Labouchère system to double or lose with 1000 units?
About the system. The math assumes your goal was to win 10 units while using up to 1000 units etc.
What if your goal was to win 1000 units risking 1000 units?
Like what if you $1000 and you had a line with a flat 2000 50cents to cross out.
Your first bet would be $1. What is the chance to double up and go bust?
It feels like a system like this betting coin flip sports events like the point spread would have a good chance to prevail.
Educate me!
Quote: ranch7It feels like a system like this betting coin flip sports events like the point spread would have a good chance to prevail.
The problem with that is, the system is meant to be used with even-money bets, and bets against the point spread only pay off 10-11.
When you lose, your losses are added to the string of numbers, so when you win, if it is an even-money bet, you gain it back. When your original numbers are crossed off as well, then your overall gain is your initial target result.
However, if your bets are 10-11, then a win after a loss does not completely "cancel out" a gain; as a result, you can go through all of the numbers and end up with an overall loss.
One alternative would be, when you bet, you bet 1.1 x what the numbers tell you to bet. (For example, your first bet would be (1 + 1) x 1.1 = 2.2.)
I ran 1 million trials of "1000 win or 1000 loss" on a game with a 1% edge (probability of winning = 0.495), which is slightly better than the pass line on craps, and I got that you win only 35.11% of the time. Apparently, you lose more than 50.5% of the time because the average number of bets in a losing trial (1202) is greater than the average number of bets in a winning trial (1937).
Quote: ThatDonGuyThe problem with that is, the system is meant to be used with even-money bets, and bets against the point spread only pay off 10-11.
When you lose, your losses are added to the string of numbers, so when you win, if it is an even-money bet, you gain it back. When your original numbers are crossed off as well, then your overall gain is your initial target result.
However, if your bets are 10-11, then a win after a loss does not completely "cancel out" a gain; as a result, you can go through all of the numbers and end up with an overall loss.
One alternative would be, when you bet, you bet 1.1 x what the numbers tell you to bet. (For example, your first bet would be (1 + 1) x 1.1 = 2.2.)
I ran 1 million trials of "1000 win or 1000 loss" on a game with a 1% edge (probability of winning = 0.495), which is slightly better than the pass line on craps, and I got that you win only 35.11% of the time. Apparently, you lose more than 50.5% of the time because the average number of bets in a losing trial (1202) is greater than the average number of bets in a winning trial (1937).[/q
OK so if I had $10000 and tried this ten times I should win only 3- 4 of them?
Quote: ranch7OK so if I had $10000 and tried this ten times I should win only 3- 4 of them?
If your betting unit is $10, then yes.
If your betting unit is $100, then you win $10,000 before you lose $10,000 45% of the time.
On the other hand, if you just flat bet $10, your probability of being $10,000 ahead before being $10,000 behind in a 1% house edge game is 1 in 485,488,766. The more bets you make, the more the house edge eats away at your bankroll.
In any game where the probability of winning is less than 1/2, your best chance of doubling your money seems to be to bet your entire bankroll at once.
That is indeed a fun, if somewhat depressing 'system'. You lose more often than you win, but you are a celebrity on those occasions where you win big. The trick is. pre-defining big enough to quit.Quote: KavourasIf you like Labouchere, it may be of interest to you to know that the exact opposite method of betting (increasing when you win) is also very interesting.
