labouchere progression

[frost]

hello everyone


i was trying to search for the "labby" topic with the bayes horror session when i came across the the wikipedia page for the labouchere progression.


link:://en.wikipedia.org/wiki/Labouch%C3%A8re_system


it states


"Theoretically, because the player is cancelling out two numbers on the list for every win, and adding only one number for every loss, the player needs to have his proposition come at least 33.34% to eventually complete the list. For example, if the list starts with seven numbers and the player wins five times and loses three (62.5% winning percentage) the list is completed and the player wins the desired amount, if the list starts with seven numbers and the player wins 43,600 times and loses 87,193 times (33.34% winning percentage) the list completes and the player wins"
[/size]
[/size]is this true?

[Tomla021]

It is very true, the only problem is that the bet sizes can become HUGE

[frost]

hmmm. i see. thank you tomla021. much appreciated....

[frost]

so done a quick test using numbers 1 2 3 4 5 6 7.  now based on the idea that


the player needs to have his proposition come at least 33.34% to eventually complete the list


If ten bets were made would need to win only four (40%) to complete the list.


From my understanding you add a loss to the end of the list an when you win you cross off both numbers from both ends of your list. is this correct?


Using the above numbers if we lose 6 times (8+9+10+11+12+13 = 63)

[frost]

but win four times (1+13 2+12 3+11 4+10 = 14*4 =56) we are still in the hole 9 units


What am doing wrong here? is my line to short/long?


There is also this formula


if the list starts with seven numbers and the player wins 43,600 times and loses 87,193 times (33.34% winning percentage) the list completes and the player wins.


A formula to understand this is as follows:


     Where x = Number of Wins
     y = Number of Losses
     Z = Numbers Originally on the List
When


     ( y + z ) / 2 ≤ X
The result is the list being completed.

[frost]

could anybody shed some light on this for me please?


Please excuse my multiple posting. I am using my mobile phone and the browser does not allow me to scroll down the message box.


Thank you

[superman]

You can start with whatever you like mate

1,2,3,4,5

You can also bet both ends or 1 end, it's up to you, the normal case would be to use the end digits 1,5 = 6

win = 2,3,4
lose = 1,2,3,4,5,6

next bet = 7 units (1+6)

win = 2,3,4,5
lose = 1,2,3,4,5,6,7

[MadMax]

Hi Frost!
I think, you have to win 33,34% of the actual list everytime, not from the original list.
So lets see your example:
1 2 3 4 5 6 7
now 6 losses, then it looks like this:
1 2 3 4 5 6 7 8 9 10 11 12 13

Now, you need to win 33,34% of your list. But now your list contains 13 numbers, not the previous 7. So you need to win the next 5 (4,33) bets to complete. It would go as follows:

1 2 3 4 5 6 7 8 9 10 11 12 13      BR -63
   2 3 4 5 6 7 8 9 10 11 12           BR -49 1. Win
      3 4 5 6 7 8 9 10 11                BR -35 2. Win
         4 5 6 7 8 9 10                     BR -21 3. Win
            5 6 7 8 9                          BR -7   4. Win
               6 7 8                             BR +7  5. Win 33,34% won->profit

I hope my thoughts about this are right, but thats the way I understand it.

[frost]

o okay. I think understand where I was going wrong. that would explain the formula then as well it.


Thank you both for help clearing that up for me

[frost]

sorry to come back to this but here is an idea...


so in order to complete a labouchere line you need to win 33% of the line.


what if you applied the same progression to say line (double street) bets?


they pay 5-1 so technically you would only need to win a fifth of the labouchere line to reach your intended target (profit amount) right?


now when we add zeros to a labby does that make it more suitable for a system with a lower strike-rate (say something like we hit our intended bet 20% of the time, 2 out of 10 bet played)


or are the zeros in there just to keep the progression down?


am i on to something here or is there a chink in my thoughts?


thank you in advance for you responses


frost

[GLC]

Frost,

We can get no advantage playing the labby on different bet locations.

If you are playing on the dozens, you will lose a lot more times than you win.  If you are betting on a single dozen and you win, you must cross out numbers on your labby line that are equal to the number of units you just won.  Every time you lose, which will be 5 out of 6 times on average, you must increase the size of your bet if you are betting the sum of the end numbers.

If you are playing a single dozen, every loss adds 1 number to your line and you will have twice as many losses as wins on average.
If you are playing a double dozen system, every loss adds 2 numbers to your line but a win only crosses out 1 number.

The Labourchere is no missing link to winning at roulette.  It has been tried a million different ways and in the end the bets always get really big and must be broken into small number lines which can take a long time to work through and even then on your  last small number line, you can go into another bad streak and wind up with really large bets again.

That's the problem even with adding the zeros.  When you get to the end of the labby line and have a rash of losses the numbers skyrocket all over again.

If you come up with a solution, you'll be "The Man".

GLC