Martingale (betting System) - Mathematical Analysis

Mathematical Analysis

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, …, X_N–1 = T, and X_N = 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 = 4. 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 (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, which will be a very small number unless tails are nearly certain on each toss. In the fair case where, we could expect to wait something on the order of 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, 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.