# How to maximize returns in this scenario

You have a machine. You can put money into it. You have $$s$$ initial budget. $$p$$ percent of the time the machine will double your investment. $$(100-p)$$ percent of the time it will just swallow your money and not return anything. You can choose a ratio $$a$$ of your money to reinvest at every turn into the machine. E.g. if $$a=1$$, then you are reinvesting all your money at every turn (and very quickly will be left with nothing).

What is the optimal ratio $$a$$ with which your money grows the fastest with respect to the number of times you put money in the machine?

Let $$p' = p/100$$ denote the probability of winning at a turn with the machine. Your initial capital is $$W_0 = s$$. After the first turn where you bet $$aW_0$$, your wealth is

$$W_1 = W_0 + aW_0X_1 = W_0(1 +aX_1),$$

where $$X_1$$ is a binary random variable such that $$P(X_1 = 1) = p'$$ and $$P(X_1 = -1) = 1-p'$$.

Assume that the outcome of a turn depends in no way on the outcomes of previous turns. After $$n$$ turns, your wealth is

$$W_n = W_0(1 +aX_1)(1+aX_2) \cdots(1+aX_n)$$

where $$X_1, X_2, \ldots, X_n$$ are independent and identically distributed binary random variables.

The compounded rate-of-growth is

$$R_n(a) = \log \left[\left(\frac{W_n}{W_0} \right)^{1/n}\right] = \frac{1}{n} \log \left(\frac{W_n}{W_0} \right)= \frac{1}{n}\sum_{k=1}^n\log(1+aX_k),$$

with the expected value

$$G_n(a) = E[R_n(a)] = \frac{1}{n}\sum_{k=1}^n E[ \log(1+aX_k)].$$

Since the random variables are identically distributed, we have for all $$k$$,

$$E[ \log(1+aX_k)] = E[ \log(1+aX_1)] = p'\log(1+a) + (1-p')\log(1-a),$$

and, hence, the expected rate-of-growth is independent of $$n$$:

$$G_n(a) = \frac{1}{n} \sum_{k = 1}^n [p'\log(1+a) + (1-p')\log(1-a)] = p'\log(1+a) + (1-p')\log(1-a)$$

If the odds are in your favor we have $$p' > 1/2$$ and setting the derivative of $$G_n$$ equal to $$0$$ determines the optimal proportion $$a^*$$ that maximizes the rate-of-growth, that is

$$G_n'(a^*) = \frac{p'}{1+a^*} - \frac{1-p'}{1 - a^*} = 0 \\ \implies a^* = 2p'-1$$

This is commonly known as the Kelly criterion.

If the odds are not in your favor we have $$p' \leqslant 1/2$$ and there is no strategy that produce a positive expected rate-of-growth -- see Gambler's ruin. We can only establish another objective like maximizing the probability of doubling initial capital and quitting. At even odds, $$p'= 1/2$$, it is best to bet everything on one turn.