ranch7
• Posts: 2
Joined: Mar 3, 2015
March 3rd, 2015 at 10:21:08 PM permalink

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!
bobsims
• Posts: 316
Joined: Apr 8, 2014
March 4th, 2015 at 6:51:35 AM permalink
I used it 30 years ago. Believe me, you WILL lose.
Baccaratfrom79
• Posts: 741
Joined: Jan 12, 2015
March 4th, 2015 at 7:26:27 AM permalink
deleted, I was wrong when I posted.
Bac79=Hazardous Material and Chemical person correcting other's mistakes. Non AP'er, I can't count cards, low intelligence. Sprinkles magical dust on the cards. Has a lucky monkey. Baby also has a green one. Sum it up: "It's okay just blame me, it's all my fault"! ( No one believes me--so I chose to stop posting)
ThatDonGuy

• Posts: 6405
Joined: Jun 22, 2011
March 4th, 2015 at 7:46:03 AM permalink
Quote: ranch7

It 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).
DeMango
• Posts: 2958
Joined: Feb 2, 2010
March 4th, 2015 at 9:39:51 AM permalink
What is it with posters who say they play with chocolate chips and don't know the difference between LaBouchere and Martingale? Even Egalite knows that!
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
Baccaratfrom79
• Posts: 741
Joined: Jan 12, 2015
March 4th, 2015 at 9:55:32 AM permalink
Don't follow the 'systems' or progressive techniques. Just assume they all mean in some shape or form to double up progressively until the win. At least the ones I read a bit about. Hopefully that explains it, if not sorry.
Bac79=Hazardous Material and Chemical person correcting other's mistakes. Non AP'er, I can't count cards, low intelligence. Sprinkles magical dust on the cards. Has a lucky monkey. Baby also has a green one. Sum it up: "It's okay just blame me, it's all my fault"! ( No one believes me--so I chose to stop posting)
ranch7
• Posts: 2
Joined: Mar 3, 2015
March 4th, 2015 at 9:07:16 PM permalink
Quote: ThatDonGuy

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).[/q

OK so if I had \$10000 and tried this ten times I should win only 3- 4 of them?

ThatDonGuy

• Posts: 6405
Joined: Jun 22, 2011
March 5th, 2015 at 6:45:27 AM permalink
Quote: ranch7

OK 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.
Kavouras
• Posts: 52
Joined: Apr 22, 2014
March 14th, 2015 at 2:30:29 PM permalink
If 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. Search for Reverse Labouchere. And it has one additional advantage: your bet increases in much better accordance to your overall bankroll than with the cancellation system.
http://www.Roulette30.com
OnceDear