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LeetQuant
Quantitative Trading

Module 2 · Betting and Expected Value

Optimal Betting & the Kelly Criterion

Why maximising expected wealth is the wrong goal for repeated bets, and how Kelly sizing follows from log growth.

20 min

Positive edge is not enough

Suppose you can repeatedly bet on a coin that wins with probability p=0.6p = 0.6 at even odds. Betting your whole bankroll maximises expected wealth each round — and goes broke the first time you lose, which happens eventually with probability 1.

The relevant quantity for repeated, compounding bets is the growth rate. If you bet a fraction ff each time, after nn bets with kk wins your wealth is W0(1+f)k(1−f)n−kW_0(1+f)^k(1-f)^{n-k}, so

1nln⁡WnW0  ⟶  g(f)=pln⁡(1+f)+qln⁡(1−f).\frac1n \ln\frac{W_n}{W_0} \;\longrightarrow\; g(f) = p\ln(1+f) + q\ln(1-f) .

The Kelly fraction

Maximising g(f)g(f) gives, for a bet paying bb-to-1,

f∗=bp−qb=edgeodds.f^* = \frac{bp - q}{b} = \frac{\text{edge}}{\text{odds}} .

At even odds (b=1b = 1) this is f∗=p−qf^* = p - q. Key properties:

  • Kelly maximises long-run growth and never risks ruin (it never bets everything).
  • Betting more than f∗f^* lowers growth; at roughly 2f∗2f^* the growth rate falls to about zero.
  • Betting less (e.g. half Kelly) gives up a little growth for much less volatility — popular in practice because your estimate of pp is uncertain.

Worked example

a 2-to-1 payout

A bet wins with probability 0.40.4 and pays 2-to-1. What fraction should a Kelly bettor stake?

Show solution

Edge per unit staked is bp−q=2(0.4)−0.6=0.2bp - q = 2(0.4) - 0.6 = 0.2. Dividing by the odds:

f∗=0.22=0.1.f^* = \frac{0.2}{2} = 0.1 .

Even though the bet loses 60% of the time, it has positive expectation and Kelly stakes 10%.

Knowledge check

A coin wins with probability 0.55 at even odds. What is the Kelly fraction?

Knowledge check

Why do many practitioners bet a fraction of Kelly (e.g. half Kelly)?

Practice problems

Finished reading?

Answer the 2 knowledge checks to complete this lesson (0 done).