Hello everyone, I have not been playing roulette very long at all but, I have learned a lot about it. Anyway, I was playing around the other day and placed only 4 chips and was covering all but 3 numbers, which are 0, 14, 23. My problem is I know it needs more chips in certain places so that each spin will either break even or profit. I am so bad with numbers and math which, I know sounds silly since I want to play roulette. I really need someone to do the math on this for me on how many chips should be placed where in order to break even if I don't land in a profit area or to just profit on every spin if it's possible. I really appreciate anyone's input on this. Thank you so much! Also, please just tell me if it's a garbage strategy or if you think a tweak to where the chips are placed should be made. Thanks
Quote from: Jkirb07 on Feb 01, 07:35 PM 2019
Hello everyone, I have not been playing roulette very long at all but, I have learned a lot about it. Anyway, I was playing around the other day and placed only 4 chips and was covering all but 3 numbers, which are 0, 14, 23. My problem is I know it needs more chips in certain places so that each spin will either break even or profit. I am so bad with numbers and math which, I know sounds silly since I want to play roulette. I really need someone to do the math on this for me on how many chips should be placed where in order to break even if I don't land in a profit area or to just profit on every spin if it's possible. I really appreciate anyone's input on this. Thank you so much! Also, please just tell me if it's a garbage strategy or if you think a tweak to where the chips are placed should be made. Thanks
Hi Jkir, looks like you are covering 30 numbers there. I advise you definitely do not attempt any kind of rapidly rising progressions - keep it simple, set stops that you do not go beyond. Don't be afraid to reset and go again. Test and run trials, learn off members here, read and research and learn what kind of strategy best suits you and your personality. Problem with starting out,e specially with covering a large chunk of the board is, you get lots of wins in a row and think you've cracked it and that you're invincible. Never get that mindset, always be cautious and don't outstay your welcome because the wheel will have your guts for garters if you hang around long enough. Enough of me droning, but best of luck and I actually like the bet you have. Maybe wait for the middle dozen to hit consecutively, and then use your bet to get one hit. Stop, retrack. Etc.
It's impossible to always be in profit or breakeven on a hit with this setting.
Your chips on dozens must have at least a third of table value each.
The chip on the EC must have half the total value.
The line must have 1/6th of the value at least.
2x 1/3 + 1/2 + 1/6 = 4/3. Bigger than overunity...
Consider changing the EC for a corner in doz2 and slide the line left or right. You will then be able to portion the values properly.
Or have a look st the Romanovsky bet.
Quote from: Jkirb07 on Feb 01, 07:35 PM 2019
Hello everyone, I have not been playing roulette very long at all but, I have learned a lot about it. Anyway, I was playing around the other day and placed only 4 chips and was covering all but 3 numbers, which are 0, 14, 23. My problem is I know it needs more chips in certain places so that each spin will either break even or profit. I am so bad with numbers and math which, I know sounds silly since I want to play roulette. I really need someone to do the math on this for me on how many chips should be placed where in order to break even if I don't land in a profit area or to just profit on every spin if it's possible. I really appreciate anyone's input on this. Thank you so much! Also, please just tell me if it's a garbage strategy or if you think a tweak to where the chips are placed should be made. Thanks
Hi Jkirb07,
It seems you are on your way to building a 100% WIN at roulette strategy. :thumbsup:
Now here is the maths
W L
Black 1 1
Dozens 1 2
Line 5 1
8 Combinations
B+D+L 1+1+5 = 7W
B+D-L 1+1-1= 1W
B-D+L 1-2+5= 4W
B-D-L 1-1-1 = 1L
-B+D+L -1+1+5= 5W
-B+D-L -1+1-5 = 5L
-B-D+L -1-2+5 = 2W
-B-D-L -1-2-1 = 4L (Lose all bets)
Now bet these to the
correct ratio and
sized correctly for each spin of your cycle in addition to another bet that will grant you a profit 100% of the time and you have yourself a HG
good Luck
Ricky
Correction
Now here is the maths
W L
Black 1 1
Dozens 1 2
Line 5 1
8 Combinations
B+D+L 1+1+5 = 7W
B+D-L 1+1-1= 1W
B-D+L 1-2+5= 4W
B-D-L 1-2-1 = 2L
-B+D+L -1+1+5= 5W
-B+D-L -1+1-1 = 1L
-B-D+L -1-2+5 = 2W
-B-D-L -1-2-1 = 4L (Lose all bets)
Hint, note how many ways you can win and how many you can lose. Very well done :thumbsup:
And you say you are just starting out at roulette. Well I think you may be going places :-X
My Apologies I have not taken my medication today :'(
Correction
Now here is the maths
W L
Black 1 1
Dozens 1 2
Line 5 1
8 Combinations
B+D+L 1+1+5 = 7W
B+D-L 1+1-1= 1W
B-D+L 1-2+5= 4W
B-D-L 1-2-1 = 2L
-B+D+L -1+1+5= 5W
-B+D-L -1+1-1 = 1L
-B-D+L -1-2+5 = 2W
-B-D-L -1-2-1 = 4L (Lose all bets)
You cannot win the Dozen and Line so these combinations are struck out
Sorry, in my excitement I thought you had something. Unfortunately this combination does not provide you an advantage. But there is one that does so I will allow others to ponder.
Cheers,
Ricky
Quote from: Bigbroben on Feb 01, 08:27 PM 2019omanovsky bet.
That's more like it Bigbroben :thumbsup:
@Buffalowizard Thank you so much for the input! I am actually covering 34 numbers because of the EC Black
@Bigbroben Thanks a lot for replying
Quote from: Ricky on Feb 01, 09:01 PM 2019
My Apologies I have not taken my medication today :'(
Correction
Now here is the maths
W L
Black 1 1
Dozens 1 2
Line 5 1
8 Combinations
B+D+L 1+1+5 = 7W
B+D-L 1+1-1= 1W
B-D+L 1-2+5= 4W
B-D-L 1-2-1 = 2L
-B+D+L -1+1+5= 5W
-B+D-L -1+1-1 = 1L
-B-D+L -1-2+5 = 2W
-B-D-L -1-2-1 = 4L (Lose all bets)
You cannot win the Dozen and Line so these combinations are struck out
Sorry, in my excitement I thought you had something. Unfortunately this combination does not provide you an advantage. But there is one that does so I will allow others to ponder.
Cheers,
Ricky
Ricky, Thank you so much for all of your replies, I really appreciate it. I finally figured out how to properly respond to people, Lol
Quote from: Ricky on Feb 01, 09:01 PM 2019
My Apologies I have not taken my medication today :'(
Correction
Now here is the maths
W L
Black 1 1
Dozens 1 2
Line 5 1
8 Combinations
B+D+L 1+1+5 = 7W
B+D-L 1+1-1= 1W
B-D+L 1-2+5= 4W
B-D-L 1-2-1 = 2L
-B+D+L -1+1+5= 5W
-B+D-L -1+1-1 = 1L
-B-D+L -1-2+5 = 2W
-B-D-L -1-2-1 = 4L (Lose all bets)
You cannot win the Dozen and Line so these combinations are struck out
Sorry, in my excitement I thought you had something. Unfortunately this combination does not provide you an advantage. But there is one that does so I will allow others to ponder.
Cheers,
Ricky
Yea, hopefully someone can figure out a good tweak to this strategy.
So this is the current amount I am getting back from these bets using only 1 unit on each place for now.. 4 units total
Returns:
7 if hits black inside split street
5 if hits red inside split street
4 if hits black in 2 dozens that are covered
2 if hits red in the 2 dozens that are covered
1 if hits black in the uncovered dozen, outside of split street
Is it possible to tweak this by putting more units in one or more of the bets I placed or more units somewhere else on the board OR switch them around all together so that I can at least breakeven on spins if I dnt hit a profit area? The last 2 I listed are, of course, the bets I am not at least breaking even on. People have already been helping out a lot with input. Just wondering if someone can really figure this out for me or if its just not a good strategy. Thanks so much!
Hi JKIRB07 :)
Well, i've coded this bet and tried all kinds of triggers/progression, and then Finally i found the solution to this Riddle with this bet. (ofc, it's a "BAD betselection" but point here now is,(How to make it profitable?) --IF you play this bet, *and imagine only the middle LINE-bet exist,* so then, you play this bet with a ordinary 6 numbers negative progression (or Use the one in the code, AND then, you RESET ONLY the negative progressionline (6 numbers or whatever progression you want to use) WHEN the 6 numbers middle LINE BET Hits.
That's what i figure, anyway,
cheers O0
TEST 1-2 RNG/LIVE
RX-code
system "4 Chips Bet"
//(copyright) ignatus 2019
method "main"
begin
while starting a new session
begin
Set List[1,2,3,5,7,9,11,12,14,16,18,20,22,25,28,30,32,35,40,45,50,55,60,70,80,90,100,
110,130,150,190,230,275,350]
to Record "progression" Data
end
put 0 to Record "Highest Bankroll" Data
Copy List [Black, 1st dozen, 3rd dozen, Line(16-21)] to Record "Bet" Layout
while on each spin
begin
if total bankroll >= 3000 each time
begin
stop session
end
if total bankroll <= -1000 each time
begin
stop session
end
{If any inside bet lost each
begin
add 1 on Record "progression" Data Index
end
}
add 1 on Record "progression" Data Index
If Any inside bet won each
begin
if Bankroll > Record "Highest Bankroll" Data
begin
clear Record "Highest Bankroll" Data
put 5000% Bankroll to Record "Highest Bankroll" Data
put 1 on Record "progression" Data Index
end
end
put 100% of Record "progression" Data on Record "Bet" Layout list
If Record "progression" Data Index >
Record "progression" Data Count
Begin
Put 1 on Record "progression" Data Index
End
end
END
What you described is a play on 34nrs. You'll make 2u on every hit if they all are balanced equally. With the overlaping color, not all nrs will give equal return. But all sums will equal zero.
Say you had 12u on doz1, 12u on doz2, 6u on line 13-18, 4u on corner 19-23. All 34 nrs will be covered equally, with 0,21,24 free.
You could guess and say (gf) : doz 3 hasnt hit in 3 spins, I'll call it sleepy and will go 11-13 on doz 1-3 or any other ponderation.
You could also say (gf): doz2 hasnt hit in 5 spins, I'll add a unit on line 13-18 and hope for the best. A hit on line 13-18 will give you 7u profit, a hit elsewhere will give you 1u, a loss will get you to -35.
Up to you, really.
Is there a reason you play color with dozens?
Quote from: Ricky on Feb 01, 09:03 PM 2019
That's more like it Bigbroben :thumbsup:
Ricky, I was thinking of you when I wrote the ...ovsky word...
Wow, did you just miss my post,...hmmok
Quote from: ignatus on Feb 01, 10:13 PM 2019
Wow, did you just miss my post,...hmmok
Lol I was responding to jkirb, we wrote simultanously!
It might be better to bet like this instead?
Look at last 5 spins to determine dominant color.
If Dominant Color is RED, then bet 2 units Red, 1 unit Column 1, 1 unit Column 2.
If Dominant Color is BLACK, then bet 2 units Black, 1 unit Column 1, 1 unit Column 3.
Tested as follows:
Profit Target 10 units
Stop Loss 20 units
# Spins Profit Balance
1-----11-----10-----10
2-----176--- 12-----22
3-----5-------12-----34
4-----11-----12-----46
5-----86-----11-----57
Session #2 was a work out but session 3 & 4 had 9 straight blacks. Riding this streak made it easy.
Quote from: Bigbroben on Feb 01, 10:04 PM 2019
Ricky, I was thinking of you when I wrote the ...ovsky word...
Thanks for the pointer on the Romanovsky bet. A very interesting concept I am sure we can do something with. 6 betting patterns each offering 4+1 numbers left open for the wheel to find. Just like a line of pigeon holes waiting for the pigeons to come home. Can it be a way to stretch time and space?
Cheers,
Ricky
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.
Testing the validity of this bet with Roulette Xtreme.
First you need to enable Le Partage:
Then proceed to test the winning bets:
At a straight-up win, we win 4 x 36 = 144 chips, minus 143 chips layed = +1
At the corner, we win 16 x 9 = 144 chips, we lost 143 chips layed = +1
At the dozen, we win 48 x 3 = 144 chips, taking in to account 143 chips layed = +1
At the even chance, we win 72 x 2 = 144 chips, substracting 143 chips layed = +1
If Zero shows up, dealer returns us 108 chips as payout for it, then takes away losing chips: while from the 72 chips layed at low by "Le Partage" rule enabled at the even chances dealer takes 36 and we get 36 chips back. As we won 108 by zero + 36 chips back from low = 144 chips - 143 layed = +1 unit profit even when zero hits!
Positively letting only 1 out of the 37 numbers as a loss, and winning +1 unit at 36 numbers.
It is important to notice at regular Roulette without Le partage, you have a net loss of 35 chips when zero hits so we really need this Le Partage splitting rule:
Of course, the downside is a losing spin costs 143 chips.
You just can't play this and expect to make a long-term profit. For the very reason that a number is expected to show up once in every 37 spins and you are laying 143 units.
The only imaginable way to use this if you really, really, REALLY have to bet 1 single "hit and run" spin.
This is the only way I know for having the numerical coverage at 36 numbers for you, and only 1 number for the bank.
As a single-spin strategy, the best way being playing it only once in your lifetime and then leaving it alone if you win, so that unit is yours forever.
I guess the probability of 19 being spun three times in a row is 50,653 so we could wait until 19 repeats twice and then bet once.
But we might die of boredom waiting for this to happen...
Or this could happen the first time you played it LOL 😂
Perhaps a more practical way to play it would be to wait for 19 to hit once then bet £143 to win £1 then wait for 19 to reappear and repeat.
Eventually 19 will repeat probability 1369 and you will lose £143.
One could then play a standard 35 spin parachute flat using 19 as the target number and betting £144 per spin.
The total bankroll needed would be...
1 x £143
35 x £144 = £5,040
Total: £5,183
A win at any stage on the standard parachute would show a profit of at least + £1 overall at which time you would stop and wait for 19 to show up again.
Probably still too slow and boring for real casino play but could be OK for an online RNG wheel with turbo mode activated.
If such a thing exists with a suitable spread.
Would have to still offer the La Partage rule too.
The probability of success has to be almost 100%
So there you have it the holy grail at last.
The parachute I refer to is the original...
So sometimes you will win 3,4,5 or more chips whilst using the parachute which will give you some juicy profits as you're betting with a chip value £144 at this point.
You might also be sh!tting yourself too at some of the later stages where the real big juice is so better to have a bot to play it whilst you sleep.
Another possibility is to start off by betting every spin as soon as you start the session.
143 units at 10p so the the total stake is £14.30
Lot's of online live wheels will let you bet 10p as long as your total stake is £1.
You win 10p every spin until 19 shows up.
Then you have to recover £14.30 but that is such a small amount there are unlimited recovery ideas available assuming you are prepared to stretch the progression all the way out to £5,00 or so....
Are you crazy enough to risk £5,000 to win 10p?
Yes Sir I am!
I've done a little research and there is a game by Evolution Gaming called Auto Roulette La Partage which is streamed directly from the well known Belgium Casino de Spa.
This game offers a minimum bet of 10p and a maximum bet of £4,000 on the Even Chances.
At least it does on the William Hill website where I have an account.
I will attempt to make an unbreakable progression that could really be played within the table limits and within a maximum bankroll loss of around £5,000.
It appears to spin about once a minute so earnings will be about £6 an hour whilst playing the static bet of £14.30 which wins 10p per spin until number 19 appears at which point we go into recovery to regain our capital of £14.30.
I should mention when I say create an unbreakable progression I mean I will create a series of events that have to exactly occur in such an order that there is an extremely unlikely chance I shall ever witness it during my lifetime.
After I Iose £14.30 because 19 came out I shall start the next spin following the chart I've attached.
19 is red so we always start the first bet with one chip on Red.
So by then following the chart and using £1 as my unit stake ONLY one of two things can happen.
I lose £100 giving me a total loss of £114.30 so far and all well within any table limits and I still have over £4,800 of bankroll left because I started with around £5000
Or I win on the chart which will give me a win of approx 20 units.
As I'm playing £1 per unit at this point it means I've won £20 which puts me over £5 in profit.
Of course I would not carry on until I had made £5 overall profit I would stop as soon as I had won £15 thus putting me up 0.70 over all.
So to lose £114.30 this HAS to happen.
19 must be the last winning number.
The wheel must now spin exactly nine reds and ten blacks over the next 19 spins.
Next part I will attempt to recover my £114.30 total loss.
.
The next part to try and recover the running loss of £114.30 is a low risk gamble that can go on for ages before a result is known.
Play as follows using £12 a chip and as already previously posted.
Look at last 5 spins to determine dominant color.
If Dominant Color is RED, then bet 2 units Red, 1 unit Column 1, 1 unit Column 2.
If Dominant Color is BLACK, then bet 2 units Black, 1 unit Column 1, 1 unit Column 3.
Profit Target 10 units
Stop Loss 20 units
So either I win 10 units @ £12 per chip winning me £120 and giving me an overall profit of £5.70
Or I lose 20 x £12 = £240 which would give me an overall loss of £354.30
Again if I were to win £115 I would stop and take the 70p profit rather than try to aim for £120.
Next part I will assume the wheel keeps on fcuk!ng me and try to recover the current running total loss of £354.30
.
I'm still not having a panic attack as the probability of even getting this far is already insanely difficult and I still have over £4,500 of bankroll available and am still comfortably within the table limits.
Because the previous session could have been extremely time consuming it's necessary to make a few adjustments so we can try to stick to something of an hourly earning rate of £6.
Any less is not worth the time and is still well below minimum wage in any case but has the advantage of being able to done at home.
So the total loss is rounded up to £360 which makes future calculations easier as breaking level now at £360 will mean an overall profit of £5.70
If you feel depressed about being £360 down just remind yourself it is the same as having just bet £10 on a single number 36 times.... No big deal right?
Also worth remembering there are maybe some left over funds from the previous stages where I was at minus 18 and as the next bet is always for 4 units then the session was written off there and then as a loss because losing the next bet would have meant be being - 22 which is over the allowed limit of - 20
There maybe are also extra funds from where zero may have hit during the stage where I was betting the colours and was written off as a full loss.
Or the extra zero money could have been used to finish the next stage early without having to reach +10 to show an overall profit.
All of these things will help keep up the overall hourly win goal rate during large periods of treading water.
But to return to the matter at hand I will assume a full loss of £360 at this point.
I shall ponder my next level of recovery some more.....
I'm going to try to win 36 units playing £10 a chip in the following manner with a stop loss of 150 chips which would be £1,500 and an overall running loss of £1,860.
I still have well over 50% bankroll of my starting bankroll of approximately £5,000 intact.
I am still comfortably within the table limits.
Don't think that because this system is simple to play that it's not a winner. Test it and you'll be surprised.
I have tested this for 3 sessions of 100+ units and only had to bet more than 10 units 1 time and that was up to 13 units. That means I've never been more than -100 units in the hole. In one of these sessions I had 2 sets of 3 zeros in a row and another 15 zeros to boot. It still overcame it. Granted, I was in the hole for a while a couple of times, but the hit rate was enough to pull back out.
This is soooo simple and effective!
I bet on an even chance and if I win, I bet what's on the table on 2 dozens.
If I win within the 1st 5 bets, I start over because I will be ahead or even. I go into +1 on loss and -2 on win from the 6th bet on. Anytime I'm at a new high, I reset to 1 unit.
Example: I bet 1 unit on Red. Red spins so I win 1 unit. I now have 2 units on the table so I split them between 2 dozens and bet 1 unit on each dozen. If I win, I have won 2 units, 1 on the even chance bet and 1 on the 2 dozens bet, so I start over.
Example: I bet 1 unit on Red and lose. I bet 2 units on Red and lose. I bet 3 units on Red and lose. I bet 4 units on Red and win. I now bet 4 units on the last 2 dozens. If I win I will be up 2 units so I start over at 1 unit.
If I lose on the two dozens, I will go on to bet 5 units on an even chance, etc...
Example: I have lost my 1st 7 bets. Very rare, but it does happen. I bet 8 units on Red and win so I bet 8 units on the last 2 dozens and win. Losing the 1st 7 bets puts me down 28 units for this attack. Winning on the 8th bet nets me 16 units. That leaves me still down 12 units. I drop back to the 6 unit bet on an even chance. If I win the even chance bet and also the 2 dozens bet, I will net 12 units which will make me even so I will start over at 1 unit.
Bet Selection same as previous stage.
Look at last 5 spins to determine dominant color.
If Dominant Color is RED, then bet Red
If Dominant Color is BLACK, then bet Black
That should give you enough info to understand the basic principle behind the bet selection to be able to play the system.
This is a consistent winner.
Why does this work so well? Because winning on an Even Chance and then on 2 dozens happens more often than winning on 2 even chances in a row.
Next I will assume I am still getting smashed relentlessly and try to recover the new running total loss of £1,860
Quote from: Let Me Win on Feb 02, 01:52 PM 2019
After I Iose £14.30 because 19 came out I shall start the next spin following the chart I've attached.
19 is red so we always start the first bet with one chip on Red.
So by then following the chart and using £1 as my unit stake ONLY one of two things can happen.
I lose £100 giving me a total loss of £114.30 so far and all well within any table limits and I still have over £4,800 of bankroll left because I started with around £5000
Or I win on the chart which will give me a win of approx 20 units.
As I'm playing £1 per unit at this point it means I've won £20 which puts me over £5 in profit.
Of course I would not carry on until I had made £5 overall profit I would stop as soon as I had won £15 thus putting me up 0.70 over all.
So to lose £114.30 this HAS to happen.
19 must be the last winning number.
The wheel must now spin exactly nine reds and ten blacks over the next 19 spins.
Next part I will attempt to recover my £114.30 total loss.
.
I seen this chart years ago but forgot what you do with it start with?
Quote from: Jkirb07 on Feb 01, 07:35 PM 2019
Hello everyone, I have not been playing roulette very long at all but, I have learned a lot about it. Anyway, I was playing around the other day and placed only 4 chips and was covering all but 3 numbers, which are 0, 14, 23. My problem is I know it needs more chips in certain places so that each spin will either break even or profit. I am so bad with numbers and math which, I know sounds silly since I want to play roulette. I really need someone to do the math on this for me on how many chips should be placed where in order to break even if I don't land in a profit area or to just profit on every spin if it's possible. I really appreciate anyone's input on this. Thank you so much! Also, please just tell me if it's a garbage strategy or if you think a tweak to where the chips are placed should be made. Thanks
You may want to look at this thread:
link:s://:.rouletteforum.cc/index.php?topic=14779.0
You'll still lose money in the long run, but some different staking strategies for you to consider.
Hi Winner
If the square is a zero you wait and watch a spin.
If it's red you go across if its black you go down.
The numbers in the square are the units to bet on that colour on that spin.
You keep going until you fall off the square either by going across or down.
When you fall off you have either won 20 units or lost 100 units.
To lose it has to spin exactly ten reds and ten blacks any other result and you win 20 units.
Quote from: Let Me Win on Feb 03, 04:31 AM 2019
Hi Winner
If the square is a zero you wait and watch a spin.
If it's red you go across if its black you go down.
The numbers in the square are the units to bet on that colour on that spin.
You keep going until you fall off the square either by going across or down.
When you fall off you have either won 20 units or lost 100 units.
To lose it has to spin exactly ten reds and ten blacks any other result and you win 20 units.
ok I remember now thanks
I also realised that I can improve my odds ever so slightly by betting either High or Low and a 6-line when I am betting two dozens at the same time.
Example....
I am betting dozen one and two with £2 on each dozen for a total stake of £4
If I win I will get £6 returned for £2 overall profit.
If zero shows I will lose £4
If I however bet £3 on low and £1 on the 6-line 19 - 24
Then my total stake is still only £4 and I still get £6 back if either dozen one or two should hit.
However if zero shows now I get £1.50 returned to me so I only lose £2.50 instead of £4 for the same bet.
Only a very tiny adjustment in my favour but you need everything you can get in this game!
Quote from: ignatus on Feb 01, 09:51 PM 2019
Hi JKIRB07 :)
Well, i've coded this bet and tried all kinds of triggers/progression, and then Finally i found the solution to this Riddle with this bet. (ofc, it's a "BAD betselection" but point here now is,(How to make it profitable?) --IF you play this bet, *and imagine only the middle LINE-bet exist,* so then, you play this bet with a ordinary 6 numbers negative progression (or Use the one in the code, AND then, you RESET ONLY the negative progressionline (6 numbers or whatever progression you want to use) WHEN the 6 numbers middle LINE BET Hits.
That's what i figure, anyway,
cheers O0
TEST 1-2 RNG/LIVE
RX-code
system "4 Chips Bet"
//(copyright) ignatus 2019
method "main"
begin
while starting a new session
begin
Set List[1,2,3,5,7,9,11,12,14,16,18,20,22,25,28,30,32,35,40,45,50,55,60,70,80,90,100,
110,130,150,190,230,275,350]
to Record "progression" Data
end
put 0 to Record "Highest Bankroll" Data
Copy List [Black, 1st dozen, 3rd dozen, Line(16-21)] to Record "Bet" Layout
while on each spin
begin
if total bankroll >= 3000 each time
begin
stop session
end
if total bankroll <= -1000 each time
begin
stop session
end
{If any inside bet lost each
begin
add 1 on Record "progression" Data Index
end
}
add 1 on Record "progression" Data Index
If Any inside bet won each
begin
if Bankroll > Record "Highest Bankroll" Data
begin
clear Record "Highest Bankroll" Data
put 5000% Bankroll to Record "Highest Bankroll" Data
put 1 on Record "progression" Data Index
end
end
put 100% of Record "progression" Data on Record "Bet" Layout list
If Record "progression" Data Index >
Record "progression" Data Count
Begin
Put 1 on Record "progression" Data Index
End
end
END
IGNATUS, thank you so much for doing all of that testing and figuring the math for these bets. I really appreciate it! Like I said, I am terrible at math and still very new to roulette so, to be extremely honest I do not quite understand the 6 number negative progression that you are describing. I assume when you say 6 numbers you are talking about the split street I have covered? I can see the sequence you have written in the code but, I still just don't quite understand what you are saying here.. "IF you play this bet, *and imagine only the middle LINE-bet exist,* so then, you play this bet with a ordinary 6 numbers negative progression (or Use the one in the code, AND then, you RESET ONLY the negative progressionline (6 numbers or whatever progression you want to use) WHEN the 6 numbers middle LINE BET Hits." Could you please explain it just a little clearer, I am so sorry but, I am still learning. Also, are you saying to leave only 1 unit on the other places except the 6 Line during the session? Thank you so much!
Quote from: Let Me Win on Feb 02, 12:13 PM 2019
The parachute I refer to is the original...
So sometimes you will win 3,4,5 or more chips whilst using the parachute which will give you some juicy profits as you're betting with a chip value £144 at this point.
You might also be sh!tting yourself too at some of the later stages where the real big juice is so better to have a bot to play it whilst you sleep.
Let Me Win, thank you so much for all of your posts and feedback. Could you please explain a little more about having a bot play for you? How much is one and where can I get one? Thanks!
Quote from: Let Me Win on Feb 03, 03:51 PM 2019
I also realised that I can improve my odds ever so slightly by betting either High or Low and a 6-line when I am betting two dozens at the same time.
Example....
I am betting dozen one and two with £2 on each dozen for a total stake of £4
If I win I will get £6 returned for £2 overall profit.
If zero shows I will lose £4
If I however bet £3 on low and £1 on the 6-line 19 - 24
Then my total stake is still only £4 and I still get £6 back if either dozen one or two should hit.
However if zero shows now I get £1.50 returned to me so I only lose £2.50 instead of £4 for the same bet.
Only a very tiny adjustment in my favour but you need everything you can get in this game!
You can improve your odds even more if you don't make the 19-24 bet at all. As you discovered, even chance bets are better value due to partage on zero. But layout bets and dozens pay twice the tax.
Your expected loss on this bet on low and 19-24 is
£3 x 1.35% + £1 x 2.7% = 6.75p
Whereas place that £4 on low alone and your expected loss is
£4 x 1.35% = 5.4p
So with another tiny adjustment, you're saving yourself losing a bit more. In fact 1.35p per spin!