Quote:daveylbraTo everyone:

From wikipedia, labouchere system - "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."

Similarly, I understood Labouchere to require one third plus one bets, out of the total bets, to win.

BUT, suppose I start with 1,1,1,1,1. I lose the first 9 bets.

I have 1,1,1,1,1,2,3,4,5,6,7,8,9, 10. I have lost 54 units. (2+3+4+5+6+7+8+9+10)

I then win 6 bets. I have 3,4. I have won 52 units in those 6 bets. (11+10+9+8+7+7)

I have won 6 out of 15 bets. This is a third plus one. But I am still clearly down!

Can someone please explain?? Is there a certain number of total bets required before the one third plus one law 'kicks in?'

I don't think the "one-third plus one" rule is entirely accurate. If you add one number when you lose and remove two when you win, then you will have the same number as when you started if you have one-third wins; one additional win removes an additional two numbers, but if the remaining numbers (caused by earlier losses) total more than the sum of your initial numbers (as happened in your case; you started with five 1s, and are now left with 3,4), then you are still behind.

Quote:daveylbraTo everyone:

From wikipedia, labouchere system - "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."

Similarly, I understood Labouchere to require one third plus one bets, out of the total bets, to win.

BUT, suppose I start with 1,1,1,1,1. I lose the first 9 bets.

I have 1,1,1,1,1,2,3,4,5,6,7,8,9, 10. I have lost 54 units. (2+3+4+5+6+7+8+9+10)

I then win 6 bets. I have 3,4. I have won 52 units in those 6 bets. (11+10+9+8+7+7)

I have won 6 out of 15 bets. This is a third plus one. But I am still clearly down!

Can someone please explain?? Is there a certain number of total bets required before the one third plus one law 'kicks in?'

I'm not very familiar with the labouchere system, I've heard of it but never really crunched the numbers. So, I played around with it a little bit, and it is truly a bogus system.

Let's say after losing 9 in a row, you don't immediately win 9 in a row back, but instead you win 1 lose 1 for a while. You would end up clearing the line, but you wouldn't make a profit.

1-1-1-1-1-2-3-4-5-6-7-8-9-10-11 -54 units.

Win 1 +12 units

1-1-1-1-2-3-4-5-6-7-8-9-10 -42 units

Lose 1 -11 units

1-1-1-1-2-3-4-5-6-7-8-9-10-11 -53 units.

Win 1 +12 units

1-1-1-2-3-4-5-6-7-8-9-10 -41 units.

Lose 1 -11 units

1-1-1-2-3-4-5-6-7-8-9-10-11 -52 units.

You've cleared 2 of the 1's from your line, but only made +2 units (from -54 units to -52). If you continue to win 1 lose 1 you would clear the line, but you would not make a profit.

In other words, simply clearing the line DOES NOT guarantee you will make a profit, as they say it will. That is completely bogus as well.

Just really a bad system all together.

"While he made great strides in mathematics and physics, d'Alembert is also famously known for incorrectly arguing in Croix ou Pile that the probability of a coin landing heads increased for every time that it came up tails. In gambling, the strategy of decreasing one's bet the more one wins and increasing one's bet the more one loses is therefore called the D'Alembert system, a type of martingale."

https://en.wikipedia.org/wiki/Jean_le_Rond_d'Alembert

( Number of losses + Numbers originally on list ) / 2 <= Number of wins

results in the list being completed.

My idea is that we could have a long list of 1111...s, to go for a 'worst case scenario' number of wins, but we stop whenever

we are up. But I cannot work out the formula for how many wins to losses for x numbers on list would be required, or even if

it is calculatable.

Can anyone help? Wizard, perhaps??