|
The Labouchère system, also called the cancellation system or split martingale, is a gambling strategy used in roulette. The user of such a strategy decides before playing how much money they want to win, and writes down a list of positive numbers that sum to the predetermined amount. With each bet, the player stakes an amount equal to the sum of the first and last numbers on the list. If only one number remains, that number is the amount of the stake. If bet is successful, the two amounts are removed from the list. If the bet is unsuccessful, the amount lost is appended to the end of the list. This process continues until either the list is completely crossed out, at which point the desired amount of money has been won, or until the player runs out of money to wager. ==Description of the Gameplay== The theory behind this Labouchère system is that since the player is crossing two numbers off of the list (win) for every number added (loss) that the player can complete the list, (crossing out all numbers) thereby winning the desired amount even though the player does not need to win as much as expected for this to occur. It should be mentioned that the Labouchère System is meant to be applied to even money Roulette propositions such as Even/Odd, Red/Black or 1-18/19-36. When any of these bets are made in the game of Roulette, a spin resulting in a, "0," or, "00," results in a loss, so even though the payout is even money, the odds are clearly not 50/50. The Labouchère System attempts to offset these odds. If a player were to play any one of the above propositions, there are eighteen individual results which result in a win for that player and (for an American Roulette wheel) twenty individual results that result in a loss for that player. The player has an 18/38 chance of success betting any of the above propositions, which is around 47.37%. 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. 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. Assuming a player bets nothing but black (red/black proposition) and black can be expected to hit 47.37% of the time, but the system only requires that it hit 33.34% of the time, it can be said that black only need hit approximately 70.38% of the time (33.34/47.37) it can generally be expected to in order for the system to prevail. An obvious downfall to the system is bankroll, because the more losses sustained by the player, the greater the amount being bet on each turn (as well as the greater the amount lost overall) is. Consider the following list: 10 10 20 20 20 10 10 If a player were to bet black and lose four times in a row, the amounts bet would be: $20, $30, $40, and $50. By taking these four consecutive losses, the player has already lost $140 and is betting $60 more on the next bet. Consecutive losses, or an inordinate amount of losses to wins can also cause table limits to come into play. Occasionally, a player following this system will come to a point where he can no longer make the next bet as demanded by the system due to table limits. One work-around for this problem is simply to move to a higher limit table, or a player can take the next number that should be bet, divide it by two and simply add it to the list twice. The problem with the latter option is that every time a player commits such a play, it will infinitesimally increase the percentage of spins a player must win to complete the system. The reason this is so is because the player is adding two numbers (which both will be crossed out in the event of wins) where only one loss was sustained. To prove this, if a player were to play the Labouchère System the same way with the exception being that the player always added half of the wager lost to the bottom of the list twice for every wager lost 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. The player would actually have to win in excess of 50% of the time (the actual percentage of wins necessary, given x and y, being dependent on z) in order to complete the list, or more than the player could actually be expected to win. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Labouchère system」の詳細全文を読む スポンサード リンク
|