**Gambling man**

Although apparently new to economists, gamblers know this as the Martingale, a class of betting strategies popular in 18th century France. The simplest was designed for a game in which Neil wins his stake if our coin comes up heads and loses it if the coin comes up tails. The strategy has Neil 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.

Because a gambler with infinite wealth will, almost surely, eventually flip heads, the Martingale betting strategy is a sure thing. Of course, Neil is not possessed of infinite wealth, and the exponential growth of the bets would eventually bankrupt those who chose to use the Martingale, and will do so to our friend Neil.

One round of the idealized Martingale without time or credit constraints can be formulated mathematically. 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.

We’ll let N be the time of appearance of the first H. So, in other words, X0, X1, …, XN–1 = T, and XN = 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 one unit over the first N tosses. For example, suppose the first four coin tosses are tails, tails, tails, heads, making N = 3. The player loses one, two and four units on the first three tosses, for a total loss of seven units, then wins eight units on the fourth toss, for a net gain of one unit. As long as the coin eventually shows heads, the player realizes a gain. But, always of only one unit.

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 non-negative number less than or equal to qk for all values of k is zero. It follows that N is finite with probability one. Therefore, with probability of one, the coin eventually will show heads and the player will realize a net gain of one unit. How does this work if the probability of a winning trade is just slightly less than of a losing trade? What if it is even a "winning" strategy?

The law of large numbers is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed. However, the more trials, the more likely the "unlikely" events also will occur. If you perform an infinite number of tests, then all possible events will occur, even improbable consecutive runs of losers.

**Perfect system**

This brings us back to Neil, who has started to follow Chicago trading guru Martin Gale. Martin Gale boasts of phenomenal results. Even though he has never heard of Taleb or the law of large numbers, Martin has it all figured out. For a price, he will tell you exactly what that is. Neil wants to know, so he pays Martin $5,000 for a four-day training session and flies to Chicago.

Martin explains this wondrous system using the E-mini S&P index where each tick is equal to $12.50. If the hourly, daily, weekly and monthly bars are all above their moving average and the five-minute bar trades down from a high and touches the moving average, go long one contract. It is going to bounce. Take a 10-tick profit, so Martin will be up $125. If the market goes against Martin by 20 ticks, he buys three more contracts and will exit on a five-tick increase. Observe that Martin loses $187.50 on his original purchase and gains $187.50 so he comes out flat; less four commissions, of course.

If the market goes down another 20 ticks, Martin adds 12 more contracts to his existing four and will exit on a five-tick bounce. That gives him 12 x 5 or 60 ticks profit for $750. Of the original contracts, Martin loses 3 x 15= 45 and 1 x 35, for a total of 80 ticks, so he is down $250 plus commissions on all 16 contracts. Assume Martin holds out for six ticks. He would gain 72 and lose 76. To post a profit, the market must bounce seven ticks. But, what if it doesn’t?

If it goes down another 20 ticks, Martin now adds 48 more contracts, giving him 64 in play. Again if the market bounces five ticks, he wins 240 on the new, but loses 12 x 15 for 180 plus 3 x 35 for 105 plus 1 x 55, or a total of 340 ticks. At a six tick goal, Martin loses 36 ticks before slippage and commissions. At seven ticks, he makes $350. If he pays only a $12.50 round turn commission and slippage, Martin must make an "eight-tick" bounce to net $350.

And, Martin tells Neil it always does in the S&P, Russell 2000 or the Euro FX contract. It never gets beyond 64 contracts and always has made the eight-tick bounce. Martin claims the ‘method’ will work on virtually anything.