Kelly Criterion — Cheatsheet Derivation

Step 1 One Period Expectancy

E = pB − q

where p = win probability, q = 1−p = loss probability, B = net odds (win B per unit staked, lose 1).

Why: Weighted average of outcomes. Win B with probability p, lose 1 with probability q.

Bet fraction f of current wealth W:

Why: You keep the unbet portion W(1−f) regardless. On a win you collect B times your stake Wf on top. On a loss your stake Wf is gone.

After h wins and (n−h) losses:

Wₙ = W₀(1+Bf)^h(1−f)^n−h

Why: The flips compound — each one rescales whatever the previous left. That means multiply, not add. Order doesn’t matter, only h and n−h.

Take the nth root of total growth to extract the per-period rate:

G = (Wₙ/W₀)^1/n = (1+Bf)^h/n · (1−f)^(n−h)/n

As n → ∞, law of large numbers: h/n → p, (n−h)/n → q.

G = (1+Bf)^p(1−f)^q

Why nth root: Same logic as extracting r from (1+r)ⁿ = total growth. Geometric mean, not arithmetic, because the process is multiplicative.

Define g = ln(G). Since ln is monotonically increasing, maximizing g gives the same f* as maximizing G.

Apply two log rules:

g = p·ln(1+Bf) + q·ln(1−f)

Why: Differentiating a product of powers is a mess. A sum of logs is trivial. Valid because ln is monotone — same maximum, easier math.

Rule: ddx[ln(x)] = 1x. Chain rule: multiply by derivative of the inside.

Set dg/df = 0:

pB1+Bf − q1−f = 0

Cross-multiply:

pB(1−f) = q(1+Bf)

Expand:

pB − pBf = q + qBf

Collect f terms:

pB − q = pBf + qBf = Bf(p+q)

Since p+q = 1:

f* = pB−qB = p − qB

f* = pB − qB

Read as: edge / odds

Special case B=1 (even money): f* = p−q

Your optimal bet equals your raw edge.

What	Why it matters
Multiplicative wealth function	One bad bet can’t be offset by other bets — sizing matters
Geometric mean not arithmetic	Compounding processes need per-period rates, not averages
ln transform	Turns product into sum without moving the maximum
Chain rule on ln(1−f)	The −1 derivative is what creates a finite optimum
p+q=1	The simplification that closes the algebra cleanly