The First Black-Scholes Term Is a Share Count, Not a Probability

A share at 100, a one-year call struck at 100, a zero rate and a twenty percent swing return 0.539828 and 0.460172. The first is the number of shares in the replicating portfolio, and reading it as the chance of finishing in the money hands you the complement of the right answer, since at the money with a zero rate the two numbers sum to one. The article carries the density identity that makes the share count exact, the measure under which the first number is a probability after all, and the four cents the straight line misses over a two-dollar move.

A share trades at 100. A European call on it is struck at 100 and runs for a year. Take the rate to be zero and the volatility twenty percent, feed that into the pricing formula, and two numbers fall out on the way to the answer: 0.539828 and 0.460172. The interview question is what they are.

The tempting answer says the bigger one is the chance the call finishes above the strike. It is not, and at these inputs the error has a cruel shape: the chance is the other number. Read the first term as a probability and you report 54 percent when the answer is 46.

The two numbers, and why the confusion hides

Both numbers are values of the same function, the standard normal distribution function $\Phi$ , read at two arguments that differ by the volatility times the root of the time left:

d_1 = \frac{\ln(S/X) + \left(r + \tfrac12\sigma^2\right)\tau}{\sigma\sqrt{\tau}}, \qquad d_2 = d_1 - \sigma\sqrt{\tau}

(1)

At the money the logarithm vanishes, and with a zero rate what is left is $d_1 = \tfrac12\sigma\sqrt{\tau} = 0.1$ and $d_2 = -0.1$ . The two arguments are exact opposites, so by the symmetry of the normal distribution the two numbers sum to one. That is the whole reason the confusion survives: a pair of numbers between zero and one that add up to one looks like a probability and its complement, and 0.539828 sits close enough to a half to pass for a coin.

The sum is a coincidence of this parameter set, not a structural fact. Put the rate at five percent and the two arguments become 0.35 and 0.15, giving 0.636831 and 0.559618, which add to 1.1965. Nothing partitions anything.

What the formula is telling you to trade

The call price is a difference of two terms, and each is a position you can hold:

c = \underbrace{S\,\Phi(d_1)}_{\text{shares}} \;-\; \underbrace{Xe^{-r\tau}\Phi(d_2)}_{\text{borrowing}}

(2)

The replicating portfolio

A portfolio that holds $\Phi(d_1)$ shares and owes $Xe^{-r\tau}\Phi(d_2)$ in cash. Its value changes exactly as the option does for small moves in the share, so no-arbitrage forces the option to cost the same. The share count $\Phi(d_1)$ is what a desk actually trades against the option, which is why it is also called the hedge ratio.

At our numbers that reads: hold 0.539828 shares, which at 100 a share is 53.9828 of stock, and borrow 46.0172 against it. The difference is 7.965567, the price of the call.

53.9828 - 46.0172 = 7.9656

(3)

Fig. 1 — The formula read as a trade. Nothing here is a probability: the top bar is stock and the middle bar is a loan.

The second term deserves a closer look, because it is not simply the discounted strike. It is the discounted strike multiplied by the chance of paying it. You borrow against the shares only in the states where the option is exercised, and 0.460172 is how often that happens.

Why the share count is exactly the first term

A sceptic should object here. Both $d_1$ and $d_2$ depend on $S$ , so differentiating (2) ought to produce three terms, not one. It does, and two of them cancel:

\frac{\partial c}{\partial S} = \Phi(d_1) + \Big[S\varphi(d_1) - Xe^{-r\tau}\varphi(d_2)\Big]\frac{\partial d_1}{\partial S}

(4)

The bracket is empty. Both arguments have the same derivative in $S$ , namely $1/(S\sigma\sqrt{\tau})$ , since they differ by a constant, and the two normal densities are related by an identity that does the rest of the work:

\frac{\varphi(d_2)}{\varphi(d_1)} = e^{\,d_1\sigma\sqrt{\tau} - \tfrac12\sigma^2\tau} = \frac{S}{X}e^{r\tau} \;\Longrightarrow\; Xe^{-r\tau}\varphi(d_2) = S\varphi(d_1)

(5)

So the share count is $\Phi(d_1)$ on the nose, with no correction. That cancellation is the reason the formula can be read as a portfolio at all. Without it the first term would be a coefficient in an expression rather than a number of shares you can pick up the phone and buy.

The probability you actually wanted

Under the measure the price is computed in, the terminal share price is $S_\tau = S\exp\!\big((r - \tfrac12\sigma^2)\tau + \sigma\sqrt{\tau}Z\big)$ with $Z$ standard normal. Finishing above the strike is an event about $Z$ alone, and rearranging it gives $Z > -d_2$ :

\mathbb{P}\left(S_\tau > X\right) = \Phi(d_2) = 0.460172

(6)

At the money with a zero rate that is $\Phi(-\tfrac12\sigma\sqrt{\tau})$ , which is strictly below a half for every positive volatility and every positive maturity. The share is a martingale here, so its average finish is exactly where it started, and yet it finishes below where it started more often than not. Both statements are true because the average of a lognormal sits above its median, and the gap in the exponent is exactly the $\tfrac12\sigma^2\tau$ that separates $d_1$ from $d_2$ .

The qualifier about which measure is doing the work is not decoration. The real-world chance of finishing above 100 depends on the drift of the share, which nobody knows and which the price does not contain. Change the drift and 0.460172 moves while the option price does not.

The first number is a probability too, of a different thing

There is a sentence about all this that goes too far, which is that the first term is not a probability at all. It is one. Divide the share-weighted expectation of the exercise event by the expected share price and the result is exactly the first number:

\Phi(d_1) = \frac{\mathbb{E}\big[S_\tau \mathbf{1}\{S_\tau > X\}\big]}{\mathbb{E}\big[S_\tau\big]}

(7)

That ratio is the probability of the same event under a different measure, the one that uses the share rather than the money as its unit of account. Weighting each path by its terminal share price loads up precisely the paths that finish in the money, which is why 0.539828 exceeds 0.460172. A four-million-path simulation measured the pricing-measure figure at 0.46026 and the share-measure figure at 0.539882, against closed forms of 0.460172 and 0.539828, so which number belongs to which measure was checked rather than asserted.

The honest short version: both numbers are probabilities of the same event, computed under two different measures, and only one of them is the one an interviewer means by the chance of finishing in the money.

Where the two-line story stops working

The portfolio in (2) replicates the option at one share price and one instant. It is a derivative, not a position you can hold and forget. The curvature is $\Gamma = \varphi(d_1)/(S\sigma\sqrt{\tau}) = 0.019848$ , so over a two-dollar move the quadratic correction is:

\tfrac12\Gamma(\Delta S)^2 = \tfrac12 (0.019848)(2)^2 = 0.0397

(8)

Four cents on an option worth 7.97, and the sign is the same on both sides, because a long call is convex in the share. A hedge left alone loses that amount whichever way the share moves.

Fig. 2 — The share count is the slope of the tangent, so it is right at exactly one point. Two dollars away the straight line is already four cents low.

Two further caveats. With a continuous dividend yield $q$ the share count becomes $e^{-q\tau}\Phi(d_1)$ , strictly smaller, because holding the share pays you something the option does not, so fewer shares are needed. And $\Phi(d_2)$ is the chance of exercise only for a European call. Give the holder the right to exercise early and the exercise event is no longer the terminal event that (6) describes.

Sources and further reading

The formula that (1) and (2) come from — Black-Scholes model
The share count among the other sensitivities — Greeks (finance)
The change of unit of account behind (7) — Numéraire
Why the median sits below the mean — Log-normal distribution
The measure that (6) is computed under — Risk-neutral measure

Every number here was checked three ways: symbolically, by a four-million-path Monte Carlo that measured both probabilities without evaluating $\Phi$ , and by rebuilding the price from the replicating portfolio and sweeping a hedging grid to confirm that the first term really is the local share count. The below-a-half result was also confirmed monotone over forty volatilities, so it is a structural fact rather than an artefact of the twenty percent chosen here.