An alternative mathematical analysis for the Martingale betting system
The previous analysis calculates expected value, but we can ask another question: what is the chance that one can play a casino game using the Martingale strategy, and avoid the losing streak long enough to double one's bankroll.
As before, this depends on the likelihood of losing 6 roulette spins in a row assuming we are betting red/black or even/odd. Many gamblers believe that the chances of losing 6 in a row are remote, and that with a patient adherence to the strategy they will slowly increase their bankroll.
In reality, the odds of a streak of 6 losses in a row are much higher than the many people intuitively believe. Psychological studies have shown that since people know that the odds of losing 6 times in a row out of 6 plays are low, they incorrectly assume that in a longer string of plays the odds are also very low. When people are asked to invent data representing 200 coin tosses, they often do not add streaks of more than 5 because they believe that these streaks are very unlikely.[1] This intuitive belief is sometimes referred to as the representativeness heuristic.
The odds of losing a single spin at roulette are q = 20/38 = 52.6316%. If you play a total of 6 spins, the odds of losing 6 times are q6 = 2.1256%, as stated above. However if you play more and more spins, the odds of losing 6 times in a row begin to increase rapidly.
- In 73 spins, there is a 50.3% chance that you will at some point have lost at least 6 spins in a row. (The chance of still being solvent after the first six spins is 0.978744, and the chance of becoming bankrupt at each subsequent spin is (1-0.526316)x0.021256 = 0.010069, where the first term is the chance that you won the (n-6)th spin - if you had lost the (n-6)th spin, you would have become bankrupt on the (n-1)th spin. Thus over 73 spins the probability of remaining solvent is 0.978744 x (1-0.010069)^67 = 0.49683, and thus the chance of becoming bankrupt is 1-0.49683 = 50.3%.)
- Similarly, in 150 spins, there is a 77.2% chance that you will lose at least 6 spins in a row at some point.
- And in 250 spins, there is a 91.1% chance that you will lose at least 6 spins in a row at some point.
To double the initial bankroll of 6,300 with initial bets of 100 would require a minimum of 63 spins (in the unlikely event you win every time), and a maximum of 378 spins (in the even more unlikely event that you win every single round on the sixth spin). Each round will last an average of approximately 2 spins, so, 63 rounds can be expected to take about 126 spins on average. Computer simulations show that the required number will almost never exceed 150 spins. Thus many gamblers believe that they can play the Martingale strategy with very little chance of failure long enough to double their bankroll. However, the odds of losing 6 in a row are 77.2% over 150 spins, as above.
We can replace the roulette game in the analysis with either the pass line at craps, where the odds of losing are lower q=(251:244, or 251/495)=50.7071%, or a coin toss game where the odds of losing are 50.0%. We should note that games like coin toss with no house edge are not played in a commercial casino and thus represent a limiting case.
- In 150 turns, there is a 73.5% chance that you will lose 6 times in a row on the pass line.
- In 150 turns, there is a 70.7% chance that you will lose 6 times in a row at coin tossing.
In larger casinos, the maximum table limit is higher, so you can double 7, 8, or 9 times without exceeding the limit. However, in order to end up with twice your initial bankroll, you must play even longer. The calculations produce the same results. The probabilities are overwhelming that you will reach the bust streak before you can even double your bankroll.
The conclusion is that players using Martingale strategy pose no threat to a casino. The odds are high that the player will go bust before he is able even to double his money.
Table limits are not specifically designed to prevent players from using Martingale strategy. The table limits exist so that the casino is not gambling more money than they can afford to lose. (E.g., a casino that takes in an average of $1000 a day on a roulette table might not accept a $7000 bet on black at that table; that bet would have a 18/38 chance of negating an entire week's profits.)
References
- ^ (wizardofodds.com/askthewizard/images/streaks.pdf)
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Mathematical analysis of a single round for the Martingale betting system
Let one round be defined as a sequence of consecutive losses followed by either a win, or bankruptcy of the gambler. After a win, the gambler "resets" and is considered to have started a new round. A continuous sequence of martingale bets can thus be partitioned into a sequence of independent rounds. We will analyze the expected value of one round.
Let q be the probability of losing (e.g. for roulette in America, with two zeroes, it is 20/38 for a bet on black or red). Let B be the amount of the commencing bet. Let n be the finite number of bets you can afford to lose.
The probability that you lose all n bets is qn. When you lose all your bets, the amount of money you lose is
ni=1Σ B . 2i-1 = B (2n - 1)
The probability that you do not lose all n bets is 1 − qn. If you do not lose all n bets, you win B amount of money (the initial bet amount). So the expected profit per round is
(1 - qn) . B - qn . B (2n - 1) = B (2n - 1)
Whenever q > 1/2, the expression 1 − (2q)n < 0 for all n > 0. That means for any game where you are more likely to lose than to win each bet (e.g. all chance gambling games) you are expected to lose money on average per round. Furthermore, the more times you are able to afford to bet, the more you will lose.
As an example, suppose you have 6,300 available to bet. You bet 100 on the first spin. If you lose, you bet 200 on the second spin, then 400 on the third, 800 on the fourth, 1,600 on the fifth, and 3,200 on the sixth.
If you win 100 on the first spin, you make 100, and the martingale starts over.
If you lose 100 on the first spin and win 200 on the second spin, you make a net profit of 100 at which point the martingale would start over.
If you lose on the first five spins, you lose a total of 3,100 (3,100 = 100 + 200 + 400 + 800 + 1,600). On the sixth spin you bet 3,200. If you win, you again make a profit of 100.
If you lose on the first six spins, you have lost a total of 6,300 and with only 6,300 available, you do not have enough money to double your previous bet. At this point the martingale cannot be continued.
In this example the probability of losing 6,300 and being unable to continue the martingale is equal to the probability of losing 6 times or (20/38)^6 = 2.1256%. The probability of winning 100 is equal to 1 minus the probability of losing 6 times or 1 - (20/38)^6 = 97.8744%.
The expected amount won is (100 x .978744) = 97.8744 . The expected amount lost is (6,300 x .021256)= 133.9118 . So overall you can expect to lose (133.9118 - 97.8744) = 36.0374 .
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Mathematical analysis for the Martingale system
One round of the idealized martingale without time or credit constraints can be formulated mathematically as follows. Let the coin tosses be represented by a sequence X0, X1, … of independent random variables, each of which is equal to H with probability p, and T with probability q = 1 – p. Let N be time of appearance of the first H; in other words, X0, X1, …, XN–1 = T, and XN = H. If the coin never shows H, we write N = ∞. N is itself a random variable because it depends on the random outcomes of the coin tosses.
In the first N – 1 coin tosses, the player following the martingale strategy loses 1, 2, …, 2N–1 units, accumulating a total loss of 2N − 1. On the Nth toss, there is a win of 2N units, resulting in a net gain of 1 unit over the first N tosses. For example, suppose the first four coin tosses are T, T, T, H making N = 3. The bettor loses 1, 2, and 4 units on the first three tosses, for a total loss of 7 units, then wins 8 units on the fourth toss , for a net gain of 1 unit. As long as the coin eventually shows heads, the betting player realizes a gain.
What is the probability that N = ∞, i.e., that the coin never shows heads? Clearly it can be no greater than the probability that the first k tosses are all T; this probability is qk. Unless q = 1, the only nonnegative number less than or equal to qk for all values of k is zero. It follows that N is finite with probability 1; therefore with probability 1, the coin will eventually show heads and the bettor will realize a net gain of 1 unit.
This property of the idealized version of the martingale accounts for the attraction of the idea. In practice, the idealized version can only be approximated, for two reasons. Unlimited credit to finance possibly astronomical losses during long runs of tails is not available, and there is a limit to the number of coin tosses that can be performed in any finite period of time, precluding the possibility of playing long enough to observe very long runs of tails.
As an example, consider a bettor with an available fortune, or credit, of 243 (approximately 9 trillion) units, roughly the size of the current US national debt in dollars. With this very large fortune, the player can afford to lose on the first 42 tosses, but a loss on the 43rd cannot be covered. The probability of losing on the first 42 tosses is q42, which will be a very small number unless tails are nearly certain on each toss. In the fair case where q = 1 / 2, we could expect to wait something on the order of 242 tosses before seeing 42 consecutive tails; tossing coins at the rate of one toss per second, this would require approximately 279,000 years.
This version of the game is likely to be unattractive to both players. The player with the fortune can expect to see a head and gain one unit on average every two tosses, or two seconds, corresponding to an annual income of about 31.6 million units until disaster (42 tails) occurs. This is only a 0.0036 percent return on the fortune at risk. The other player can look forward to steady losses of 31.6 million units per year until hitting an incredibly large jackpot, probably in something like 279,000 years, a period far longer than any currency has yet existed. If q > 1 / 2, this version of the game is also unfavorable to the first player in the sense that it would have negative expected winnings.
The impossibility of winning over the long run, given a limit of the size of bets or a limit in the size of one's bankroll or line of credit, is proven by the optional stopping theorem.[1]
References
- ^ a b Michael Mitzenmacher; Eli Upfal (2005), Probability and computing: randomized algorithms and probabilistic analysis, Cambridge University Press, p. 298, ISBN 9780521835404
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Intuitive analysis for Martingale
When the expected value of the stopping time is finite (which is true in practice), the following argument explains why the betting system fails: Since expectation is linear, the expected value of a series of bets is just the sum of the expected value of each bet. Since in such games of chance the bets are independent, the expectation of each bet does not depend on whether you previously won or lost. In most casino games, the expected value of any individual bet is negative, so the sum of lots of negative numbers is also always going to be negative.
The martingale strategy fails even with unbounded stopping time, as long as there is a limit on earnings or on the bets (which are also true in practice).[1] It is only with unbounded wealth, bets and time that the martingale strategy can succeed.
References
- ^ a b Michael Mitzenmacher; Eli Upfal (2005), Probability and computing: randomized algorithms and probabilistic analysis, Cambridge University Press, p. 298, ISBN 9780521835404
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Martingale
Originally, martingale referred to a class of betting strategies popular in 18th century France. The simplest of these strategies was designed for a game in which the gambler wins his stake if a coin comes up heads and loses it if the coin comes up tails. The strategy had the gambler double his bet after every loss, so that the first win would recover all previous losses plus win a profit equal to the original stake. Since a gambler with infinite wealth will, with probability 1, eventually flip heads, the Martingale betting strategy was seen as a sure thing by those who advocated it. Of course, none of the gamblers in fact possessed infinite wealth, and the exponential growth of the bets would eventually bankrupt those who chose to use the Martingale. It is widely believed that casinos instituted betting limits specifically to stop Martingale players, but in reality the assumptions behind the strategy are unsound. Players using the Martingale system do not have any long-term mathematical advantage over any other betting system or even randomly placed bets.
Effect of variance
As with any betting system, it sometimes happens that one achieves a better result than the expected negative return, by temporarily avoiding a losing streak. Furthermore, a straight string of losses is the only sequence of outcomes that results in a loss of money, so even when a player has lost the majority of his bets, he can still be ahead overall, since he always wins 1 unit when a bet wins, regardless of how many previous losses.[1]
Anti-martingale
In a classic martingale betting style, gamblers will increase their bets after each loss in hopes that an eventual win will recover all previous losses. The anti-martingale approach instead increases bets after wins, while reducing them after a loss. The perception is that in this manner the gambler will benefit from a winning streak or a "hot hand", while reducing losses while "cold" or otherwise having a losing streak. As the single bets are independent from each other (and from the gambler's expectations), the same conclusions as above apply.
References
- ^ "Martingale Long Term vs. Short Term Charts". Blackjackincolor.com.
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Reverse Labouchere
The Labouchere system can also be played as a positive progression betting system, this is known as playing the reverse Labouchere. In this version after a win, instead of deleting numbers from the line the player adds the previous bet amount to the end of the line. You continue building up your Labouchere line until you hit the table maximum. After a loss, the player deletes the outside numbers and continues working on the shorter line. The player starts their line again if they run out numbers to bet.[1]
The Reverse Labouchere system is often used because where the Labouchere list represents how much the player wants to win, a reverse Labouchere line represents the most that the player will lose during the betting cycle. It is with this that a player with a bankroll of x can create their own line, or lines, representative of the maximum amount that they can sustain in losses.
Additionally, a player does not necessarily have to continue the system until the table limit is met or exceeded, but could instead pick a single bet that the player does not wish to exceed and make that bet their own personal limit.
Unlike the Labouchere system which (when adhered to strictly) requires a winning percentage of at least 33.34% to complete, the winning percentage needed to complete a Reverse Labouchere line is going to be dependent on both the table limit (or the maximum single bet a player is willing to make) as well as the numbers on the initial line in relation to the table limit.
For instance, if a table had a limit of $500 and a player composed a Labouchere line as follows:
50, 50, 50, 50, 50
Nine consecutive wins (100, 150, 200, 250, 300, 350, 400, 450, 500) would cause the next bet in the system to exceed the table limit, and thus the line would be completed with a player profit of $2700.
In contrast, if a player composed a Labouchere line such as:
25, 25, 25, 25, 25
Nineteen consecutive wins (50, 75, 100, 125, 150, 175, 200, 225, 250, 275, 300, 325, 350, 375, 400, 425, 450, 475, 500) would cause the next bet in the system to exceed the table limit, thus the line would be completed with a player profit of $5,225.
The length of the line in the Reverse Labouchere system is also important as it relates to the percentage of wins necessary to complete the system. For instance, if a line of:
50, 50, 50, 50, 50
Suffers three consecutive losses as soon as the system begins, then the line is completed and a new line must be started, or the player may choose to quit.
In contrast, if a line of:
50, 50, 50, 50, 50, 50, 50, 50, 50, 50, 50, 50
Suffers three consecutive losses, then there are still six numbers remaining on the list. In the line immediately above, it would take an opening streak of six consecutive losses for the line to be completed.
All other things remaining the same, the longer a player's line, the more the player is risking losing, however, the longer the player's line, the better winning percentage the casino need have in order to break the player's line.
Advocates of this system point out that when a player uses the Labouchere System, where a streak in the casino's favor, or many mini-streaks in the casino's favor, will cause the player to sustain a huge loss, a single streak, or a few streaks in the player's favor using the Reverse Labouchere system will cause the player to have a huge gain.
A formula that can be used to determine how this system could fail is as follows:
Where:
x = Number of Wins
y = Number of Losses
Z = Numbers on original List
When:
x + z ≤ y * 2
The system has failed, and all numbers on the line are crossed completely out.
Given an infinite line, the Labouchere System when played by the player requires a winning percentage of at least 33.34% to complete. In contrast, for the Reverse Labouchere to fail requires only that the player lose 33.34% of the time.
Once again, the winning percentage necessary for the system completing to success depends upon a number of variables.
References
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Labouchère system
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.[1]
The theory behind this strategy 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 Labouchere 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 Labouchere 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 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 canceling 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 Labouchere 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.
References
- ^ Burrell, Brian. Merriam-Webster's Guide to Everyday Math. Merriam-Webster.
- Tijms, Henk (2004). "Probabilities in everyday life". Understanding probability: chance rules in everyday life. Cambridge University Press. pp. 91–93. ISBN 0-521-54036-4.
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Horse racing systems
Horse racing betting systems are based on a number of criteria, some of which include analysis of the horses' form.
Often horse racing systems are based on financial systems such as hedging (betting on multiple outcomes in a race) and arbitrage (lay the horse a low price and back it at a high price). Other horse racing systems exist which are based on items such as horse name, jockey form, trainer form, and lane draw. Modern horse racing systems can rely on specific betting possibilities only offered on betting exchanges.
Loss recovery systems such as Martingale can also be applied to horse racing.
This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia.
Independent events for betting strategies
The following betting strategies apply to games which operate on independent events. For such games, the odds of a particular outcome are identical for every bet played. No such strategy can beat the house edge (if any) in the long run, and all of them trade off many small wins for a big loss or vice versa.
- Martingale - doubling bet after each loss until a win is achieved (or fails when the amount of the bet becomes excessive)
- Kelly criterion
- Split martingale
- Anti-martingale
- d'Alembert
- Contra d'Alembert
- Regression
- Paroli of Three
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Betting strategy
A betting strategy or betting system is a structured approach to gambling intended to counter the inherent bias held by the house in casino and card games and by bookmakers in horseracing and sports betting. A successful strategy should increase the odds of winning in order to produce long term profits from a pursuit which under normal circumstances will only ever result in a long term loss.
All betting systems are predicated on statistical analysis, seeking to exploit the rare circumstances when the odds are in the favour of the player. Though the basis of all risk is fundamentally the same, betting systems vary in relation to the rules and circumstances of each particular game. The most established betting systems include:
- Card games - Card counting
- Roulette - Martingale
- Horse racing - Hedging, Arbitrage
- Sports - Handicapping[citation needed]
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