5/7 EC's....
In the first 4 rows, no matter which combination of EC roulette produces, 1 column WILL have 4 hits.....
Considering over 7 spins 1 column MUST emerge the winner with 5 hits, would it be fair to say that over the next 3 spins the column that had 4 hits would be a better selection because we only need 1 hit to make it 5, whereas any other column needs 2 hits....
A A A A A A A A
A A A A B B B B
A A B B A A B B
A B A B A B A B
A A B B B B A A
A B A B B A B A
A B B A B A A B
No, because future results are unrelated to previous outcomes.
Each spin is always exactly the same.
Why do so many people struggle with this?
Even in the old roulette textbooks written last century the authors knew, fully accepted and understood that waiting for things to happen is completely pointless.
Fair call....
But this is something I found written centuries ago too...
How about dozens...
1 1 1 2 2 2 3 3 3
1 2 3 1 2 3 1 2 3
2 3 1 3 1 2 1 2 3
2 3 1 1 2 3 3 1 2
In 4 spins, of the combinations, 1 column must have 3 hits...
But instead of just going out and playing every spin what if on the third spin we play against the column that initially hits twice, betting it won't make a 1,2,3 treble... Then if win, parlay those winnings back onto that column we avoided to make it a win on spin 1,2 and 4....
A 7/2 shot paying 8/1... but of course that would depend on the result of the 3rd spin...
why must it hit? sorry cant follow
Quote from: Anastasius on May 20, 12:53 AM 2019
why must it hit? sorry cant follow
Sorry mate forgot to mention,
Because in the 9 sets of 4 we are covering all 81 combinations... so in 4 spins something must hit 3 times excluding zero
A good topic, as I understand it, if in the first two columns we have
11 - bet against 11 - 23?
21
32
22 - bet against 22 - 13?
Quote from: Person S on May 20, 03:19 AM 2019A good topic, as I understand it, if in the first two columns we have
11 - bet against 11 - 23?
Hi mate,
If the initial 2 results were 1,1
You would look to see which column in the grid of 9 has that double as the first 2 results...
In this case it is column 1...
The 4 results for column 1 are 1122...
All i was saying was, you COULD play for column 1 not to win in the first 3 spins...
So bet against its third outcome...
Which would mean playing for dozen 1 and 3...
From there, if that won, you could then put your 3 units back on column 1 to be the winning column with 3 hits on spin 1,2 and 4...
Which would mean playing its 4th result - a single dozen 7/2 shot that pays 8/1...
Though there are many many ways to play... Muck around with it mate let me know what you come up with
Quote from: mickavelli on May 19, 08:10 PM 2019
5/7 EC's....
In the first 4 rows, no matter which combination of EC roulette produces, 1 column WILL have 4 hits.....
Considering over 7 spins 1 column MUST emerge the winner with 5 hits, would it be fair to say that over the next 3 spins the column that had 4 hits would be a better selection because we only need 1 hit to make it 5, whereas any other column needs 2 hits....
A A A A A A A A
A A A A B B B B
A A B B A A B B
A B A B A B A B
A A B B B B A A
A B A B B A B A
A B B A B A A B
This is a bit like the even chance cycle argument beloved by the non random gurus.
Once one chance defines a cycle, there is a 75% chance that chance will complete it. Unfortunately it's a fallacy argument and offers no betting advantage. You'll lose to the House edge.
My version of the example of AB:
You can try to play 2 games
We have 4 sets of 9 combinations. As is well known, a combination of A or B will have 4 repeats AAAA / BBBB
1. Play VDW in 4 columns before AAAA / BBBB appear. If the winnings start over. If you lose, switch to step 2.
2. We are looking for the dominant A or B, for example, AABBA and we put on the fact that A will become AAAA.
The problem is in good money management. I'm not very good at it ...
We can also try to create pairs and get 1/3.
Quote from: Firefox on May 20, 07:19 AM 2019You'll lose to the House edge.
Take the Ec's for what they are mate...
But what are your thoughts on the dozens??
Don't forget when doing your joint probability calculations the first 2 spins ARENT played...
So the 106 combinations are reduced after the first spin, and further reduced after 2 spins....
And if the third spin matches the column that won the first 2 spins we lose and our session has ended we don't have a bet on spin 3 and 4...
But if the double dozen bet wins on spin 3 we have now reduced it to how many options? A 3/1 chance that when parlayed pays 7/2
Quote from: mickavelli on May 20, 08:12 AM 2019the 106 combinations are reduced
My bad - 108 combinations...
The third and fourth rows are no different to the first 2, they contain all possible doubles for 2 spins - NOT including zero..
If you choose 2 , you have 7 against you, hence - 7/2 chance.. Which when parlayed onto spin 4, pays 8/1... But because we have already bet 2, the payout is reduced to 7/2...
Over 72 spins we can expect to bet 36 x 2 points = 72 and expect to win 8 for a payout of 9 for each win..
So 8×9 = 72 - Break even
Except we forgot the zero so,
over 37 bets we can expect to lose 2
But the strategy is to not bet the fourth spin if the third is lost - Which means the third and fourth of the first 2 can't be considered as 1 of 9 available possibilities! Leaving only 8 possibilities reducing the odds to 2/8 for a 7/2 payout!
Can you give an example. That moment when we miss the moves? Tnx!
Played in combinations, play is not perfect.
ONLY E / C.
Play money
Slightly changed the game, now the bet on the cycle.
I wonder who can analyze how many times in a row can get CL4?
Quote from: mickavelli on May 20, 08:12 AM 2019Take the Ec's for what they are mate...
But what are your thoughts on the dozens??
The dozens also lose in a similar manner.
But they lose twice as quick, as the house edge on dozens is -2.7% whereas on the even chances it is only -1.35% assuming partage or en prison rule.
It's not possible to get an advantage on an Even Chance cycle without zero and the same with dozens. Just play a cycle out through all outcomes and you'll see they are break even.
Add in zero and it's a losing game, the dozens losing twice as much for the same value bets.
Quote from: Firefox on May 21, 06:37 PM 2019It's not possible to get an advantage on an Even Chance cycle without zero and the same with dozens.
What was suggested is that we bet against the third in the column that initially hits twice because it has only a 1 in 3 chance of being the winning column. So for the third we are betting two dozens on that single spin. If one of them wins the problem then becomes "Which dozen do we bet on the next (4th ) spin? We have three options here , haven't we -
dozen 1 - dozen 2 - dozen 3
Which one you choose is up to you.
It was suggested betting the one below the first two winners but that is a personal choice. No matter which one you choose you get odds of 7/2 when you bet the 3 chips from the previous win and thus collect 9 chips.
It can be seen that these are correct odds when discounting the zero.
You mentioned non random but this relies on random to be in our favour to beat the HE
Basically, if the first bet looses, play is over, there is no bet on spin 4 - The 9 options are reduced to 8 !
Having now only 8 "Pairs" to contend with after putting ONE chip on each of the TWO dozens on the first bet and if one of them wins bet THREE chips on ONE dozen on the next spin and collect 9 chips - if won - we win with fair odds of 7/2.
We had 2 chances to win "1 pair" of NOW - 8 "pairs" - a 1 in 4 chance.
Fair odds for which is 3/1 ( 3+1 =4 )
2/8 is one in four chance so - payout should be 3/1 if successful.....
Here, we are getting 7/2, even after allowing for zero!
Nobody is claiming that advantage must win - people like yourself do !
Nothing is guaranteed except for the fact that when you put 3 numbers against each other using the grid 1 column is guaranteed more hits
How we use it is up to us!
Dozens aren't the only thing with 3 numbers on the table for example if you played columns at the same time there are straights that overlap.. Or 4 numbers that intersect...
See what you come up with!
Quote from: Person S on May 21, 06:25 PM 2019
Slightly changed the game, now the bet on the cycle.
I wonder who can analyze how many times in a row can get CL4?
What game are you playing here mate?
the topic has reminded me of this one. so fyi.
cheers
hans :)
This is exactly it !! Thanks a million mate I have been looking everywhere for this
Quote from: mickavelli on Jun 01, 09:10 PM 2019
What game are you playing here mate?
We have 4 possible outcomes of RR-RB-BB-BR.
I'm looking for an opportunity to repeat, if you take the grid, it will look like this:
RB - if 9 spin comes out R, then on spin 13 we can play for the combination of RB. The rest is identical ...
RR
RR
RB