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    All Forums | Systems & Strategies

    What's the break-even win% for a labouchere system?

    BarcsHasBite
    kevnbs
    kkirby99
    psic711
    Leprechaun
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    BarcsHasBite
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    Posted: Jul. 22, 2010 - 6:30 PM ET #1

    Quick question - I know that you need a 33.33% winning percentage on even money bets using the labourchere system, but what is it with an average juice of -110?
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    To remove first post, remove entire topic.
    Quick question - I know that you need a 33.33% winning percentage on even money bets using the labourchere system, but what is it with an average juice of -110?
     
    kevnbs
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    Posted: Jul. 23, 2010 - 1:52 PM ET #2

    Since you roll your losses into the next line on the labourchere system the 33.3% winning percentage would still be the break even point.

    Example, first line is

    10-10-10-10

    you lose your first bet of $22 ($22 to win $20). Your new line is now

    10-10-10-10-22

    making your next bet $32.

    Big Happy
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    Since you roll your losses into the next line on the labourchere system the 33.3% winning percentage would still be the break even point.

    Example, first line is

    10-10-10-10

    you lose your first bet of $22 ($22 to win $20). Your new line is now

    10-10-10-10-22

    making your next bet $32.

    Big Happy
     
    kkirby99
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    Posted: Jul. 23, 2010 - 2:50 PM ET #3

    Happy, that's not true since you would never clear the line. All you would do is just prepetually increase your wagers and never be even. you need to string a couple of  wins together to maintain the dollar amount of the line at 40. To clear the line and make any money you have to have around a 40% win rate. So in answer to the question the B/E is going to be somewhere between 35 and 38%. Im guessing 36.63% since that is 110% of 33.3.

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    Happy, that's not true since you would never clear the line. All you would do is just prepetually increase your wagers and never be even. you need to string a couple of  wins together to maintain the dollar amount of the line at 40. To clear the line and make any money you have to have around a 40% win rate. So in answer to the question the B/E is going to be somewhere between 35 and 38%. Im guessing 36.63% since that is 110% of 33.3.

     
    psic711
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    Posted: Jul. 23, 2010 - 3:11 PM ET #4

    Technically in the original roulette version of the labourchere it takes 39% winners at a 5% house edge to break even but it is very dependent on what you're playing.   I've hit 47% on underdogs in NHL and made a ton of money.   The BEST methods for using the labourchere are those that avoid long losing streaks.   It's a great way to manage your money but you can't stare at your account balance every day....you need to think long term with this.
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    Technically in the original roulette version of the labourchere it takes 39% winners at a 5% house edge to break even but it is very dependent on what you're playing.   I've hit 47% on underdogs in NHL and made a ton of money.   The BEST methods for using the labourchere are those that avoid long losing streaks.   It's a great way to manage your money but you can't stare at your account balance every day....you need to think long term with this.
     
    kevnbs
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    Posted: Jul. 23, 2010 - 3:59 PM ET #5

    Your right, it is higher than 33.3%. If you started with the 10-10-10-10 line and lost 100 in a row at -110 you would have 100 22's. Now if you won 50 in a row you would clear the 4 10's and 96 of the 22's, laving 4 22's or 48 in the hole. if you won 1 more, that would be 51 wins, + another 44 and only 4 in the hole. 4 in the hole is basically even so 51/151 =33.77%. So I will concede that 34% will keep you even. 

    To me it doesn't matter if your betting at -110 or minus -200. As long as you add the total loss to the end of the line and have a never ending bankroll the percentages hold up.

    Good discussion though.

    Big Happy

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    Your right, it is higher than 33.3%. If you started with the 10-10-10-10 line and lost 100 in a row at -110 you would have 100 22's. Now if you won 50 in a row you would clear the 4 10's and 96 of the 22's, laving 4 22's or 48 in the hole. if you won 1 more, that would be 51 wins, + another 44 and only 4 in the hole. 4 in the hole is basically even so 51/151 =33.77%. So I will concede that 34% will keep you even. 

    To me it doesn't matter if your betting at -110 or minus -200. As long as you add the total loss to the end of the line and have a never ending bankroll the percentages hold up.

    Good discussion though.

    Big Happy

     
    kevnbs
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    Posted: Jul. 23, 2010 - 4:01 PM ET #6

    Good to see PSIC back on here.
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    Good to see PSIC back on here.
     
    kkirby99
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    Posted: Jul. 23, 2010 - 5:52 PM ET #7

    Quote Originally Posted by kevnbs:

    Your right, it is higher than 33.3%. If you started with the 10-10-10-10 line and lost 100 in a row at -110 you would have 100 22's. Now if you won 50 in a row you would clear the 4 10's and 96 of the 22's, laving 4 22's or 48 in the hole. if you won 1 more, that would be 51 wins, + another 44 and only 4 in the hole. 4 in the hole is basically even so 51/151 =33.77%. So I will concede that 34% will keep you even. 

    To me it doesn't matter if your betting at -110 or minus -200. As long as you add the total loss to the end of the line and have a never ending bankroll the percentages hold up.

    Good discussion though.

    Big Happy

    If you lost 100 in a row you wouldn't have 100 22's since after your first losing bet you would have to lay enough to win your 2 outside numbers  10-10-10-10-22. You lay enough to win 32, say 35. Then if you won your third bet (thus attaining your 33.3%) your line still has 10-10-10-22 and you're down $12. In your example of losing 100 in a row and then winning 51 you can imagine what your remaining two numbers are that make up your line. You would be down big. So you have to win one more to clear your line. 52 wins out of 152 plays (34.2%) nets you $40 so it looks like your 34% will keep you even minus the cost of Pepto to lay that kind of dough.

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    Quote Originally Posted by kevnbs:

    Your right, it is higher than 33.3%. If you started with the 10-10-10-10 line and lost 100 in a row at -110 you would have 100 22's. Now if you won 50 in a row you would clear the 4 10's and 96 of the 22's, laving 4 22's or 48 in the hole. if you won 1 more, that would be 51 wins, + another 44 and only 4 in the hole. 4 in the hole is basically even so 51/151 =33.77%. So I will concede that 34% will keep you even. 

    To me it doesn't matter if your betting at -110 or minus -200. As long as you add the total loss to the end of the line and have a never ending bankroll the percentages hold up.

    Good discussion though.

    Big Happy

    If you lost 100 in a row you wouldn't have 100 22's since after your first losing bet you would have to lay enough to win your 2 outside numbers  10-10-10-10-22. You lay enough to win 32, say 35. Then if you won your third bet (thus attaining your 33.3%) your line still has 10-10-10-22 and you're down $12. In your example of losing 100 in a row and then winning 51 you can imagine what your remaining two numbers are that make up your line. You would be down big. So you have to win one more to clear your line. 52 wins out of 152 plays (34.2%) nets you $40 so it looks like your 34% will keep you even minus the cost of Pepto to lay that kind of dough.

     
    kevnbs
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    Posted: Jul. 23, 2010 - 6:15 PM ET #8

    Thanks for straightening that out.  In the middle of a long vacation and I start drinking way to early every day. It's amazing how logic is affected by alcohol.

    Hate to think of losing 100 in a row though.
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    Thanks for straightening that out.  In the middle of a long vacation and I start drinking way to early every day. It's amazing how logic is affected by alcohol.

    Hate to think of losing 100 in a row though.
     
    Leprechaun
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    Posted: Jul. 24, 2010 - 2:59 PM ET #9

    Guys...you are asking a much more complicated question than you think, mathematically.  A very small amount of leg work gives a definitive answer to this question but before I get into the nuts and bolts, you should understand that you cannot calculate the break even point for Labouchere!!.  Labouchere is about clearing lines, NOT counting units.  The reason you can't count units is that the order of the wins and losses matters if you are counting units.  Further, the number of possible outcomes increases exponentially, and the betting odds matter.  Let's say you start with a 10-10-10-10 line (meaning $20 units).  Now, let x= possible outcomes (that is a 2 in this case, you either win or you lose) and let y= the number of games you play before stopping.  To keep it simple and illustrate my point, lets say y=3 games.  In this case, there are a total of 8 possible outcomes (determined by raising x to the y power...in this case 2^3=8):   WWW, WWL, LWW, WLW, WLL, LWL, LLW, LLL.  Your outcome depends on the order, as shown (example assumes -110 odds and losses rounded to nearest dollar):

    WWW: +3 units
    WWL: +1.9 units
    LWW: +1.5 units
    WLW: +1.5 units
    WLL:  -1.85 units
    LWL: -0.6 units
    LLW: -0.6 units
    LLL: -5.35 units

    Well, as you add on games, the number of possible outcomes increases exponentially.  If you look at 4 games, you have 16 different outcomes, 5 games, 32 outcomes and so on.  Also, your units fluctuate drastically with a change in odds.  You're talking about an infinite number of possibilities.  If you pass the 4 win mark and you have not cleared your lines, there is no telling what numbers you are playing from that point forward.

    What you can calculate with the Labouchere is the percentage of wins it will take to clear your lines.  The reason for this is that you clear 2 numbers for every 1 win, but only add one number for a loss.  That's all that matters.  This number depends on 3 factors only:  The number of wins, the number of losses, and the number of figures you start your line with.  It doesn't matter what your odds are, and the order of wins and losses doesn't matter.  The equation for understand this is given as [( Y + Z ) / 2 = X,].  Here x = Number of Wins, Y = Number of Losses and Z= Number of figures you started with.  I'm not some sort of genius who came up with this equation...it's all over the internet.  A very simple google search would have answered your question pretty easily.

    To borrow from your example of losing 100 in a row, Kirby is right, it will take 52 games to clear your lines...but that is NOT to break even...it is to clear you original lines of 10-10-10-10.  What kirby fails to mention is that as long as you have numbers on your line, if you play that many games and have exactly that many losses and wins, you'll end up with $40 no matter what.  No matter how you draw it out, your last win will clear the line.  Order is not important!  That's the reason you can calculate this percentage and you can't calculate the break even percentage.  If you don't increase the numbers on your labby lines, and you never clear your line at anytime in the process, it doesn't matter when the 100 losses out of 152 come.  Furthermore, you should now understand something else important...the win percentage needed to clear your line goes down as you add more games.  Therefore, 34% is only accurate if you play 152 games.  If you played 20 games, you would need to win 40%.  Every time you add a game, the percentage goes down, approaching 1/3 or 33%.  That's the best you could ever do as no matter what, you must win 1 game out of 3 to maintain the lines as even.  It should be pointed out that the longer you go without clearing your lines, the bigger your bets will become.  This also underscores that if you extrapolate too much, it is very deceiving.  

    Using 152 games to clear one labby line is foolish and unrealistic.  Yeah, you could win 34% with labby and profit in that case, but you'd have to have some pretty deep pockets to play 152 games.  Even 20 games without clearing a line is too many and at that point you're already looking at needing 40% wins to clear your lines.  Realistically, you're probably talking about 12 games at most in an effort to clear a line, in which case you would have to win about 46%.  If you were playing favorites, you probably wouldn't even get that far.  That's a far cry from the theoretical 33%.

    Another important thing...the number of figures you start with matters...especially in a real world scenario like a talked about above.  If you start with 10-10-10-10-10 (five tens instead of four), then you need to win 45% to clear your line, or 9 out of 20 rather than 40% (8 of 20) when starting with 4 figures.

    The whole point of Labouchere is to clear your lines without the units getting too out of hand.  If you focus on the units and not the clearing, you'll get in trouble quick.  I've been guilty of doing that myself.  If you're going to use these methods, it is very important that you understand them fully and know what you're getting into.

    Leprechaun
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    Guys...you are asking a much more complicated question than you think, mathematically.  A very small amount of leg work gives a definitive answer to this question but before I get into the nuts and bolts, you should understand that you cannot calculate the break even point for Labouchere!!.  Labouchere is about clearing lines, NOT counting units.  The reason you can't count units is that the order of the wins and losses matters if you are counting units.  Further, the number of possible outcomes increases exponentially, and the betting odds matter.  Let's say you start with a 10-10-10-10 line (meaning $20 units).  Now, let x= possible outcomes (that is a 2 in this case, you either win or you lose) and let y= the number of games you play before stopping.  To keep it simple and illustrate my point, lets say y=3 games.  In this case, there are a total of 8 possible outcomes (determined by raising x to the y power...in this case 2^3=8):   WWW, WWL, LWW, WLW, WLL, LWL, LLW, LLL.  Your outcome depends on the order, as shown (example assumes -110 odds and losses rounded to nearest dollar):

    WWW: +3 units
    WWL: +1.9 units
    LWW: +1.5 units
    WLW: +1.5 units
    WLL:  -1.85 units
    LWL: -0.6 units
    LLW: -0.6 units
    LLL: -5.35 units

    Well, as you add on games, the number of possible outcomes increases exponentially.  If you look at 4 games, you have 16 different outcomes, 5 games, 32 outcomes and so on.  Also, your units fluctuate drastically with a change in odds.  You're talking about an infinite number of possibilities.  If you pass the 4 win mark and you have not cleared your lines, there is no telling what numbers you are playing from that point forward.

    What you can calculate with the Labouchere is the percentage of wins it will take to clear your lines.  The reason for this is that you clear 2 numbers for every 1 win, but only add one number for a loss.  That's all that matters.  This number depends on 3 factors only:  The number of wins, the number of losses, and the number of figures you start your line with.  It doesn't matter what your odds are, and the order of wins and losses doesn't matter.  The equation for understand this is given as [( Y + Z ) / 2 = X,].  Here x = Number of Wins, Y = Number of Losses and Z= Number of figures you started with.  I'm not some sort of genius who came up with this equation...it's all over the internet.  A very simple google search would have answered your question pretty easily.

    To borrow from your example of losing 100 in a row, Kirby is right, it will take 52 games to clear your lines...but that is NOT to break even...it is to clear you original lines of 10-10-10-10.  What kirby fails to mention is that as long as you have numbers on your line, if you play that many games and have exactly that many losses and wins, you'll end up with $40 no matter what.  No matter how you draw it out, your last win will clear the line.  Order is not important!  That's the reason you can calculate this percentage and you can't calculate the break even percentage.  If you don't increase the numbers on your labby lines, and you never clear your line at anytime in the process, it doesn't matter when the 100 losses out of 152 come.  Furthermore, you should now understand something else important...the win percentage needed to clear your line goes down as you add more games.  Therefore, 34% is only accurate if you play 152 games.  If you played 20 games, you would need to win 40%.  Every time you add a game, the percentage goes down, approaching 1/3 or 33%.  That's the best you could ever do as no matter what, you must win 1 game out of 3 to maintain the lines as even.  It should be pointed out that the longer you go without clearing your lines, the bigger your bets will become.  This also underscores that if you extrapolate too much, it is very deceiving.  

    Using 152 games to clear one labby line is foolish and unrealistic.  Yeah, you could win 34% with labby and profit in that case, but you'd have to have some pretty deep pockets to play 152 games.  Even 20 games without clearing a line is too many and at that point you're already looking at needing 40% wins to clear your lines.  Realistically, you're probably talking about 12 games at most in an effort to clear a line, in which case you would have to win about 46%.  If you were playing favorites, you probably wouldn't even get that far.  That's a far cry from the theoretical 33%.

    Another important thing...the number of figures you start with matters...especially in a real world scenario like a talked about above.  If you start with 10-10-10-10-10 (five tens instead of four), then you need to win 45% to clear your line, or 9 out of 20 rather than 40% (8 of 20) when starting with 4 figures.

    The whole point of Labouchere is to clear your lines without the units getting too out of hand.  If you focus on the units and not the clearing, you'll get in trouble quick.  I've been guilty of doing that myself.  If you're going to use these methods, it is very important that you understand them fully and know what you're getting into.

    Leprechaun
     
    PicKnRole
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    Posted: Jul. 25, 2010 - 1:40 AM ET #10

    Great Info
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    Great Info
     
    cisco
    cisco
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    Posted: Jul. 25, 2010 - 12:33 PM ET #11

    Damm Lep, you put a lot of work into this.

    Thanks-

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    Damm Lep, you put a lot of work into this.

    Thanks-

     
     
    kevnbs
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    Posted: Jul. 25, 2010 - 2:11 PM ET #12

    Great Work Leprechaun.

    In summary in order to stay even or slightly ahead of the game when betting into an average -110 line you need to hit about 34%.

    If you were to play a single game a day for a year your record would be 123 and 242 (33.7%). If on the last day of betting you managed to clear your lines you would be even (4 numbers + 242 losees equal 246 and 123 x 2 = 246), however if you ended up 122 and 243 (33.4%) you would be down (leaving 3 numbers on your line).

    So with 365 bets you need to hit 33.7% to stay even for one year of betting at one bet a day.

    I'm exhausted.
    Big Happy


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    Great Work Leprechaun.

    In summary in order to stay even or slightly ahead of the game when betting into an average -110 line you need to hit about 34%.

    If you were to play a single game a day for a year your record would be 123 and 242 (33.7%). If on the last day of betting you managed to clear your lines you would be even (4 numbers + 242 losees equal 246 and 123 x 2 = 246), however if you ended up 122 and 243 (33.4%) you would be down (leaving 3 numbers on your line).

    So with 365 bets you need to hit 33.7% to stay even for one year of betting at one bet a day.

    I'm exhausted.
    Big Happy


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