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Thus, for all games where a gambler is more likely to lose than to win any given bet, that gambler is expected to lose money, on average, each round. Increasing the size of wager for each round per the martingale system only serves to increase the average loss. Suppose a gambler has a 63 unit gambling bankroll. The gambler might bet 1 unit on the first spin. On each loss, the bet is doubled. Thus, taking k as the number of preceding consecutive losses, the player will always bet 2k units.
With a win on any given spin, the gambler will net 1 unit over the total amount wagered to that point. Once this win is achieved, the gambler restarts the system with a 1 unit bet. With losses on all of the first six spins, the gambler loses a total of 63 units. This exhausts the bankroll and the martingale cannot be continued. In this example, the probability of losing the entire bankroll and being unable to continue the martingale is equal to the probability of 6 consecutive losses: The probability of winning is equal to 1 minus the probability of losing 6 times: Thus, the total expected value for each application of the betting system is 0.
In a unique circumstance, this strategy can make sense. Suppose the gambler possesses exactly 63 units but desperately needs a total of Eventually he either goes bust or reaches his target. This strategy gives him a probability of 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 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 coin tosses, they often do not add streaks of more than 5 because they believe that these streaks are very unlikely.