Disliked{quote}As yet, I have not seen any martingale system that can survive a big run, and still deliver gains that are worthwhile for me. I would love to be proved wrong, but for now, i will stick to my own manual trading.Ignored
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. Following is an analysis of the expected value of one round.
Let q be the probability of losing (e.g. for American double-zero roulette, it is 10/19 for a bet on black or red). Let B be the amount of the initial bet. Let n be the finite number of bets the gambler can afford to lose.
The probability that the gambler will lose all n bets is qn. When all bets lose, the total loss is
http://upload.wikimedia.org/math/b/f...072ba02201.png
The probability the gambler does not lose all n bets is 1 − qn. In all other cases, the gambler wins the initial bet (B.) Thus, the expected profit per round is
http://upload.wikimedia.org/math/6/e...62b6a9f5fa.png
The probability that the gambler will lose all n bets is q^n, where q is your loss rate and n is the number of bets. I have made a quick graph, it is really easy you can do it yourself too, just replace the value of q by the loss rate of your strategy:
http://fooplot.com/#W3sidHlwZSI6MCwi...ItMCIsIjIiXX1d
As you can see the value of y approaches 0 but never actually reaches it unless q is 0 (win rate is 100%): this would be the chance of you getting a margin call with a Martingale system. 7 losses in a row with 0.5 loss rate still has a 0.78% chance of happening: which is 1 in every 128 trades!