### Author Topic: Scep' s roulette strategies .  (Read 105472 times)

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#### Herby

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##### Re: Scep' s roulette strategies .
« Reply #510 on: March 28, 2018, 08:06:46 AM »
What difference does it make about their origin ?
Hi scepticus,
This is not an attack, my English language knowledge is too worse to find better words:

There is a big difference in reading the mathematical origin compared to the deviated thoughts in a roulette forum.

The word deviated is meant in a neutral way:

e.g. I see a difference to bet on the outcomes of 1 - 12 soccer games and
to bet on the outcome of a dozen using 2 virtual bets.
To bet on a soccer game I see just one chance for a bet.

Sorry if it sounds negative.

#### scepticus

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##### Re: Scep' s roulette strategies .
« Reply #511 on: March 28, 2018, 08:44:33 AM »
herby
Think of dividing your 12 soccer games into 4 soccer games each having 3 possible results and each of these results having the same Probability.
Home win - Away win - Draw .
How would you choose  the results from  3 of them  - in any order?
That is a simple maths question isn't it? Gamblers in the UK are faced with this type of problem during the soccer season. If they studied a block they might be able to eliminate permutations that are UNLIKELY to be profitable. UNlikely due to their knowledge of soccer. Knowledge is a help in any gambling environment .
Anyway, Hopefully Mike  has begun to grasp what I am on about  and I await his reply .
No need to excuse your grasp of English . It seems that I have difficulty in expressing my thoughts  !
Calling a column a line for instance
I think I will call my system . The Crossword Puzzle System
. 4 across 3 down !

#### scepticus

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##### Re: Scep' s roulette strategies .
« Reply #512 on: March 28, 2018, 08:51:20 AM »
as for the maths , herby.
I have already said  that E.L Figgis has given it in  his book " Focus on Gambling "
No maths is really needed . Merely check ANY 4 numbers against ANY block and you will find that 3 will be correct..
Even Mr. Perfect's clever little 4 year old  daughter could do this

#### Herby

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##### Re: Scep' s roulette strategies .
« Reply #513 on: March 28, 2018, 12:20:43 PM »
How would you choose  the results from  3 of them  - in any order?
That is a simple maths question isn't it?

First we look at all possible results.

scepticus blocks are allways exactly 1 element in one row and one column (sudoku similar)
« Last Edit: March 28, 2018, 12:24:38 PM by Herby »

#### scepticus

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##### Re: Scep' s roulette strategies .
« Reply #514 on: March 28, 2018, 12:58:27 PM »
hefrby
if you look at ALL POSSIBLE RESULTS in any block you will find that there are 36 Sets of THREE- THREE ! .
What I claim is that . in the absence of a zero ONE of those COLUMNS will contain at least 3 correct.
If 4 people look for this " needle in  haystack " one  will find it. Won't they. ? And WHY ? Because it is there  !
As for proof I have given it with Mikes 74 numbers and his chosen block. One of the 4 options WINS on each and
every bet !
What more can I do ?
It is up to you guys to PROVE different !
You need to open your minds  !

#### Herby

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##### Re: Scep' s roulette strategies .
« Reply #515 on: March 28, 2018, 01:19:16 PM »
Hi scepticus,
don't be shy, now you can convince that 27 sets of three are enough...

but as you allways wanted more so I began with 81 of four.

#### scepticus

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##### Re: Scep' s roulette strategies .
« Reply #516 on: March 28, 2018, 02:41:54 PM »
And WHERE did I say that 27 sets of three was ENOUGH  herby ?
and your 81 are Sets of  FOUR not THREES - just like Mikes .
Don't you guys know the difference between Threes and Fours ?
« Last Edit: March 28, 2018, 02:45:09 PM by scepticus »

#### scepticus

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##### Re: Scep' s roulette strategies .
« Reply #517 on: March 29, 2018, 11:54:45 PM »

THE LIGHTBULB  MOMENT. For
THE CROSSWORD PUZZLE  SYSTEM .

I have shown that ONE Of the 4 options MUST win - in the absence of zero.
I have also shown that
1-2-3   needs only a 1 unit bet. Pays 2/1 on a win
The other  3
1-2-4
1-3-4 and 2-3-4 each need 2 units bet.  Each pays 7/2 on a win.
Total bets on a certain win are therefore 7 while there are 9 available.
If it were possible to bet  all  7  on the same spin  then 7x37 units  would be required = 259 units..
Payoffs ?
Each has a 1 in 4 chance of winning so each is expected to win  9 in 37 spins. So payoffs would be ;

1-2-3 = 9x 3- 27
1-2-4-- 9x 9= 81
1-3-4 = 9x9 =81
2-3-4 = 9x 9 =81
Total            - 270

270 - 259 - 11
11 / 259 = appr =4.2%

As it is impractical to bet all 7 on the one spin we need to choose either 1 or 2 .
But which  ?
Guys ,  it is only maths applied to roulette   but  - never forget - VARIANCE  RULES !
« Last Edit: March 29, 2018, 11:59:18 PM by scepticus »

#### scepticus

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##### Re: Scep' s roulette strategies .
« Reply #518 on: March 31, 2018, 04:23:28 PM »
Some of the points I have made in this forum.
Not all of them popular LOL
1 )
I have argued that the HE is not the main obstacle to winning . It is lack of knowledge .
2 )
I have questioned WHY Even- Chance Bettors get half their stake back when zero occurs. Why should they when they did not bet it ?
3)
a )
The probability when you bet 1 unit on all 3 EC is 1 in 8.
Some have 5 numbers running for them while others have only  4 .
b)
I have argued that any 1 from each of the three means that , theoretically, each has  one -in 8 chance of winning- odds  7/1- chance yet if you bet some of them  on the  layout you get paid odds of 8/1 on a win .
4 )
Introduced the 9 blocks of 9 numbers which guarantees 3 in one of the 9 lines no matter the 4 results - in the absence of a zero. I argue that it is an aid / tool to help the Bettor .

This will be my last post in this forum. I may look in from time to time but wonâ€™t post - even if I read snide posts directed at  me .
Good Luck To You All !