I'm fascinated if such a system exists... You wouldn't win anything if you had exactly 1 chip on each number, of course, but perhaps something magical could happen with differing bets per number? If you had a tally showing the appearance totals of each number after so many spins, you could bet a different amount on each number based on how far they are individually behind or in-front one another, relatively. But you would not have enough time to make such a bet, though what you can do is maybe add chips as numbers get cold and/or reduce chips as numbers get hot. Eventually, you would have a different amount of chips on each number based on their trends, and you could work the system simply by using the "repeat bet" button. To make it work would mean good bet selection for the initial bet only, followed by excellent timing and pacing of your progression.
link:://wizardofodds.com/gambling/ten-commandments/ (link:://wizardofodds.com/gambling/ten-commandments/)
7.Thou shalt not hedge thy bets
Hedge bets usually carry a high house edge. For example, never take insurance in blackjack and never bet the any craps or any seven in craps. Exceptions can be made for insuring life changing amounts of money.
with the info your now beginning to get on ds's, why not back the first 3 ds's group (1) and the last3 ds's group(2), 1st 18 (1) 2nd 18 (2), 1 unit on Mr Green.
Now if Mr Green comes bingo, if not you've lost1 unit,so restake and the group that lost double up. So 1st bet 7 units 2nd bet 10 units, if double group hits you'd be +1. Just hope you dont get a long run of doubling up,when stake is going above the return of Mr Green add another unit on it
Quote from: falkor on Oct 05, 07:51 PM 2014
You wouldn't win anything if you had exactly 1 chip on each number, of course, but perhaps something magical could happen with differing bets per number?
That tells a lot about your thinking. Not only you wouldn´t win anything,
you would lose 1 unitQuoteNow if Mr Green comes bingo, if not you've lost1 unit,
Nottophammer, why then not betting only Mr. Green?
QuoteThat tells a lot about your thinking
Yeah I thought that too.
QuoteNot only you wouldn´t win anything, you would lose 1 unit
He did say INCLUDING the 0
@falkor - if you are saying cover the whole table then start adding, why not leave the table empty then start adding, starting at no chips as opposed to 1 chip, that would keep this astronomical waste of chips/time thought of yours a bit cheaper, good luck with the tests.
Nope if the doubled group not hit you'll be plus, only problem is a long run on doubling up could get you to table limit.
(1) (2) (Zero) (1) (2) (Zero)
2 2 L L 3 3 1 = 7 units return 6 units
17 17 L L 3 6 1 = 10units return 6 units
15 15 L L 3 12 1 = 16 units return 6 units
34 L 34 L 3 24 1 = 28 units return 48 units
= 61 units return 66 units +5
Winkel
Nottophammer, why then not betting only Mr. Green?
I do play the 6 ds's and sometimes .20p on zero because in uk betting shops they want £1.00 to spin the wheel and i like to have 10 spins to see how many left on the mat . Most times leave the .20p off thou as if Mr Green dont come in the 1st 10 spins thats £2.00 wasted,if green comes once still .80p better off
I would find it beneath my dignity to enter a casino and cover every single number. I am better off joining an old ladies knitting circle on a Thursday afternoon.
Tam can you think of any other way of doing what the topic asked.
Like winkel said why not just go for the green,but then theres only 1 chip not 37 at the start, i've only gone to 6 doubles as thats the max in the betting shops and lost it but once you start the bet why stop, if you've not got the nerve to play it, then thats fine.
I saw a system once where 30 numbers were covered. 15 numbers had 2 units 15 numbers had 1 unit
yes Rich but he asked for the whole board to be covered, you could use the double up method on your mid ds's but as said you wont want to be doubling for to long.
okay lets have 1st 18 as 3 DS' and top 18 as 3 DS's, lets call them (1) and (2).
if you see (1) 1st 18 miss for 11 spins would you not fancy backing both hoping that (1) misses again, then restake doubling (2) and comes in giving you +1, if it didnt but say green dropped in you be up,you could relay the bet and carry on doubling (2) remember we're saying missed for 11 spins, i'd play for another 6 spins with the old foot tapping,worst case be -1 on the 1st bet,its only 1
Quote from: iggiv on Oct 05, 09:16 PM 2014
link:://wizardofodds.com/gambling/ten-commandments/ (link:://wizardofodds.com/gambling/ten-commandments/)
7.Thou shalt not hedge thy bets
Hedge bets usually carry a high house edge. For example, never take insurance in blackjack and never bet the any craps or any seven in craps. Exceptions can be made for insuring life changing amounts of money.
Off-topic but now that you mention it:
Quote4. Thou shalt trust the odds, not hunches
If you want to maximize your odds then believe in mathematically proven strategies, not hunches. If hunches are so great why are there so many psychics working the Boardwalk in Atlantic City as opposed to playing?
Can anyone take this seriously?
If you always cover the zero while betting on say DSs, after a bad run you may need more wins than usual to brake even, because of the wages lost on "insurance".
If we cover numbers 1 thru 35 with 1 unit and we leave number 36 and the zero uncovered. What will the outcome be if we were to make this bet for the next 37 spins and each number hit exactly once starting with the zero and ending with #36?
On the zero you would lose 35 units which equals -35 units.
On numbers 1 thru 35 we would win 1 unit each which equals +35 units.
On number 36 you would lose 35 units which equals -35 units.
That means we net -35 units playing this bet method.
If we only bet 1 unit on 1 number for our 37 spins, what would be the outcome if every number hits exactly once in the next 37 spins?
On 35 of the numbers we would lose 1 unit which equals -35 units.
On the covered number we would win 35 units which equals +35 units.
On the uncovered zero we would lose 1 unit which equals -1 units.
That means we net -1 unit with this bet method.
I've asked this before and I ask it again. How can we play multiple numbers knowing that the zero will have such devastating results on a hit?
I am not a math guy so there may be something wrong with my logic. If so, can someone please point out my error? If I'm right, how can we entertain the idea of betting more than 1 number at a time?
GLC
Falkor asks the above, well dont reply 5 make +5 units, I know and you all know it would eventually hit a loseing run, but it does what was asked
Well, if zero hits on spin #4 instead of nr.34, return is 35 units, when 61 units were already spent. Doesn´t that kind of defeat the point of betting on zero?
no as in reply2 said if stake goes above return add to mr green who would return 70 +your 2,
Anyway it was just 2 show it would work for 99.99% that .01% would catch you sometime. Just now in corals won £11.00 in under 30 mins as there warning never showed, playing granpa way on wheel layout not like method says use mat, heres the numbers psimoes () the group
27(2)
19-2
18- 1
14-1
16-1
16-1
20-2
10-1
35-2
33-2
27-2
7-1
23-2
33-2
35-2
3-1
33-2
14-1
1-1
31-2
22-2
4-1
5-1
32-2
18-1
22-2
15-1
12-1
28-2
ther you go doubling up was okay psimoes
see you all later of to upton park
COYI
Quote from: nottophammer on Dec 20, 05:49 AM 2014
no as in reply2 said if stake goes above return add to mr green who would return 70 +your 2,
[...]
Of course, but then the amount wagered on zero starts to rise really fast...
Plus, on spin #1 you don´t need to bet 3+3 on the Ecs, just 1u on zero will be enough. If zero hits by first spin profit is 35 against 29u. Same -1 in case of a loss. Then on spin #2 you´ll bet 0+3+1 and so forth. Same 4 units lost or +2 right away, or 32u profit against 25...
One more thing: when you bet on both opposing even chances (which is the same as covering the entire board), you will gain or lose the difference between them. If there are 6 chips on red and 3 chips on black, if red hits you win the difference (3 chips); if black hits you lose the difference). If an equal number of chips is placed on each EC, you lose nothing and win nothing, because there is no difference. So any number of chips common to both chances is superfluous. You just place the difference and leave the other chance unbet. Instead of 6+3, bet 3+nothing. Instead of 12+3, bet 9+nothing. Results will be the same whether win or lose.
Now betting on zero in this much more efficient situation starts to make as much sense as betting on any other number that corresponds to the EC you didn't bet on. Say you bet 3 chips on red and no chip on black. Zero has as much chance to cause damage as all the black numbers: 1/19. By this time, betting for insurance on zero has the same effect as betting on #2, #10, #26, etc.
It's like when you see someone betting on dozens one and two and on zero for insurance. In that case why on zero and not on any other number from 25 to 36?
Quote from: GLC on Dec 19, 08:46 PM 2014
If we cover numbers 1 thru 35 with 1 unit and we leave number 36 and the zero uncovered. What will the outcome be if we were to make this bet for the next 37 spins and each number hit exactly once starting with the zero and ending with #36?
On the zero you would lose 35 units which equals -35 units.
On numbers 1 thru 35 we would win 1 unit each which equals +35 units.
On number 36 you would lose 35 units which equals -35 units.
That means we net -35 units playing this bet method.
If we only bet 1 unit on 1 number for our 37 spins, what would be the outcome if every number hits exactly once in the next 37 spins?
On 35 of the numbers we would lose 1 unit which equals -35 units.
On the covered number we would win 35 units which equals +35 units.
On the uncovered zero we would lose 1 unit which equals -1 units.
That means we net -1 unit with this bet method.
I've asked this before and I ask it again. How can we play multiple numbers knowing that the zero will have such devastating results on a hit?
I am not a math guy so there may be something wrong with my logic. If so, can someone please point out my error? If I'm right, how can we entertain the idea of betting more than 1 number at a time?
GLC
Im no math guy either, but many times I came to the conclusion that bettng less numbers is better.So you are on the rigth track here in my opinion.
Anyone knows of any systems that cover 2 boards together -- double zero and zero roulette? And also a couple of baccarah and black jack tables?
I think those questions should be in a Humour section what do you think :)
Quote from: iggiv on Dec 20, 04:21 PM 2014
I think those questions should be in a Humour section what do you think :)
I don't think so. We're here to help each other out, not to poke fun at other's mistakes or naiveness. We've all gone through the same, and sometimes still do.
If someone posts a simple system based on Gambler's Fallacy with no pretensions other than fun play in the short run, it's fine. The moment there's some math involved to support HG claims, there will be criticism. I think this particular thread would fall in the first category if it wasn't for the recent history of said HG claims.
In all honesty, we're all fools into thinking the game can be beaten with math; "experts" or not. The best we can do is to minimize costs.
Just MHO.
OK sorry
Falkor here you have a method that cover 36 number and win +1 unit each spin.
link:://betselection.cc/math-statistics/cover-36-numbers-and-yet-win-1-unit-on-a-hit/ (link:://betselection.cc/math-statistics/cover-36-numbers-and-yet-win-1-unit-on-a-hit/)
There's a mathematical curiosity for having a 36-number layout coverage and yet win +1 unit on a hit:
How can it be?
Targetting #19 as the only number to leave open, the bet goes like this:
3 chips go to number 0
72 chips to low (Covers 1 to 18 )
48 chips to 3rd dozen (Covers 25 to 36)
16 chips to corner 20/24 (Covers 20-21-23-24)
4 chips to straight-up number 22
If any number other than 19 is spun, you win +1.
You need a table with "Le Partage" rule for this bet to work.