Author Topic: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2  (Read 5347 times)

BlueAngel

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On the first part I've proved with calculations how a progression can turn a negative expectation in to positive.
This time, on the sec
ond part of the mini series, I'm going to show how a bet selection without any progression can result in positive results in contrast with the conventional negative expectation.

Take a close look at the following groups of numbers

ROL    1  3  5  7  9         BEH   20  22  24  26  28
ROH  19 21 23 25 27      BEL      2  4  6  8  10
REL   12  14  16  18       BOH   29   31   33   35
REH   30  32  34  36       BOL     11  13  15  17

These are the groups of all 3 EC's combinations, each group contains 4 up to 5 numbers.
Let's say I bet simultaneously Black,Even,High what would make me lose all 3 bets?

The numbers 1,3,5,7 and 9, a total of five numbers which all of them belong to 1st dozen...
The 1st dozen contains also numbers which I could win with my EC bets, such as: 2,4,6,8,10

By betting 3 EC's simultaneously there are more chances to lose or win 1 unit, rather than win or lose 3 units and that's because more numbers are resulting in 2 out of 3 EC's to change or remain the same.

Except from the Even,Black,High there is also equally good the Red,High,Odd in combination with 1st dozen.
A total of 2 winning combinations, each one contains 4 flat bets with 1 unit each.

There is 1 number resulting in -4 (0)                                                                                   -4
There are 5 numbers resulting in -1 (1,3,5,7,9 for BEH or 2,4,6,8,10 for ROH)                        -5
There are 5 numbers resulting in +3 (1,3,5,7,9 for ROH or 2,4,6,8,10 for BEH)                    +15
There are 5 numbers resulting in +2 (20,22,24,26,28 for BEH or 19,21,23,25,27 for ROH)    +10
There are 4 numbers resulting in 0 (30,32,34,36 for BEH or 29,31,33,35 for ROH)                   0
There are 2 numbers resulting in +1 (11,12 for both BEH and ROH)                                      +2
There are 6 numbers resulting in -2 (13,14,15,16,17,18 for both BEH and ROH)                    -12

Calculating all totals we find: (15+10+2)-(4+5+12)= 27 - 21 = +6 grand total


This is what I call positive expectation and if you are not convinced so far, then nothing and nobody could ever convinced you!

« Last Edit: February 01, 2016, 10:26:14 AM by BlueAngel »


 

december

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #1 on: February 01, 2016, 10:57:00 AM »
What to say, plenty of material to analyze...
Thanks one more time, Blue!
 

Bebediktus

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #2 on: February 01, 2016, 11:17:02 AM »

Calculating all totals we find: (15+10+2)-(4+5+12)= 27 - 21 = +6 grand total


This is what I call positive expectation and if you are not convinced so far, then nothing and nobody could ever convinced you!


Interesting how much millions you won on such your mathematic :)
 

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #3 on: February 01, 2016, 11:35:53 AM »

Calculating all totals we find: (15+10+2)-(4+5+12)= 27 - 21 = +6 grand total


This is what I call positive expectation and if you are not convinced so far, then nothing and nobody could ever convinced you!


Interesting how much millions you won on such your mathematic :)

Can you prove the opposite??

I've calculate every possible outcome, the only issue is variance which could deviate results from the mean, but variance is a double edged sword, thus not necessarily bad if you get what I mean...
The way I've proved with simple calculations how positive expectation is possible, the same way all those mathematicians have calculated the house edge of -2.7% negative expectation.

So by making an argument with me you are arguing also with basic probability theory!
 

Bebediktus

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #4 on: February 01, 2016, 11:52:17 AM »
Quote
Can you prove the opposite??
I can .

You really  have such knowledges in mathematic, as you want to show ?
I ask you -  HOW MUCH MILLIONS YOU WON ON THIS ?
Why you not run to casino to take "your money "?
 

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #5 on: February 01, 2016, 12:01:13 PM »
Quote
Can you prove the opposite??
I can .

You really  have such knowledges in mathematic, as you want to show ?
I ask you -  HOW MUCH MILLIONS YOU WON ON THIS ?
Why you not run to casino to take "your money "?

Unfortunately for you, you cannot prove me wrong.

Who says I didn't, actually I practice what I preach EVERY single day of my life!
When you posted your first message I had just finished another successful session, guys like you make me laugh so hard!:-D LOL

Bebedictus I would like to thank you for your contribution to my entertainment!
 

december

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #6 on: February 01, 2016, 12:14:34 PM »
Bebedictus I would like to thank you for your contribution to my entertainment!

I am sure there are more entertainers to come...
 

rovait

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #7 on: February 01, 2016, 12:39:38 PM »

Hi, BlueAngel!

I don't see how you get +1 in bellow statement: if we bet BEH/ROH + first dozen we don't get eny profit. +1 for first dozen + 1 for color or even/odd depending on combination of EC we bet on. -1 for high and -1 for for color or even/odd
There are 2 numbers resulting in +1 (11,12 for both BEH and ROH)          +2

where is my misstake?
Thanks.
Stas
« Last Edit: February 01, 2016, 12:46:26 PM by rovait »
 

rovait

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #8 on: February 01, 2016, 12:45:40 PM »

I also do not see how you get here +3:
There are 5 numbers resulting in +3 (1,3,5,7,9 for ROH or 2,4,6,8,10 for BEH)    +15

when we win on color, dozen and even/odd +3 we also loose on high. so, the profit is 3-1=2
 

rovait

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #9 on: February 01, 2016, 12:52:16 PM »

in bellow case we loose -2 not -1 (by my calculation)

There are 5 numbers resulting in -1 (1,3,5,7,9 for BEH or 2,4,6,8,10 for ROH)
 

Mike

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #10 on: February 01, 2016, 12:57:29 PM »
LOL, BA demonstrates nicely that systems are for players who can't do math.
 

rovait

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #11 on: February 01, 2016, 01:13:20 PM »
according to my calculation the expectation is -6 which is negative.

I wish it was positive.
 

Mike

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #12 on: February 01, 2016, 01:27:53 PM »
Of course it's negative. Do the math for ANY roulette system and the result will be negative. It cannot be otherwise and why BA persists with this nonsense is beyond me.
 

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #13 on: February 01, 2016, 01:45:54 PM »

I also do not see how you get here +3:
There are 5 numbers resulting in +3 (1,3,5,7,9 for ROH or 2,4,6,8,10 for BEH)    +15

when we win on color, dozen and even/odd +3 we also loose on high. so, the profit is 3-1=2

Every time the total bet is 4 units, on the above situation we win from Odd 2 units plus from Red 2 units plus from 1st dozen 3 units for a total return of 7 units in our 4 units bet, deduct four from seven and you have 3 units net gain from 5 numbers (1,3,5,7,9)
The same goes for Black and Even which are winning the same amounts with 5 numbers (2,4,6,8,10).
« Last Edit: February 01, 2016, 01:54:42 PM by BlueAngel »
 

BlueAngel

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Re: POSITIVE EXPECTATION ON ROULETTE CAN BE PROVED MATHEMATICALLY! part 2
« Reply #14 on: February 01, 2016, 01:47:13 PM »

Hi, BlueAngel!

I don't see how you get +1 in bellow statement: if we bet BEH/ROH + first dozen we don't get eny profit. +1 for first dozen + 1 for color or even/odd depending on combination of EC we bet on. -1 for high and -1 for for color or even/odd
There are 2 numbers resulting in +1 (11,12 for both BEH and ROH)          +2

where is my misstake?
Thanks.
Stas

Here there is not any mistake.