The Black-Scholes Formula, Explained From Scratch

Most explanations of Black-Scholes throw the formula at you first and the intuition second, if at all. That's backwards — the formula is just bookkeeping once you understand the two questions it's answering. Let's do it in the right order.

The problem the formula is solving

A call option gives you the right to buy a stock at a fixed price (the strike, K) on a fixed date (expiration, T). Its value at expiration is simple: if the stock ends above K, the option is worth stock price − K. If it ends below K, it's worth nothing. The hard part is pricing it today, before you know which of those happens.

Black-Scholes prices it by answering two questions:

1. What's the expected payoff, if it pays off? — roughly, the stock price times the probability the option finishes in the money.
2. What do you have to pay to guarantee that payoff, discounted to today? — roughly, the strike price times that same probability, discounted at the risk-free rate.

The call price is question 1 minus question 2.

The formula

C = S · N(d1) − K · e^(−rT) · N(d2)

• S — current stock price
• K — strike price
• r — risk-free interest rate (annualized)
• T — time to expiration, in years
• σ (sigma) — volatility, the annualized standard deviation of the stock's returns
• N(x) — the standard normal cumulative distribution function: the probability that a standard normal random variable is less than x

N(d1) and N(d2) are both probabilities (values between 0 and 1), which is why the formula looks like "stock price times a probability, minus strike times a probability" — because that's exactly what it is.

Where d1 and d2 come from

d1 = [ln(S/K) + (r + σ²/2)·T] / (σ·√T)
d2 = d1 - σ·√T

d1 measures how far in-the-money the option is, in standard-deviation units, adjusted for drift and time. ln(S/K) is the log-moneyness — how far the current price is from the strike, on a log scale (log scale because stock returns are roughly log-normal, not normal). The (r + σ²/2)·T term adjusts for the fact that the stock is expected to drift upward at the risk-free rate under the no-arbitrage assumption the model makes, plus a volatility correction. Dividing by σ·√T converts that into standard deviations of the stock's expected move over the life of the option.

d2 is d1 shifted down by one volatility-adjusted standard deviation — it's the term that actually corresponds to "probability the option finishes in the money," while d1 is closer to "probability, weighted by how much it finishes in the money." That distinction is why the formula needs both instead of just one.

Pricing a call in Python

import math

def norm_cdf(x):
    return 0.5 * math.erfc(-x / math.sqrt(2))

def black_scholes_call(S, K, T, r, sigma):
    d1 = (math.log(S / K) + (r + 0.5 * sigma**2) * T) / (sigma * math.sqrt(T))
    d2 = d1 - sigma * math.sqrt(T)
    return S * norm_cdf(d1) - K * math.exp(-r * T) * norm_cdf(d2)

# A $100 stock, $100 strike, 1 year to expiry, 5% risk-free rate, 20% vol
print(black_scholes_call(100, 100, 1, 0.05, 0.20))  # ≈ 10.45

norm_cdf uses math.erfc — the complementary error function — instead of scipy's norm.cdf, because it's one line of standard library math instead of a dependency, and it's exact, not an approximation. That's the same trick the Greeks all lean on: they're each a derivative of this same C, with respect to one of its five inputs (S, K, T, r, σ), and every one of them reuses d1/d2.

What the model assumes — and gets wrong

Black-Scholes assumes constant volatility, no dividends, frictionless trading, and log-normal stock returns. Real markets violate all four, which is exactly why implied volatility exists: instead of trusting the model's volatility input, traders back-solve it from real option prices, effectively asking "what volatility would the market have to believe for this price to be correct?" That's a different lesson, but it only makes sense once this one does.