What do d1 and d2 actually represent in the BSM call formula?

They are not arbitrary — both are standardised distances that measure how far in-the-money the option is, but they differ by exactly one standard deviation of the stock's log return.

The intuition:

Under the risk-neutral measure $Q$, the log of the stock at expiry is normally distributed:

In other words, $S_T$ is lognormal. The standard deviation of $\ln(S_T)$ is $\sigma\sqrt{T}$.

$d_2$ in plain English:

$d_2$ is the number of standard deviations between today's expected log-return and the log-strike. Specifically:

So $N(d_2)$ = the risk-neutral probability that $S_T > K$, i.e., the probability the call finishes in-the-money. This is the single most useful interpretation to remember.

$d_1$ in plain English:

So $d_1$ is $d_2$ shifted one standard deviation higher. $N(d_1)$ is also a probability — specifically, the risk-neutral probability that the call finishes ITM under a different probability measure (the "stock-numeraire" measure rather than the risk-neutral measure).

You can also think of $N(d_1)$ as the delta of the call (it is, for a non-dividend stock; for a dividend-paying stock it is $e^{-qT} \cdot N(d_1)$).

Why two probabilities and not one?

The BSM formula $c = S_0 \cdot N(d_1) - K \cdot e^{-rT} \cdot N(d_2)$ has two terms, and each term has its own probability weight:

The reason $N(d_1)$ and $N(d_2)$ are different is technical: when you "expect a stock value conditional on the call being ITM" you have to weight by the stock's own distribution, which is a shift of one $\sigma\sqrt{T}$ from the risk-neutral one.

For the exam:

• $N(d_2) \approx$ probability the call expires ITM (risk-neutral)
• $N(d_1) \approx$ delta of the call (for $q=0$)
• Both are between 0 and 1
• $d_1 > d_2$ always, and the gap is $\sigma\sqrt{T}$
• For deep-ITM calls, both $\to 1$. For deep-OTM calls, both $\to 0$.

If you can give the $d_2$ probability interpretation cold, you will outperform 90% of candidates.