35) - 210] + [(22 x 35) - 231] + [(23 x 35) - 253] + [(24 x 35) - 276] + [(25 x 35) - 300] + [(26 x 35) - 325] + [(27 x 35) - 351] + [(28 x 35) - 378] + [(29 x 35) - 406] + [(30 x 35) - 435] =>

(245 - 21) + (280 - 28) + (315 - 36) + (350 - 45) + (385 - 55) + (420 - 66) + (455 - 78) + (490 - 91) + (525 - 105) + (560 - 120) + (595 - 136) + (630 - 153) + (665 - 171) + (700 - 190) + (735 - 210) + (770 - 231) + (805 - 253) + (840 - 276) + (875 - 300) + (910 - 325) + (945 - 351) + (980 - 378) + (1015 - 406) + (1050 - 435) =>

224 + 252 + 279 + 305 + 330 + 354 + 377 + 399 + 420 + 440 + 459 + 477 + 494 + 510 + 525 + 539 + 552 + 564 + 575 + 585 + 594 + 602 + 609 + 615 = +11080

We deduct 9139 from 11080 and we find 1941 net profit, therefore the negative expectation has been turned into positive!
But wait, there are more good news!
Since always the first losing cycle/round costs us 36 units instead of 703, that's why you don't need the exact probability to happen (2 winning cycles for every 1 losing cycle), you need approximately equal winning/losing cycles.
I say approximately because you never know in which spin your number is going to appear, thus the net gain differs.
Let me put it this way,if after 2 losing cycles you have 2 winning ones, then you are in profit and re-start from 1 number flat bet with 1 unit for 36 spins.

But after 3 or more consecutive losing cycles you would need the same amount of winning cycles plus 1 more.
Even in such situation, you can be in the positive with worst results than what probability theory supports.
For example in order to overcome 3 losing rounds, I just need 4 instead of 6 which probability dictates as average.
Of course,the results could also be better than average!:-)

Just remember that in the first phase, while we flat bet, we try to find the possibility to win by frequent repeaters.
We build slowly and safely a net gain and we don't limit the possibility for more profits (sky is the limit!) but we limit the loss.
During any stage of the 1st phase (flat bet) the maximum amount we risk is 36 units or less, if you don't lose during the 1st (1x36) or 2nd stage (2x18) of the first phase, then you would be in the positive.
The longer you keep on winning with the flat bets, the more would be the profit, it's all about what happens first.

Under any circumstance you will know what to do next, I'd recommend a total bankroll of 3000 minimum, 5000 average and 10000 units maximum in order to overcome occasional distribution anomalies.

Angelo A.
« Last Edit: December 19, 2015, 12:37:11 AM by BlueAngel »

scepticus

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #1 on: December 19, 2015, 03:51:29 AM »
Congratulations on your method , Angelo, but it is not for me .
My table bank is now down to [/size] 80 units Maximum 90 and I expect to profit within 37 spins at each table and my results so far indicate a Long Term Bank would be no more than 10x 90 .Call it 1,000.   Your contribution here is just what the forum was designed for - actual methods for discussion and not carping critics . Congrats.  again.

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #2 on: December 19, 2015, 05:21:37 AM »
Congratulations on your method , Angelo, but it is not for me .
My table bank is now down to 80 units Maximum 90 and I expect to profit within 37 spins at each table and my results so far indicate a Long Term Bank would be no more than 10x 90 .Call it 1,000.   Your contribution here is just what the forum was designed for - actual methods for discussion and not carping critics . Congrats.  again.

Thanks, such recognition motivates me to strive for better and more!
I do understand that the bankroll required is not for the average gambler, but if you are feeling bold enough and start with less than 2000 units,then a bad start could ruin your budget.

I have tested extensively and I've seen up to 4 consecutive losing rounds,if those rounds happen later,after you gain some serious profits it would be fine,but what if you happen in the worst possible start??

december

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #3 on: December 19, 2015, 09:14:35 AM »
Thanks for help, Angelo.

Am I wrong if I don't see that my number must come 24 times in 37 cycles?
What if it hit only 7 times, or 5, or 10?

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #4 on: December 19, 2015, 10:12:47 AM »
Thanks for help, Angelo.

Am I wrong if I don't see that my number must come 24 times in 37 cycles?
What if it hit only 7 times, or 5, or 10?

What you are saying is not true, I believe Reyth and Kavouras could prove or disprove my claim with some simulations and/or results.
But it's somehow subjective since we can select different number with different results.
However,five,seven or ten times within 37 cycles of 37 spins each is simply unrealistic.
If you asked me what will come the next 37 spins,I could not determine with certainty,but I would have a high degree of certainty that I would win AT LEAST 18 cycles out of 37.

december

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #5 on: December 19, 2015, 11:56:36 AM »
This means that if someone was betting one number for 37 spins, 24 times he/she would have won and 13 times he/she would have lost.

Sorry Angelo, I don't know what was with my mathematics this morning...
Yes, of course by average I should've win 37 times time from 37 cycles...
It can be from Glühwein?
« Last Edit: December 19, 2015, 04:02:31 PM by december »

Real

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #6 on: December 20, 2015, 05:56:35 PM »
The law of the third will not help anyone beat the random game of roulette.  It will not provide any player an advantage over the casino.  Not even a tiny one.

The math shown above is strange.

Since the same numbers remain on the wheel from one spin to the next, the the odds do not change in the random game of roulette.
« Last Edit: December 20, 2015, 05:59:14 PM by Real »

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #7 on: December 20, 2015, 08:38:47 PM »
This means that if someone was betting one number for 37 spins, 24 times he/she would have won and 13 times he/she would have lost.

Sorry Angelo, I don't know what was with my mathematics this morning...
Yes, of course by average I should've win 37 times time from 37 cycles...
It can be from Glühwein?

You seem confused, let me clarify that what I call cycle or round is 37 spins and 37 small cycles make 1 big cycle,therefore the "big" cycle contains 37 x 37 = 1369 spins
I've never said,nor anyone could win 37 times in 37 cycles! Perhaps only God!

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #8 on: December 20, 2015, 08:41:51 PM »
The law of the third will not help anyone beat the random game of roulette.  It will not provide any player an advantage over the casino.  Not even a tiny one.

The math shown above is strange.

Since the same numbers remain on the wheel from one spin to the next, the the odds do not change in the random game of roulette.

You seem completely ignorant of statistics, hmmm...strange, are you supposed to be an expert?

Real

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #9 on: December 20, 2015, 09:12:49 PM »

Quote
5 numbers flat bet with 1 unit each for the next 7 spins
6 numbers flat bet with 1 unit each for the next 6 spins
7 numbers flat bet with 1 unit each for the next 5 spins
8 numbers flat bet with 1 unit each for the next 4 spins
9 numbers flat bet with 1 unit each for the next 4 spins
10 numbers flat bet with 1 unit each for the next 3 spins
11 numbers flat bet with 1 unit each for the next 3 spins-Blue Angel

Perhaps you can explain the following.

How can past numbers in any way influence what will happen on future spins?  What's the physical effect?

Quote
each and every time our number fails to hit we would add some interest by adding 1 unit.-Blue Angel

Regardless of whether you bet the neighboring numbers or numbers scattered about the wheel, the results will be the same in the random game of roulette.  Furthermore, adding the "up as you lose progression" in no way, whatsoever, changes the house edge.  The expectation is still NEGATIVE.
« Last Edit: December 20, 2015, 09:22:05 PM by Real »

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #10 on: December 20, 2015, 11:34:23 PM »

Quote
5 numbers flat bet with 1 unit each for the next 7 spins
6 numbers flat bet with 1 unit each for the next 6 spins
7 numbers flat bet with 1 unit each for the next 5 spins
8 numbers flat bet with 1 unit each for the next 4 spins
9 numbers flat bet with 1 unit each for the next 4 spins
10 numbers flat bet with 1 unit each for the next 3 spins
11 numbers flat bet with 1 unit each for the next 3 spins-Blue Angel

Perhaps you can explain the following.

How can past numbers in any way influence what will happen on future spins?  What's the physical effect?

Quote
each and every time our number fails to hit we would add some interest by adding 1 unit.-Blue Angel

Regardless of whether you bet the neighboring numbers or numbers scattered about the wheel, the results will be the same in the random game of roulette.  Furthermore, adding the "up as you lose progression" in no way, whatsoever, changes the house edge.  The expectation is still NEGATIVE.

It's a fact that some numbers and segments/sectors of the wheel appear more than others.

You may explain this fact due to physical elements according your perception, while others might perceive it as statistical distribution anomalies.

Personally I don't care why it happens, but I do care that it HAPPENS.

Equilibrium or balance if you like is only in the very long term and even then it's not EXACT balance.
What probability theory dictates is only a mean to measure the level of deviations, there is no practical value in equilibrium and probability theory.
A lot of short term deviations create a long term balance.

This is the way I interpret the game, you don't have to agree, neither my intention is such.

dobbelsteen

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #11 on: December 22, 2015, 09:30:37 AM »
The  House edge is only true for long run events. AP games can not ignore the House edge even.

Real

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #12 on: December 22, 2015, 08:35:31 PM »
The  House edge is only true for long run events. AP games can not ignore the House edge even.

Dobbelsteen,

Fyi...AP means advantage player.  In other words the player has the edge over the casino.

Blue Angel,

I don't know why you brought up equilibrium, and it appears that you don't know either.

-Real
« Last Edit: December 22, 2015, 08:58:17 PM by Real »

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #13 on: December 23, 2015, 01:21:23 AM »
Real,

Equilibrium is another way to say balance.

Do you consider risk of ruin a valid and sound test?
For example if a method with 1000 units capital doubles it and reaches 2000 and then doubles it again to 4000 without losing the capital on the way, would this method be valuable or lucky??

Real

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY!
« Reply #14 on: December 23, 2015, 02:55:57 AM »
Blue Angel,

Regarding equilibrium...I still don't know what your point is.

The risk of ruin in relation to the money returned is poor for up as you lose progressions.

« Last Edit: December 23, 2015, 02:58:11 AM by Real »