Martingales and fair games

Idea: A strategy that wins $75\%$

Take profit at $+2$ ticks, stop at $-6$, on a fair tick walk. It wins $75\%$ of trades.

• Expected profit: $\left(\frac{3}{4}\right)\times 2-\left(\frac{1}{4}\right)\times 6=0$
• The win rate comes from where the barriers sit

Formula: Martingale

$E[M_{t+1}\mid F_t]=M_t$

Given all you know, the best forecast of the next value is the current one: a fair game.

Idea: Check it in one step

Condition on today and take a single step: the mean must stay put.

• The fair walk $S_t$
• $S_t^2-t$, since $E[(S_t+ϵ{)}^2\mid F_t]=S_t^2+1$
• $r^{S_t}$ for a biased walk, with $r=\frac{q}{p}$

Formula: Optional stopping

$P(+b\text{first})=\frac{a}{a+b}$

Stop a fair walk at barriers +b and −a; a fair game gives the chance of the top one.

Example: Hitting the target first

A fair walk starts at $0$ with barriers at $+2$ and $-6$. What is the chance it hits $+2$ first?

1. Fair game: $P\times 2-(1-P)\times 6=0$
2. $P=\frac{6}{6+2}=\frac{3}{4}$

Quick check: How long a trade lasts

A fair walk from $0$ stops at $+2$ or $-6$. How many ticks does it last on average?

Common mistake: Doubling after each loss

It seems to always finish $1$ up. With $5$ bets you lose $31$ with chance $\frac{1}{32}$.

Instead Still fair: $\left(\frac{31}{32}\right)\times 1-\left(\frac{1}{32}\right)\times 31=0$. Unbounded stakes broke the stopping conditions.

Common mistake: A high win rate as an edge

A stopping rule on a fair game can raise the win rate as high as you like.

Instead Only expected profit matters. With bounded stakes and finite time no exit moves it from $0$.

Idea: Say it out loud

Name the martingale, check it in one step, name the stopping time and its condition.

• Then solve the linear equation
• Edge comes from a price that is not fair, never from a clever exit