Expected monetary value and preference theory

Quick refresher

EMV $= \sum_j p_j\,\mathrm{NPV}_j$ over outcomes that exclude each other and together cover all the possibilities ( $\sum_j p_j = 1$ ). All the NPVs are at one date, at one rate and in the same kind of money. The risk of failure is included through the probabilities, and the discount rate is not raised for it.
For an exploration well, EMV $= P_g V - C$ : the chance of success times the value of the discovery, minus the cost of the well. On average, drilling pays as long as $P_g > C/V$ . In the example of the lesson the break-even chance of success is 0.13.
EMV is an average, not a forecast: one well gives either $V - C$ or $-C$ . The expected values of independent wells add up, but the chance that all $n$ of them are dry is $(1 - P_g)^n$ .
In a farm-out, a partner pays for the well and gets a share of the project. If both sides assess the chances equally, a farm-out does not create EMV. It only divides the EMV, together with the risk, between the two sides. The point of such a deal lies in the different attitudes of the sides to risk, in different assessments, or in a shortage of money.
A risk-averse company (one that dislikes risk) compares the options by expected utility, that is, by the certainty equivalent CE: a guaranteed sum that is as good as the risky option. CE is smaller than EMV. The difference is the risk premium. It grows with the spread of the outcomes and falls with the risk tolerance $R$ .
A large company with many independent projects can almost ignore the risk of a single well (CE ≈ EMV). For a small company the same well can be a stake whose loss threatens ruin. Then it is wiser to take a smaller share in a larger number of projects.

Key relationships

\mathrm{EMV} = \sum_j p_j\,\mathrm{NPV}_j

the outcomes exclude each other, $\sum_j p_j = 1$

\mathrm{EMV} = P_g V - C

exploration well: a discovery worth $V$ , or a dry well

P_g^{*} = C/V

break-even chance of success

\mathrm{CE} = -R\,\ln \sum_j p_j\,\e^{-x_j/R}

certainty equivalent for an exponential utility function

P(k) = \frac{n!}{k!\,(n-k)!}\,P_g^{k}\,(1 - P_g)^{n-k}

probability of $k$ discoveries from $n$ independent wells

Common pitfalls

Taking EMV for the likely result of one well.
Comparing outcomes valued at different dates, or forgetting the cost of the well in the outcome “success”, or subtracting it twice.
Counting a dry well by its probability and also raising the discount rate for the same risk.
Including costs that have already been spent, for example for a seismic survey that is already shot.
Believing that when EMV > 0 we must always drill, and that a farm-out with a smaller EMV is always a mistake.

After this lesson you can

calculate the EMV of a decision from the NPV and the probabilities of all its outcomes, and check that the outcomes exclude each other, cover all the possibilities and are valued at one date
calculate the EMV of an exploration well, find the break-even chance of success, and compare the options “drill alone”, “farm out” and “walk away”
explain why EMV is an average over many decisions, and estimate the spread of the results of a programme of several wells
describe risk-neutral, risk-averse and risk-seeking attitudes through the utility function, and calculate a certainty equivalent
explain why companies do not always choose the highest EMV: the size of the stake compared with their resources, the risk of ruin, the working interest (the share of a project that the company takes)

Expected monetary value: the average over the outcomes

Many decisions in the oil and gas business are made before we know how they will end. Will the well find oil? Will the accumulation turn out large or small? Will a new technology work? We describe such a decision by a set of outcomes. For each outcome we need two numbers: a probability and a money result, usually the NPV (Net present value (NPV)). EMV combines them into one number.

Expected monetary value

\mathrm{EMV} = \sum_{j=1}^{m} p_j\,\mathrm{NPV}_j, \qquad \sum_{j=1}^{m} p_j = 1

where:

$\mathrm{EMV}$ expected monetary value of the decision, US$ million
$p_j$ probability of the $j$ -th outcome, fraction
$\mathrm{NPV}_j$ net present value of the $j$ -th outcome at the date of the decision (the index is the number of the outcome, not a rate), US$ million
$m$ number of outcomes, —

Units: the money units of the NPV, for example US$ million in constant prices

Holds when

the outcomes exclude each other and together cover all the possibilities, so the probabilities add up to one
all the NPVs are calculated at one date (the date of the decision), at one rate and in the same kind of money
the NPV includes only the future flows that depend on the decision; costs already spent are not included (what belongs to a decision)
the risk of failure is counted by the probabilities; the discount rate is not raised for the same risk (Choosing the discount rate)

EMV is the average value of NPV over all the outcomes, as if the decision were taken many times in the same conditions. By this rule we choose, among several options, the one with the largest EMV. The option “do nothing” has an EMV of zero.

To drill or not to drill: EMV of an exploration well

Take a prospect (a possible drilling target) whose geological chance of success is estimated at 0.30 (how it is estimated). The exploration well costs US$12 million, and this money is spent whatever the outcome. If the well finds an accumulation, the field will be developed as in the library's example project (Net present value (NPV)). The NPV of that project at a rate of 10% is US$100.13 million, counted at the middle of project year 1. That moment comes one year after the decision to drill. So at the date of the decision the discovery is worth $1.1$ times less.

EMV of an exploration well

\mathrm{EMV} = P_g\,(V - C) + (1 - P_g)(-C) = P_g V - C

where:

$P_g$ chance of success of the well, fraction
$V$ value of the discovery at the date of the decision: the NPV of its development, not counting the cost of the exploration well, US$ million
$C$ cost of the exploration well, spent whatever the outcome, US$ million

Units: US$ million at the date of the decision

Holds when

there are two outcomes: a discovery that is developed, and a dry well; the size of the discovery is taken as known (different sizes are covered in the section below)
$V$ and $C$ refer to one date; if the moment to which the NPV of the project is brought comes $t$ years after the decision, the NPV is multiplied by $(1 + i)^{-t}$ (moving the time origin)
the tax effects of the cost of the well are not counted (or $C$ is already taken after tax)

The two ways of writing give the same result. The first lists the outcomes: success brings $V - C$ , and failure brings a loss of $C$ . The second is shorter: the company pays $C$ for certain and receives $V$ with probability $P_g$ . When we list the outcomes, it is easy to forget $C$ in the outcome “success”, or to subtract it twice. The check is that both ways must give the same number.

Worked exampleEMV of an exploration well

Given (assumed data)

Chance of success $P_g$: 0.30
Cost of the well $C$: 12 million dollars
NPV of the development at 10%: 100.13 million dollars at the middle of project year 1
From the decision to the middle of project year 1: 1 year

Solution

Value of the discovery at the date of the decision: $V = 100.13/1.1 = 91.03$ million dollars.
NPV of the outcomes: discovery, $V - C = 91.03 - 12 = 79.03$ million dollars; dry well, $-12$ million dollars.
Contributions of the outcomes: $0.30\cdot79.03 = 23.71$ and $0.70\cdot(-12) = -8.40$ . EMV $= 15.31$ million dollars.
Check with the short form: $0.30\cdot91.03 - 12 = 15.31$ million dollars.
Break-even chance: $P_g^{*} = 12/91.03 = 0.132$ .

Answer: EMV ≈ 15.31 million dollars > 0: by the EMV criterion, drilling is worth it. On average, drilling pays when the chance of success is above 0.132.

What it means. On average the well returns 15.31 million dollars on top of the 10% return, but this is the average over many such wells. One particular well gives either 79.03 million dollars or a loss of 12 million, and in this example the second is 2.3 times more likely than the first. The margin in probability is large: the estimate of $P_g$ may be more than twice too high, and drilling will still pay on average. The data in this example are assumed for illustration.

Break-even chance of success

P_g^{*} = \frac{C}{V}

where:

$P_g^{*}$ chance of success at which the EMV of drilling is zero, fraction
$C,\ V$ cost of the well and value of the discovery at the date of the decision, US$ million

Units: dimensionless

Holds when

the same as for the EMV of a well: two outcomes, with $V$ and $C$ at one date.

The more expensive the well is compared with the value of the discovery, the more sure we must be of success. The ratio $V/C$ shows how many times the prize is larger than the stake. It tells at what chances drilling still pays on average: at $V/C = 10$ , a chance $P_g > 0.1$ is enough.

CalculatedEMV for the company against the chance of success: drill alone, farm out a half share to a partner who pays for the well, or walk away (EMV = 0). The break-even chance for drilling alone is 0.13. At 0.26 both ways of taking part give the same EMV.

Drill alone, farm out or walk away

The owner of an area does not have to carry the whole risk alone. A common option is a farm-out: a partner pays for the whole exploration well and for this gets a share $w$ of the project. If the well turns out dry, the owner loses nothing. If it finds an accumulation, the owner keeps the share $1 - w$ and takes part in the development in proportion to it.

EMV of a farm-out to a partner who pays for the well

\mathrm{EMV}_w = P_g\,(1 - w)\,V, \qquad P_{g,w} = \frac{C}{w V}

where:

$\mathrm{EMV}_w$ EMV of the owner of the area after giving away the share $w$ , US$ million
$w$ share of the project given to the partner, fraction
$P_{g,w}$ chance of success at which drilling alone and a farm-out give the owner the same EMV, fraction
$P_g,\ V,\ C$ chance of success, value of the discovery and cost of the well, as above, —, US$ million

Units: US$ million at the date of the decision

Holds when

the partner pays for the whole well, and the sides carry the development costs in proportion to their shares
both sides assess $P_g$ and $V$ in the same way

The difference between the EMV of drilling alone and $\mathrm{EMV}_w$ is $P_g w V - C$ , which is exactly the EMV of the partner. Above the chance $P_{g,w}$ , drilling alone gives the higher EMV. Below it, a farm-out does. But below $P_{g,w}$ the EMV of the partner is negative, and with the same assessments the partner will not accept the deal.

Worked exampleHalf of the share for paying for the well

Given (assumed data)

Prospect: as in the example above: $P_g = 0.30$ , $V = 91.03$ million dollars, $C = 12$ million dollars
Terms of the deal: the partner pays for the well and receives $w = 0.5$ of the project

Solution

Owner: $0.30\cdot0.5\cdot91.03 = 13.65$ million dollars, against 15.31 million when drilling alone.
Partner: $0.30\cdot0.5\cdot91.03 - 12 = 1.65$ million dollars. This is exactly the EMV that the owner gave away: $15.31 - 13.65 \approx 1.65$ .
The owner is indifferent (and the partner's EMV is zero) at $P_{g,w} = 12/(0.5\cdot91.03) = 0.264$ .

Answer: By EMV, the owner does better by drilling alone: 15.31 million dollars against 13.65. The deal passes 1.65 million dollars of EMV to the partner, and with it the whole risk of a dry well.

What it means. The deal does not create value: with equal assessments, whatever EMV the partner gets, the owner loses. Still, such deals are made all the time. The reasons are different assessments of the chances, the partner being able to drill more cheaply, the owner's shortage of money, and above all the different attitudes to risk: for a small company a loss of 12 million dollars can mean much more than for a large one. How to account for this in numbers is shown in the section on the attitude to risk. The data in this example are assumed for illustration.

More than two outcomes: how large the discovery will be

A discovery is not one number: the accumulation may turn out smaller or larger than expected. If it is too small, the development will not pay and will not be started. For the economics, such a discovery is the same as a failure, although geologically the well is a success. The limit is set by the minimum economic field size (MEFS): the volume at which the NPV of the development is zero. The probability of finding an accumulation not smaller than that is the commercial chance of success $P_c$ . It is not larger than $P_g$ (from geological to commercial success).

Worked exampleA dry well and three sizes of discovery

Given (assumed data)

Geological chance of success: 0.30; cost of the well 12 million dollars
Sizes of the discovery (as a fraction of the base production of the project): ×0.5 with probability 0.3; ×1 with 0.5; ×1.5 with 0.2 (if the well succeeds)
Development model: the library's example project: the production of each year is multiplied by the fraction, the costs stay the same

Solution

NPV of the development at the middle of project year 1: −26.40, 100.13 and 226.51 million dollars. NPV becomes zero at a production of 0.60 of the base, that is about 7.2 million barrels instead of 12. This is the minimum economic field size. The small discovery is below it, and it is not developed.
Outcomes at the date of the decision: the dry well ( $p = 0.70$ ) and the small discovery ( $p = 0.09$ ) each give $-12$ million dollars. The medium one ( $p = 0.15$ ) gives $100.13/1.1 - 12 = 79.03$ . The large one ( $p = 0.06$ ) gives $226.51/1.1 - 12 = 193.92$ million dollars.
EMV $= 0.70\cdot(-12) + 0.09\cdot(-12) + 0.15\cdot79.03 + 0.06\cdot193.92$ $= 14.01$ million dollars.
Commercial chance of success: $P_c = 0.30\cdot(0.5 + 0.2) = 0.21$ .

Answer: EMV ≈ 14.01 million dollars, with a commercial chance of success of 0.21 and a geological one of 0.30.

What it means. The four outcomes exclude each other, and their probabilities add up to one. This is the first thing to check in such a calculation. The small discovery is not developed: the company loses only the well, and does not add the negative NPV of a development to the loss. If it were developed “since we found it”, EMV would be 2.16 million dollars lower. For the same reason we cannot replace the distribution of sizes by one average discovery. The average size ×0.95 would give an EMV of 11.86 million dollars, which is less than the correct sum over all the outcomes. The data in this example are assumed for illustration. In this simplified model, the size of the discovery changes only the production. In reality the costs depend on it too: a small discovery would be developed more cheaply, and a large one at a higher cost.

EMV is an average over many wells

The meaning of EMV shows itself over many decisions. If we drill many similar wells, the share of discoveries comes close to the chance of success, and the average result per well comes close to EMV. But a company drills not hundreds of wells but a few, and then not only the average matters. It also matters how far the result can differ from the average. The number of discoveries among $n$ independent wells with the same chance of success follows the binomial distribution.

Probability of

k

discoveries from

n

independent wells

P(k) = \frac{n!}{k!\,(n-k)!}\;P_g^{k}\,(1 - P_g)^{n-k}, \qquad k = 0, 1, \ldots, n

where:

$P(k)$ probability that exactly $k$ of the $n$ wells are successful, fraction
$n$ number of wells in the programme, —
$k$ number of discoveries, —
$P_g$ chance of success of one well, the same for all, fraction
$\frac{n!}{k!\,(n-k)!}$ the number of ways to choose $k$ successful wells out of $n$ ( $n! = 1\cdot2\cdots n$ , $0! = 1$ ), —

Units: dimensionless

Holds when

the outcomes of the wells are independent: the prospects share no play risk (a play is a group of prospects in the same geological setting: if a key ingredient is missing there, all of them fail)
the chance of success is the same for all wells; with different chances the distribution is found by going through the outcomes
the whole programme is drilled without revising it after the first results

The mean number of discoveries is $nP_g$ , and the EMV of the programme is $n$ times the EMV of one well. The chance that all the wells are dry is $P(0) = (1 - P_g)^n$ . It falls quickly as $n$ grows, but for a small programme it stays noticeable.

Worked exampleA programme of five wells

Given (assumed data)

Wells in the programme: 5 similar prospects in different plays
Each well: $P_g = 0.30$ , $C = 12$ million dollars, the discovery is worth $V = 91.03$ million dollars

Solution

Cost of the programme: $5\cdot12 = 60$ million dollars. NPV of the programme with $k$ discoveries: $k\cdot91.03 - 60$ .
No discoveries: $P(0) = 0.7^5 = 0.168$ , NPV $= -60$ million dollars.
One discovery: $P(1) = 5\cdot0.3\cdot0.7^4 = 0.360$ , NPV $= 31.0$ . Two: $P(2) = 10\cdot0.3^2\cdot0.7^3 = 0.309$ , NPV $= 122.1$ million dollars.
Three discoveries or more: 0.163 in total.
EMV of the programme: $\sum_k P(k)\,\mathrm{NPV}_k = 76.55$ million dollars, exactly $5\cdot15.31$ . The standard deviation of the result (a measure of its spread) is 93.3 million dollars.

Answer: The EMV of the programme is 76.55 million dollars, but with a probability of 17% the company loses all 60 million, and with a probability of 53% it gets no more than 31.0 million dollars.

What it means. The average result of the programme is good, but the spread around it is larger than the EMV itself. In more than half of the cases the programme brings less than half of its expected value, and in one case out of six it brings only a loss. For a company that has no other projects, such a programme is a serious stake. If all five prospects belonged to one untested play, EMV would stay the same, but the probability of a complete failure would be higher (dependent prospects). The data in this example are assumed for illustration.

CalculatedPossible results of a programme of 5 independent wells: the NPV with 0, 1 … 5 discoveries and their probabilities. The vertical line is the EMV of the programme (US$76.55 million), which is not equal to any outcome.

Attitude to risk: the utility function and the certainty equivalent

Compare two options for the same company: to receive a guaranteed US$15 million, or to drill the well of the example, whose EMV is US$15.3 million. By EMV they are almost equal. But with the second option, in seven cases out of ten the company loses US$12 million. For a small company this can mean a reduced work programme or a missed debt payment. The EMV criterion does not see this difference: it treats all the options with the same average in the same way.

Risk-neutral attitude: Options are compared only by the average, that is, by EMV. This is reasonable when each single stake is small compared with the resources and there are many stakes.
Risk aversion: A guaranteed sum is valued more than a risky option with the same average. A company behaves like this when it sells oil at a fixed price below the expected one (the example with the forward sale). A customer does the same when paying a contractor more for a fixed “turnkey” price (one agreed sum for the whole job).
Risk seeking: A risky option is valued more than a guaranteed one with the same average, for the sake of a chance of a big win. This is rare in company decisions, but it happens, for example when a company needs a big discovery at any price.

The attitude to risk is described in numbers by the utility function $u(x)$ . It gives each sum of money its “worth” for the decision-maker. We compare options by the expected utility $\sum_j p_j\,u(x_j)$ . For a risk-averse company the utility curve flattens as it rises: each next million adds less utility than the previous one, and losing a million takes away more than winning the same million brings.

It helps to turn the expected utility back into money. The certainty equivalent (CE) is the guaranteed sum whose utility equals the expected utility of the risky option: the company does not mind whether it gets this sum for certain or takes the risk. For a risk-averse company, CE is smaller than EMV. The difference $\mathrm{EMV} - \mathrm{CE}$ is called the risk premium. It is the part of the average value that the company is ready to give up for certainty. It is a sum of money, not an addition to the discount rate (the premium on the rate and the certainty equivalent).

Exponential utility function and the certainty equivalent

u(x) = 1 - \e^{-x/R}, \qquad \mathrm{CE} = -R\,\ln \sum_{j} p_j\,\e^{-x_j/R}

where:

$u(x)$ utility of the sum $x$ (dimensionless; the scale of the value does not matter), —
$x,\ x_j$ money result, the NPV of the $j$ -th outcome, US$ million
$R$ risk tolerance is a parameter of the company: the larger it is, the closer the behaviour is to risk-neutral, US$ million
$p_j$ probability of the $j$ -th outcome, fraction
$\mathrm{CE}$ certainty equivalent of the option, US$ million
$\ln$ natural logarithm, —

Units: $x$ , $R$ and CE are in the same money units

Holds when

the attitude to risk is the same at any wealth of the company: adding the same guaranteed sum to every outcome adds that sum to CE too
$R$ is set for the company as a whole and is revised together with its size and financial position
when $R$ is much larger than the spread of the outcomes, the function is almost a straight line on this interval, and CE almost equals EMV

The second formula comes from the condition $u(\mathrm{CE}) = \sum_j p_j\,u(x_j)$ . The exponential function is often used in the valuation of exploration projects: it is set by one number $R$ , and it is easy to explain. Here is what $R$ means in practice. The bet “win $R$ or lose $R/2$ with equal chances” leaves such a company almost indifferent. It would accept a slightly larger win and reject a smaller one.

Worked exampleOne well through the eyes of a small and a large company

Given (assumed data)

Options: drill alone (+79.0 or −12 million dollars) or farm out a half share (+45.5 or 0)
Chance of success: 0.30
Risk tolerance: small company, $R = 30$ million dollars; large company, $R = 1000$ million dollars

Solution

Small company, drilling alone: the expected utility is $\mathrm{EU} = 0.3\cdot u(79.0) + 0.7\cdot u(-12) = -0.0658$ . This is less than the utility of a guaranteed zero. From $u(\mathrm{CE}) = \mathrm{EU}$ we get $\mathrm{CE} = -R\ln(1 - \mathrm{EU})$ $= -30\cdot\ln 1.0658 = -1.91$ million dollars.
The same company, with a farm-out: $\mathrm{CE} = 8.01$ million dollars, with an EMV of 13.65.
Large company: drilling alone has a CE of 14.45 million dollars, and a farm-out has 13.44 million dollars.
A large partner that takes the share (+33.5 or −12 million dollars): EMV 1.65, CE 1.44 million dollars.
The risk tolerance at which drilling alone stops being acceptable (CE = 0): $R \approx 38$ million dollars.

Answer: For the small company, drilling alone is unacceptable (CE −1.91 million dollars), and a farm-out is attractive (CE 8.01). The large company will prefer to drill alone (CE 14.45 against 13.44).

What it means. One and the same prospect, with the same EMV of 15.31 million dollars, can sensibly lead to different decisions. For the small company, the risk premium of drilling alone is 17.2 million dollars, more than the whole EMV. For the large one it is about 0.9 million, and it behaves almost as a risk-neutral company. The deal from the previous section, which is not attractive by EMV, is attractive to both sides by the certainty equivalent: the risk passes to the one for whom it costs less. The data in this example are assumed for illustration.

CalculatedExponential utility function of a small company (

R = 30

million dollars) and drilling the well alone: the two outcomes on the curve, the expected utility on the straight line between them at an EMV of US$15.3 million, and the certainty equivalent of US$−1.9 million. The distance between them is the risk premium.

Why companies do not simply maximise EMV

The EMV criterion is right for someone who makes many independent bets, each small compared with their resources. Then random wins and losses cancel each other out, and the total is close to the sum of the EMVs. A real company departs from this ideal for several reasons.

The size of the stake. The same well is a small stake for a company with an annual budget of billions of dollars. For a company that has set aside a few tens of millions for exploration, it is an all-in stake.
Ruin. The money runs out before the law of large numbers has time to work. A series of failures can stop the programme before the first discovery, even if every well has a positive EMV. In probability theory this is called gambler's ruin.
Decisions that do not repeat. A large development project or the purchase of a company happens once in several years, and “the average over many attempts” is an abstraction for it.
The interests of those who provide the money. Lenders and shareholders treat the company's risk differently, and the terms of loan agreements and budget limits themselves restrict the loss that is allowed.

Worked exampleThree whole wells or nine one-third shares

Given (assumed data)

Exploration budget: 36 million dollars: enough for three wells of 12 million each
Prospects: as in the example above: $P_g = 0.30$ ; the prospects are from different plays, and the outcomes are independent
Option B: instead of three whole wells, a third of the share in nine wells (partners take the rest)

Solution

Option A: all three wells are dry with a probability of $0.7^3 = 0.343$ . EMV: $3\cdot15.31 = 45.9$ million dollars.
Option B: the money at risk is the same, 36 million. No discoveries at all: $0.7^9 = 0.040$ . EMV: $9\cdot15.31/3 = 45.9$ million dollars.

Answer: With the same EMV of 45.9 million dollars, the probability of being left with no discoveries falls from 34 to 4%.

What it means. Splitting the stakes does not change the average, but it sharply reduces the probability of a catastrophe. That is why exploration projects are usually run by consortia: each company takes a share in several prospects and not everything in one. Partnership, insurance and other ways to reduce risk are covered in the topic Risk reduction and diversification. The data in this example are assumed for illustration.

Here is what this means in practice. Large companies with a portfolio of many independent projects take EMV as the main criterion: for them the risk premium of a single well is small. The utility function and the certainty equivalent are used for stakes comparable with the size of the company, and for choosing the working interest. For the same reason, small companies more often take shares and bring in partners, even if they give away part of the EMV.

Terms and symbols

Outcome: One of the consequences of a decision that exclude each other. Each has its own probability and its own NPV.
EMV: The average NPV of the outcomes, weighted by their probabilities (expected monetary value).
Break-even chance $P_g^{*}$: The chance of success at which the EMV of drilling is zero. For two outcomes it is $C/V$ .
Farm-out: A deal in which a partner pays for the work (for example, an exploration well) and gets a share of the project for it.
Utility function $u(x)$: The worth of a sum of money for the decision-maker. For a risk-averse decision-maker the curve flattens as it rises.
Certainty equivalent (CE): The guaranteed sum that is as good as the risky option by expected utility.
Risk premium: The difference $\mathrm{EMV} - \mathrm{CE}$ : the part of the average value that is given up for certainty.
Risk tolerance $R$: A parameter of the exponential utility function. The larger it is, the closer the behaviour is to risk-neutral.
Gambler's ruin: The loss of all the money through a series of failures before the positive average has time to show itself.

Common mistakes and misconceptions

Wrong

“The EMV of the well is US$15 million, so it will bring about US$15 million.”

In fact

The well will bring either US$79 million or a loss of US$12 million. EMV is the average over many similar wells. For one well it only shows whether such a decision pays on average.

Wrong

“The chance of a dry well is already in EMV, but to be safe we will also discount at 20% instead of 10%.”

In fact

This counts the same risk twice. The rate reflects the cost of money and the general market risk. The probability of failure is counted in the outcomes, and the attitude to the size of the loss is counted through utility.

Wrong

“The NPV of the successful outcome is the NPV of the development, and that of the dry well is minus the well.”

In fact

The well is paid for in every outcome: the NPV of success is the value of the discovery minus the cost of the well, and both quantities are at the date of the decision. The check: $P_g(V - C) - (1 - P_g)C$ must equal $P_g V - C$ .

Wrong

“EMV is positive, so we always drill; a farm-out with a smaller EMV is a mistake.”

In fact

For a small company the same well can have a negative certainty equivalent. In the example of the lesson, at $R = 30$ million dollars, drilling alone gives a CE of −1.9 million, and a farm-out of a half share gives 8.0 million.

Wrong

“US$5 million has already been spent on the seismic survey, so it is included in the EMV of drilling.”

In fact

Costs that are already spent do not depend on the decision and are not part of it. We calculate the EMV of drilling from the future flows, from the cost of the well to the income of the development.

In depth: advanced detail

Certainty equivalent: an approximation and its limits

If the spread of the outcomes is small compared with $R$ , we can expand the exponential in a series (a sum of powers) up to the quadratic term: $\e^{-x/R} \approx 1 - x/R + x^2/(2R^2)$ . Putting this into the formula for CE gives a simple approximation.

Certainty equivalent through the mean and the variance

\mathrm{CE} \approx \mathrm{EMV} - \frac{\sigma^2}{2R}, \qquad \sigma^2 = \sum_j p_j\,(x_j - \mathrm{EMV})^2

where:

$\sigma^2$ variance of the NPV over the outcomes (a measure of the spread); $\sigma$ is the standard deviation, (US$ million)²
$R$ risk tolerance, US$ million
$\mathrm{EMV},\ \mathrm{CE}$ expected value and certainty equivalent, US$ million

Units: US$ million

Holds when

exponential utility function
the spread of the outcomes is small compared with $R$ ( $\sigma \ll R$ ); for a normally distributed NPV the formula is exact at any spread

The risk premium is proportional to the variance and inversely proportional to the risk tolerance: doubling the spread of the outcomes makes the premium four times larger, and doubling $R$ halves it. For two outcomes $\sigma^2 = P_g(1 - P_g)V^2$ : the spread is set by the value of the discovery, not by the cost of the well.

Worked exampleWhen the approximation works and when it does not

Given (assumed data)

Well: EMV 15.31 million dollars, $\sigma = 41.7$ million dollars
Risk tolerance: 1000 and 30 million dollars

Solution

$R = 1000$ : the approximation is $15.31 - 1740/(2\cdot1000) = 14.44$ million dollars; the exact CE is 14.45.
$R = 30$ : the approximation gives −13.7 million dollars, and the exact CE is −1.9.

Answer: At

R = 1000

million dollars the approximation is almost exact; at

R = 30

it is wrong by 12 million dollars.

What it means. The standard deviation of the outcomes of the well is larger than the risk tolerance of the small company, and the expansion in a series stops working. Besides, the distribution of the outcomes is strongly skewed (lopsided): a rare large win and a frequent small loss. The variance punishes a win in the same way as a loss, but the exponential utility does not, so the approximation overstates the risk premium. For the exploration wells of small companies we calculate CE by the exact formula. The data in this example are assumed for illustration.

What share of a project to take

If a company can choose its working interest $\alpha$ (it pays the share $\alpha$ of the cost of the well and receives the same share of the value of the discovery), its EMV grows in proportion to $\alpha$ . The risk premium, by the approximate formula, grows in proportion to $\alpha^2$ . So with a small share the risk costs almost nothing, and with a large share it can outweigh the gain. The certainty equivalent then has its highest value at some share smaller than one.

CalculatedCertainty equivalent against the working interest in the well, for a small (

R = 30

million dollars) and a large (

R = 1000

million dollars) company. The dashed line is the EMV of the share. For the small company, CE is highest at a share of about 34%.

Connections to other topics

Net present value (NPV) — the NPV of the outcomes: the value of a discovery and the library's example project.
Geological risk and chance of success — where the chance of success comes from, and why the prospects of one play depend on each other.
Sensitivity analysis and scenarios — what belongs to a decision, the payoff matrix and the distribution of the project's NPV.
Choosing the discount rate — why the risk of a dry well is not allowed for by a premium on the rate.
Probabilistic volumetrics by Monte Carlo simulation — the distribution of the volumes if the well succeeds, from which the outcomes of a discovery are estimated.
Decision trees and the value of information — sequential decisions and the value of extra information.
Risk reduction and diversification — partnership, farm-out, insurance and the portfolio as ways to reduce risk.

Check yourself

1
The chance of success of a well is 0.2, the value of the discovery at the date of the decision is 60 million dollars, and the cost of the well is 9 million dollars. What is the EMV of drilling?
1. A.12 million dollars
2. B.3 million dollars
3. C.10.2 million dollars
4. D.−9 million dollars
Show the answer
Correct answer: B. 3 million dollars
$\mathrm{EMV} = 0.2\cdot60 - 9 = 3$ million dollars. By outcomes: success gives $0.2\cdot(60 - 9) = 10.2$ , failure gives $0.8\cdot(-9) = -7.2$ , and the sum is 3. The answer 12 forgets the cost of the well, and 10.2 forgets the dry well.
2
With a chance of success of 0.25, a discovery is worth 80 million dollars at the date of the decision. Up to what cost of the well does drilling keep a non-negative EMV?
1. A.60 million dollars
2. B.20 million dollars
3. C.80 million dollars
4. D.16 million dollars
Show the answer
Correct answer: B. 20 million dollars
EMV $= 0.25\cdot80 - C \ge 0$ , so $C \le 20$ million dollars. The same limit from the other side: at $C = 20$ the break-even chance is $C/V = 20/80 = 0.25$ .
3
The EMV of a company drilling alone is larger than its EMV with a farm-out to a partner. Can the farm-out still be a sensible decision?
Show the answer
Yes. With equal assessments the deal only moves EMV and risk to the partner, but for a small company a possible loss means more than for a large one. Its certainty equivalent for drilling alone may be smaller than for the farm-out, and even negative. Besides, the sides may have different assessments of the chances and the costs, and the owner may be short of money. We take the decision with the attitude to risk in mind, not by EMV alone.
4
Four independent wells each have a chance of success of 0.25. What is the probability that all of them are dry?
1. A.0.25
2. B.0
3. C.0.004
4. D.0.32
Show the answer
Correct answer: D. 0.32
$(1 - 0.25)^4 = 0.75^4 \approx 0.32$ . Almost a third of the programmes of four such wells give no discovery at all, although the expected number of discoveries is one. The answer 0.004 is the probability that all four succeed: $0.25^4$ .
5
Two options have the same EMV of 10 million dollars. A is a guaranteed 10 million. B is +60 million or −40 million with equal chances. Which will a company with an exponential utility function and $R = 50$ million dollars choose? Estimate the certainty equivalent of option B.
Show the answer
$\mathrm{CE}_\text{B} = -50\ln(0.5\,\e^{-60/50} + 0.5\,\e^{40/50})$ $= -50\ln(0.151 + 1.113) \approx -50\cdot0.234 \approx -11.7$ million dollars. The certainty equivalent is negative: a possible loss of 40 million weighs more for such a company than an equally likely win of 60 million. It will choose option A. The risk premium of option B is about 21.7 million dollars.
6
What is true for a programme of 10 independent wells, each with an EMV of 2 million dollars?
1. A.The EMV of the programme is 20 million dollars, but the actual result can differ greatly from this sum
2. B.The EMV of the programme equals the EMV of one well
3. C.The programme will almost surely bring about 20 million dollars
4. D.The EMV of the programme is less than 20 million dollars because of risk
Show the answer
Correct answer: A. The EMV of the programme is 20 million dollars, but the actual result can differ greatly from this sum
Expected values add up: $10\cdot2 = 20$ million dollars. But the result of the programme is the number of discoveries times the value of a discovery, minus the costs, and with a small number of wells it is widely spread around the average. A complete failure is also possible. Risk reduces not the EMV but the certainty equivalent.

Updated: 5 October 2026