Martingale (betting System) - Mathematical Analysis of A Single Round

Mathematical Analysis of A Single Round

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

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

Whenever q > 1/2, the expression 1 − (2q)n < 0 for all n > 0. 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: (10/19)6 = 2.1256%. The probability of winning is equal to 1 minus the probability of losing 6 times: 1 − (20/38)6 = 97.8744%.

The expected amount won is (1 × 0.978744) = 0.978744.
The expected amount lost is (63 × 0.021256)= 1.339118.
Thus, the total expected value for each application of the betting system is (0.978744 − 1.339118) = −0.360374 .

In a unique circumstance, this strategy can make sense. Suppose the gambler possesses exactly 63 units but desperately needs a total of 64. Assuming q > 1/2 (it is a real casino) and he may only place bets at even odds, his best strategy is bold play: at each spin, he should bet the smallest amount such that if he wins he reaches his target immediately, and if he doesn't have enough for this, he should simply bet everything. Eventually he either goes bust or reaches his target. This strategy gives him a probability of 97.8744% of achieving the goal of winning one unit vs. a 2.1256% chance of losing all 63 units, and that is the best probability possible in this circumstance. However, bold play is not always the optimal strategy for having the biggest possible chance to increase an initial capital to some desired higher amount. If the gambler can bet arbitrarily small amounts at arbitrarily long odds (but still with the same expected loss of 2/38 of the stake at each bet), and can only place one bet at each spin, then there are strategies with above 98% chance of attaining his goal, and these use very timid play unless the gambler is close to losing all his capital, in which case he does switch to extremely bold play.