Proof to go broke almost surely Suppose you play a game where you start with capital $K_0 = 1$. In each turn $i = 1,2, \dots, n$ you throw a fair coin independently of the history. If you get "heads", you get back $3/2$ of your capital, if you get "tails" you only get back $1/2$ of your capital.
Prove that $K_n \to 0$ almost surely.
My attempt:
First define $R_i$ by $$R_i(\omega) = \begin{cases} \frac{3}{2} & \omega_i = \text{heads} \\ \frac{1}{2} & \omega_i = \text{tails}\end{cases}$$
and note that $K_n = \prod_{i=1}^nR_i $,  $\mathbb E[R_i] = 1 $ and by independence $\mathbb E[K_n] = 1.$ Now the only ways to prove almost sure convergence I know of are the Strong Law of Large Numbers and the Borel-Cantelli Lemma. A way to apply the former is to look at $$\log K_n = \sum \log(R_i).$$
By the Strong Law I get $$\frac{1}{n}\sum_{i=1}^n \log(R_i) \to \log \left(\frac{\sqrt3}{2}\right)$$
but that is not quite what I want. Also I find this pretty counterintuitive.

Formal answer. One way to do this (without martingales) is by using strong law of large numbers and the law of the iterated logarithm in tandem. You are flipping a fair coin $n$ times. Let $H_i$ equal $1$ if the coin comes up heads on flip $i$, and $0$ otherwise. Then the $H_i$ are i.i.d. uniform on $\{0, 1\}$, and $S_n = H_1 + ... + H_n$ satisfies:

*

*Strong law of large numbers: $S_n/n \rightarrow 1/2$ almost surely.


*Law of the iterated logarithm: $\limsup_{n \to \infty} \frac{\pm (S_n - n/2)}{\sqrt{2n \log\log n}} = 1/2$ almost surely. (This is because $H_i - 1/2$ has mean $0$ and variance $1/4$ for all $i \geq 1$.)
If we denote your capital after bet $n$ by $K_n$, and your starting capital is $K_0 = 1$, this strategy amounts to betting half your wealth every time. We have the expression $$K_n = K_0 \prod_{i=1}^n \left( \frac{1}{2} + H_i \right) = \frac{3^{S_n}}{2^n},$$
since the denominator is always multiplied by $2$, and the numerator is multiplied by $1$ for every tails and $3$ for every heads. Therefore, we must almost surely have $$\limsup_{n \to \infty} K_n = \limsup_{n \to \infty} \frac{3^{S_n}}{2^n} \leq \limsup_{n \to \infty} \frac{3^{n/2 + \sqrt{2n \log \log n}/2}}{2^n} = \limsup_{n \to \infty} \frac{(\sqrt{3})^{(n + \sqrt{2n \log \log n})}}{2^n}= 0,$$
which shows that your wealth converges to zero with probability one.
Informal answer. Even though you are equally likely to win as to lose this bet on any given round, a loss divides your wealth by 2, while a win only multiplies it by 1.5. This means it takes, on average, $$\log_{1.5}(2) = 1.70951129135...$$ wins to cancel out a loss. But the SLLN shows that wins and losses happen with equal frequency asymptotically. Thus, $$K_n \approx K_0 \left( \frac{1}{2} \right)^{n/2} \left( \frac{3}{2} \right)^{n/2} = \left( \frac{\sqrt{3}}{2} \right)^{n},$$
and the fluctuations in the exponent are only of order $\sqrt{n \log \log n}$ by LIL, which is not enough to cancel out the downward exponential decay in the limit as $n \rightarrow \infty$.
This also shows that any strategy of the form "On bet $n$, stake $p \%$ of your wealth for fixed $p \in (0, 1)$" must eventually give you a limiting wealth of $0$, because $(1 - p)^{1/2}(1 + p)^{1/2} = (1 - p^2)^{1/2} < 1$, so asymptotically your wealth decays exponentially to zero.
