Finance-Theory · Unit 20 · Video 6 · Interactive Practice

Two Risks, Two Rates: Why Oil Exploration Risk Earns Only the Risk-Free Rate

IKey Formulas

Formula	Step	Rate used
$V_1 = \dfrac{CF_2}{1 + r_{\text{prod}}}$	Value at $t = 1$ if the exploration succeeds	Production rate (high $\beta$ )
$E[V_1] = p\,V_1 + (1 - p)\cdot 0$	Expected value over the exploration outcome	None: probabilities, not discounting
$\text{NPV} = \dfrac{p\,CF_2}{(1 + r_{\text{prod}})(1 + R_f)} - C_0$	Two-phase NPV, drilling cost $C_0$ at $t = 0$	$R_f$ for exploration ( $\beta = 0$ )
$r_t = R_f + \beta_t\,(E[R_m] - R_f)$	Each phase priced on its own beta	CAPM, phase by phase

Key Insight: A cash flow's discount rate matches the systematic risk of that cash flow. Dry-hole risk is idiosyncratic ( $\beta = 0$ ), so the year it resolves is discounted at $R_f$ ; oil-price risk moves with the market, so production earns the industry rate.

IIVisualization 1 — Working Backward Through Two Phases

Value moves back from $t = 2$ in three moves: discount at the production rate, take the expectation, discount at $R_f$ .

Success probability p Payoff at t = 2 80 Step 1

IIIVisualization 2 — Which Rate for the Exploration Year?

Only the exploration year's rate changes here; its zero beta is what pins that rate at $R_f = 3\%$ .

Year-1 rate r₁ (%) 15

IVVisualization 3 — Why Dry-Hole Risk Is Not Priced

Across many independent fields, the fraction that strikes oil closes in on $p = 0.25$ .

Fields drilled N 1

💡 Diversified investors get this averaging by holding many firms' projects, so even a single-field firm's dry-hole risk earns no premium. The oil-price risk is common to every field and never averages out, which is why production keeps its 15% rate.

VQuiz Questions

Problem 1 · Two-Rate NPV

Given: A mining firm explores for one year; with probability $p = 0.4$ it finds ore, which is mined and sold during year 2 for an expected after-tax cash flow of $69M at $t = 2$ (nothing if the search fails). Exploration risk has $\beta = 0$ , $R_f = 3\%$ , and the mining-production rate is $15\%$ . Ignoring the exploration cost, find the project's value at $t = 0$ .

✅ Correct! Production cash at

15\%

, then the expectation, then the zero-beta year at

3\%

❌ Not quite. That discounts the exploration year at

15\%

too. The success-or-failure gamble has

\beta = 0

, so year 1 earns only

R_f

❌ Not quite. That discounts the year-2 cash flow at

R_f

as well. Once ore is found, the cash carries metal-price risk, which moves with the market and is priced at

15\%

❌ Not quite. That values the project as if the ore were certain to be found. Weight the success value by

p = 0.4

; the failure branch is worth $0.

❌ Not quite. Work backward: discount at

15\%

, weight by

p

, discount at

3\%

Show solution

Step 1 (production year, $15\%$ ): value at $t = 1$ if ore is found:

V_1 = \frac{69}{1.15} = 60 \quad (\text{in millions})

Step 2 (exploration outcome): expected value at $t = 1$ :

E[V_1] = 0.4 \times 60 + 0.6 \times 0 = 24

Step 3 (exploration year, $\beta = 0$ , so $R_f = 3\%$ ):

\text{NPV} = \frac{24}{1.03} \approx 23.30

The project is worth about $23.30M before the exploration cost, which would be subtracted from this figure.

Problem 2 · The One-Rate Mistake

Given: A field succeeds with probability $p = 0.25$ at $t = 1$ ; success brings production cash of $103.5M at $t = 2$ . Exploration risk has $\beta = 0$ , $R_f = 3\%$ , production rate $15\%$ . An analyst discounts both years at $15\%$ . By how much does she understate the project's value?

✅ Correct! Same cash flows, same risks; only the rate applied to the zero-beta year differs, and that costs about $2.28M.

❌ Not quite. The risks are the same, but the analyst prices the exploration gamble as if it had production's high beta. A higher year-1 rate gives a lower value.

❌ Close, but… That multiplies

E[V_1]

by the

12

-point rate gap. The two values are

E[V_1]/1.03

and

E[V_1]/1.15

; take their difference.

❌ Not quite. That compares the two discountings of the success value without weighting it by

p = 0.25

❌ Not quite. Compute both values from the same

E[V_1]

and subtract.

Show solution

Both methods share the first two steps (in millions):

V_1 = \frac{103.5}{1.15} = 90, \qquad E[V_1] = 0.25 \times 90 = 22.5

Two rates (exploration at $R_f$ ): $\dfrac{22.5}{1.03} \approx 21.845$ .

One rate ( $15\%$ in year 1 too): $\dfrac{22.5}{1.15} \approx 19.565$ .

21.845 - 19.565 \approx 2.28

The single rate understates the value by about $2.28M, because it charges a market-risk premium for a risk that diversified investors do not bear.

Problem 3 · Break-Even Success Probability

Given: Exploration costs $10M, paid at $t = 0$ . Success at $t = 1$ leads to production cash of $96M at $t = 2$ ; failure yields nothing. Exploration risk has $\beta = 0$ , $R_f = 3\%$ , production rate $15\%$ . Find the smallest success probability $p$ at which the project's NPV is not negative.

✅ Correct! At

p \approx 12.3\%

the expected, twice-discounted payoff just covers the $10M exploration cost.

❌ Not quite. That compares $10M today with $96M two years out without discounting. Both years must be discounted before comparing.

❌ Not quite. That discounts the exploration year at

15\%

. With

\beta = 0

for exploration, year 1 uses

R_f = 3\%

, which lowers the break-even probability.

❌ Close, but… That leaves out the exploration year's discounting. The $10M is paid at

t = 0

, so

E[V_1]

must still come back one year at

3\%

❌ Not quite. Set

\text{NPV} = 0

and solve for

p

Show solution

In millions, the NPV is

\text{NPV} = -10 + \frac{p \times 96}{(1.15)(1.03)}

Set it to zero and solve for $p$ :

p^{*} = \frac{10 \times 1.15 \times 1.03}{96} = \frac{11.845}{96} \approx 0.123

Check: at $p^{*}$ , $E[V_1] = p^{*} \times \dfrac{96}{1.15} = \dfrac{11.845}{1.15} = 10.3$ , and $10.3/1.03 = 10$ , exactly the cost. ✓

The project is worth undertaking whenever the success probability is at least about 12.3%.

Problem 4 · A One-Drug Biotech

Given: A biotech with a single drug learns at $t = 1$ whether its trial succeeds ( $p = 0.4$ ); success, a matter of biology, is unrelated to the market. Its shareholders hold well-diversified portfolios. If approved, sales cash flows (which move with the economy) arrive at $t = 2$ and are priced at $15\%$ ; $R_f = 3\%$ . The CFO says: "We have only one drug, so we can't diversify the trial risk; year 1 needs at least $15\%$ ." Which rate belongs on the trial year?

✅ Correct! Systematic risk sets the price, and it is set by the firm's investors, who hold many such bets; the firm itself does not need to be diversified. This rests on the trial outcome being unrelated to the market (

\beta = 0

). Were the trial risk systematic, year 1 would take a rate matching its own beta.

❌ Not quite. Being large is not the same as being priced. The market pays only for risk that cannot be diversified away, and trial outcomes are unrelated to the market.

❌ Not quite. The diversification that matters happens in the shareholders' portfolios, not inside the firm. A risk unrelated to the market (

\beta = 0

) earns

R_f

no matter how concentrated the firm is.

❌ Not quite. Probabilities enter through the expected value

E[V_1]

, not through the discount rate. Each phase's rate comes from its own beta.

❌ Not quite. Ask what beta the trial outcome has.

Show solution

The rate for a cash flow comes from that cash flow's systematic risk: $r_1 = R_f + \beta_{\text{trial}}(E[R_m] - R_f)$ .

Trial success depends on biology, not on whether the market is up or down, so $\beta_{\text{trial}} = 0$ and $r_1 = R_f = 3\%$ . The firm has only one drug, but its shareholders hold many firms with many independent trials. Across their portfolios the law of large numbers averages trial outcomes out, so they demand no premium for bearing them.

The sales cash flows at $t = 2$ move with the economy, so they keep the $15\%$ rate. The probability $p = 0.4$ enters only through the expectation $E[V_1] = 0.4\,V_1$ .

The answer hinges on that zero beta. If the trial outcome moved with the market, the trial year would be discounted at a rate matching its own beta, $R_f + \beta_{\text{trial}}(E[R_m] - R_f)$ , and there is no reason that rate would equal the sales rate.

As in the oil example, two risks call for two rates: $R_f = 3\%$ on the trial year and $15\%$ on the sales year.

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