N(d2) worked by hand, why delta overstates the probability, how much an assumed real-world drift changes it, and the digital option that prices it.
Under Black-Scholes the risk-neutral probability that a call expires in the money is N(d2), not delta. For a one-year at-the-money call on a stock at 100 with 24 per cent volatility and a 4 per cent rate, d2 is 0.0467 and the probability is 51.86 per cent, against a delta of 0.6128. If you assume the stock drifts up at 8 per cent a year instead of the risk-free rate, it becomes 58.45 per cent.
Worked in full in Quantitative Finance by Julian R. Sterling, with every figure reproduced in a free workbook.See the book on Amazon →
The question matters to anyone selling options for income, setting strikes on a structured product, or reading a platform's "probability ITM" column. Three numbers get used for it, delta, risk-neutral N(d2) and a real-world estimate, and they differ by several points on the same option.
The option is Kestrel 100, the call that runs through Quantitative Finance. All figures are illustrative.
| Input | Value |
|---|---|
| Spot | 100 |
| Strike | 100 |
| Rate, continuous | 4% |
| Volatility | 24% |
| Expiry | 1 year |
| Assumed real-world drift, for comparison | 8% |
d2 = [ln(S ÷ K) + (r − σ² ÷ 2) × T] ÷ (σ × √T)
ln(100 ÷ 100) = 0; r − σ² ÷ 2 = 0.04 − 0.0288 = 0.0112
d2 = 0.0112 ÷ 0.24 = 0.0467 (and d1 = d2 + 0.24 = 0.2867)
P(call in the money) = N(d2) = N(0.0467) = 0.5186
P(put in the money) = N(−d2) = 0.4814
Excel: =NORM.S.DIST(d2, TRUE)
The result is above 50 per cent even though the strike equals spot. Under the model the price after a year is lognormal, and its median is S × e(r − σ²/2)T = 101.13, a little above the strike. The −σ²/2 term pulls the median below the mean of 104.08: volatility makes large gains possible, which raises the average, while most paths end nearer the start.
Delta is N(d1), and d1 is d2 plus σ√T. For a call delta is always above N(d2), here by 9.42 points: 0.6128 against 0.5186. Delta is the probability of finishing in the money measured under a different weighting, one that counts each path by the share price it ends at. That is what makes it the right hedge ratio, as in calculating delta and the shares to hedge, and the wrong probability.
N(d2) uses the risk-free rate as the stock's drift, because that is what pricing requires. If you believe the stock earns more, replace r with your drift μ in d2. At an assumed 8 per cent:
d2 (real world) = (0 + 0.08 − 0.0288) ÷ 0.24 = 0.0512 ÷ 0.24 = 0.2133
N(0.2133) = 0.5845, against 0.5186 risk-neutral
The real-world figure is the honest answer to "how likely is it?", but it depends on a drift nobody observes. The risk-neutral figure is the one the market price implies: a digital option paying 1 if the call finishes in the money is worth e−rT × N(d2) = 0.9608 × 0.5186 = 0.4983, or 4,983 on a lot of 10,000.
| Strike | Delta N(d1) | N(d2) risk-neutral | Real world, 8% drift | Put in the money N(−d2) |
|---|---|---|---|---|
| 80 | 0.8881 | 0.8356 | 0.8735 | 0.1644 |
| 90 | 0.7660 | 0.6864 | 0.7429 | 0.3136 |
| 100 | 0.6128 | 0.5186 | 0.5845 | 0.4814 |
| 110 | 0.4560 | 0.3630 | 0.4271 | 0.6370 |
| 120 | 0.3181 | 0.2379 | 0.2924 | 0.7621 |
| 130 | 0.2100 | 0.1477 | 0.1895 | 0.8523 |
Someone selling the 120 call because its delta is about 0.32 and reading that as "a one-in-three chance" has the risk-neutral answer at 23.79 per cent; the real-world answer depends on what they assume. The gap between delta and N(d2) is widest near the money and narrows in the wings.
Volatility pushes the two measures apart. For the one-year at-the-money call, delta stays between 0.61 and 0.63 whether volatility is 16, 24, 32 or 48 per cent, while N(d2) falls from 0.5675 to 0.5186, 0.4860 and 0.4378. The gap grows from 6.2 to 18.9 points: higher volatility drags the median price further below the mean, so a delta-based reading of probability is worst on exactly the volatile names where people most want it.
The probability a call expires in the money is N(d2): 51.86 per cent for Kestrel 100, not the 61.28 its delta suggests, and 58.45 per cent if you assume an 8 per cent drift. Say which measure you are quoting. The book's workbook builds d1 and d2 on their own rows and checks every Greek against a finite difference; it is on the free companion page for this book, beside the worked Black-Scholes price.
No, though traders often use it as a rough guide. Delta is N(d1); the risk-neutral probability is N(d2), which is lower for a call. On a one-year at-the-money call at 24 per cent volatility, delta is 0.6128 and N(d2) is 0.5186, a gap of 9.42 points. The gap widens with volatility and time.
N(-d2), which is 1 minus the call's N(d2) at the same strike. For the at-the-money one-year option at 24 per cent volatility, the put finishes in the money with a risk-neutral probability of 48.14 per cent. At a strike of 120 it is 76.21 per cent.
Because the median of a lognormal price is S x e^((drift - vol squared / 2)T). With a 4 per cent risk-neutral drift and 24 per cent volatility the median after a year is 101.13, above the strike of 100, so the probability is 51.86 per cent. Only a drift of 2.88 per cent would make it exactly 50.
This article is one calculation from Quantitative Finance. The book takes the same case from first principles to the decision, chapter by chapter, and every figure it prints is a live formula in the free companion workbooks.
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