# Doubling for the win.

For many years I am debating the use of martingale or doubling strategies. In this strategy the trade size get doubles with each loss until the first win happens. The hope is that after the first win the sequence makes exactly one dollar as is justified by the following formula $1 + 2 + 2^2 + 2^3 + \ldots + 2^k = 2^k-1$. In real trading such a trading strategy usually produces very smooth nice equity curves that eventually end up with a crash. Despite this fact, I know many knowledgeable people who are still using this system and defend it.

In this post I will try to prove from at least one perspective why trading the random walk in the usual way might be beneficial to trading a martingale.

We start with the definition of the underlying process. We will assume that capital of a trader is a Markov chain with the state space given by the current capital. We say we trade a random walk when transition probability is defined as follows $P(X_{t +1} = x+1|X_t = x) = p$ and $P(X_{t +1} = x-1|X_t = x) = q$ where $p>q$.

We start with the capital $x$ and we want to compute the probability of us hitting zero capital. It is not hard to prove the following proposition

Proposition. The probability to hit zero starting from the capital of $x$ is given by $(\frac{p}{q})^x$.

In particular, if $p=q=1/2$ it is well known that the probability of a bankruptcy is sure. However, what makes it possible to trade such a strategy that expected time to a crash is equal to infinity.

One important consequence from this result is when starting capital is very large the chance to go bancrupt approaches zero.

Completely different story with the doubling strategy. I can prove the following

Proposition. For a doubling strategy, the probability to go bancrupt is equal to one.

From this perspective trading martingale is a sure loss.