Martingale (betting system)

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A separate article treats the topic of martingale (probability theory).

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 practiced it. Unfortunately, none of these practitioners in fact possessed infinite wealth, and the exponential growth of the bets would eventually bankrupt those foolish enough to use the martingale.

1 Example
2 Detailed analysis
3 Simpler analysis
4 See also

[edit] Example

Suppose that someone applies the martingale betting system at an American roulette table, with 0 and 00 values; on average, a bet on either red or black will win 18 times out of 38. If the player's initial bankroll is $150 and the betting unit is $10, he can afford 4 losing bets in a row ($10, $20, $40, and $80) before he runs out of money. If any of these 4 bets wins he wins $10 and wins back any past losses. The chance of losing 4 bets in a row (and therefore losing the complete $150) is (20/38)⁴ = 7.67%. The remaining 92.3% of the time that the progression is attempted the player will win $10. One can see that if the progression is continuously played, the average amount of money made using the progression will be (0.923·$10) − (0.0767·$150) = −$2.275. Therefore the progression loses on average $2.275 each time it is attempted.

[edit] Detailed analysis

Let $q$ be the probability of losing (e.g. for roulette it is 20/38). Let $y$ be the amount of the commencing bet (e.g. $10 in the example above). Let $x$ be the finite number of bets you can afford.

The probability that you lose all $x$ bets is $q x$ . When you lose all your bets, the amount of money you lose is

$\sum_{i=1}^x y \cdot 2^{i-1} = y (2^x - 1)$

The probability that you do not lose all $x$ bets is $1 - q x$ . If you do not lose all $x$ bets, you win $y$ amount of money. So the expected profit is

$(1-q^x) \cdot y - q^x \cdot y (2^x - 1) = y (1 - (2q)^x)$

Whenever $q > 1 / 2$ , the expression $1 - (2 q) x < 0$ for all $x > 0$ . That means for any game where it is more likely to lose than to win (e.g. all chance gambling games), you are expected to lose money on average. Furthermore, the more times you are able to afford to bet, the more you will lose.

[edit] Simpler analysis

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 all bets are going to be the same, regardless of whether you had previously won or lost. In most gambling games, the expected value of any individual bet is going to be negative, so the sum of lots of negative numbers is also always going to be negative.