Decision trees and the value of information

Quick refresher

We build the tree in the order in which we take decisions and learn outcomes: a decision may depend only on what is already known at its moment. A square is a decision and a circle is a chance event. The end of a path is a money result: the sum of all the flows of the path at the date of the first decision. The probabilities of the branches of a chance node add up to one, and they depend on the path that led to the node.
We roll the tree back from right to left. A chance node gets the EMV of its branches, and a decision node gets the value of its best branch. The result is the EMV of the strategy and the strategy itself.
A sequential decision is worth more than a simultaneous one when the result of the first step can change the next. In the example of the lesson, to drill the large prospect first and to decide on the second one after its result is worth $18.47 million, against $16.83 million when both are drilled at once.
The expected value of perfect information (EVPI) is the increase of EMV if the outcome became known in advance. It is the upper limit of the price of any data about it. For a drilling decision it equals the smaller of two numbers: the expected cost of the dry wells $(1 - P_g)\,C$ and the expected gain from the discoveries $P_g\,(V - C)$ . In the example it is $8.40 million.
We turn imperfect data into probabilities by Bayes' theorem. For this we need two numbers: the detection rate $P(+\mid H)$ (how often the survey says “yes” when there is an accumulation) and the false-alarm rate $P(+\mid \bar H)$ (how often it says “yes” when there is none). The survey of the example (0.85 and 0.30) turns a probability of 0.30 into 0.55 or 0.084, and it is worth no more than $2.32 million.
Information is worth buying if two things hold. At least one of its results must change the decision. And its value, after allowing for the time needed to get it, must be larger than its price. Information is worth the most near break-even: when the decision is obvious anyway, the data will not change it.

Key relationships

x_\text{chance} = \sum_j p_j\,(c_j + x_j)

value of a chance node: $c_j$ is the cash flow on the branch, $x_j$ is the value of the next node

x_\text{decision} = \max_k\,(c_k + x_k)

value of a decision node: the best branch

P(H\mid +) = \dfrac{P(+\mid H)\,P_g}{P(+)}

Bayes' theorem

P(+) = P(+\mid H)\,P_g + P(+\mid \bar H)\,(1 - P_g)

total probability of a positive answer

\mathrm{EVPI} = \min\{(1 - P_g)\,C,\; P_g\,(V - C)\}

expected value of perfect information for a drilling decision

0 \le \mathrm{VOI} \le \mathrm{EVPI}

the value of any data about the same event

Common pitfalls

Putting an outcome in the tree before a decision that is really taken without knowing it. The tree then gets free and error-free information, and it overstates the value.
Comparing the option “with data” only with drilling without data, and forgetting the possibility to walk away from the project.
Taking the “reliability” of a survey for the probability of an accumulation after a positive answer.
Believing that data are valuable because they “raise the chances”: on average, the probability after the data equals the probability before them.
Forgetting that information does not come at once: a delay of the decision also costs money.

After this lesson you can

build a decision tree: place the decisions and the chance events in the order in which we take them and learn them, and give the branches money flows and conditional probabilities
roll the tree back by EMV and read the result as a strategy, a plan of action for each outcome and not a single action
show when sequential decisions are worth more than simultaneous ones, and choose the order of drilling for the prospects of one play (a group of prospects that depend on the same geological conditions)
calculate the expected value of perfect information and explain why it is the upper limit of the price of any data
update a probability with imperfect data by Bayes' theorem, and find the value of a survey with given detection and false-alarm rates
decide whether to buy information, taking into account its price, the time needed to get it and the company's attitude to risk

Decisions in time: why we need a tree

Decisions in exploration and development rarely stand alone. A company decides whether to buy a seismic survey. From its results it decides whether to drill an exploration well. After a discovery it decides whether to appraise the accumulation or to develop it at once. After a dry well it decides whether to drill a neighbouring prospect. Between the decisions come events that do not depend on the company: is there oil in the trap (the rock structure that can hold it), how large is the accumulation, what will the data show.

A decision tree records such a problem as a whole. On the left is the first decision, and time goes forward to the right. Each next decision comes after the events whose outcomes it will already know. The tree has no time scale, only an order. In essence it is a payoff matrix stretched out in time (the payoff matrix and the rules of choice): the matrix has one decision and one chance event, and the tree has a sequence of them.

A tree answers three questions. Which strategy is the best: not a single step, but a plan of action for all the outcomes. How much it is worth on average: its expected monetary value (Expected monetary value and preference theory). And how much it is worth to learn something before the decision: the value of information.

The elements of a tree and the rules of building it

Decision node: A square. The branches are the possible actions, and the decision-maker chooses between them. A tree starts with a decision node: if there is no choice, there is nothing to calculate.
Chance node: A circle. The branches are mutually exclusive outcomes of an event that does not depend on the decision: a discovery or a dry well, a large or a small volume, a positive or a negative answer of a survey. Each branch has a probability, and together they add up to one.
End node: A triangle. A path from the root (the first node) to an end is one possible history of the project. Its result is the sum of all the money flows along the path.
Branch: It carries the money flow that arises if we go along it: the cost of a well or of a survey with a minus sign, the value of a discovery with a plus sign.

The numbers on a tree follow the same rules as the outcomes in an EMV calculation (EMV as the average over outcomes), and a few rules of their own.

One date and one kind of money. All the flows are brought to the date of the first decision at one rate, and later events are discounted by the time that will pass before them (the time origin of NPV).
Only future flows. What is already spent is not in the tree: it does not affect the choice.
Conditional probabilities. The probability of an outcome depends on the path that led to the node. A well on a neighbouring prospect has different chances after a discovery and after a dry well on the first one. These are different nodes with different numbers.
Completeness. The branches of a chance node exclude each other and cover all the possibilities. The branches of a decision node are all the sensible actions, including walking away.
Information only from the past. A decision may depend only on the outcomes that stand to the left of it on the same path.

Rolling the tree back: from the ends to the root

We solve a tree from right to left, from the ends of the paths to the first decision. This method is called rolling back. In decision theory it is called backward induction. The logic is simple. To value a decision today, you need to know what you will do tomorrow in each possible situation, and tomorrow's decisions are best seen from the end.

The order of rolling back

1
Chance node
We replace it by its expected value. For each branch we add the flow of the branch to the value of the node that the branch leads to, and multiply the sum by the probability of the branch. Then we add up the results of all the branches.
2
Decision node
We replace it by the value of the best branch: the flow of the branch plus the value of the next node. The other branches are cut off (on a diagram they are crossed out).
3
Order
We calculate a node only after all the nodes to its right are calculated. The value of the root is the EMV of the best strategy.
4
Strategy
The chosen branches in all the decision nodes together make up the strategy: which action to choose now and what to do after each possible outcome.

Values of the nodes when rolling back

x_\text{chance} = \sum_{j} p_j\,(c_j + x_j), \qquad x_\text{decision} = \max_{k}\,(c_k + x_k)

where:

$x_\text{chance}$ value of a chance node: the expected result of everything that follows it, US$ million
$x_\text{decision}$ value of a decision node: the same with the best action, US$ million
$p_j$ probability of the $j$ -th branch of a chance node, given the path that led to the node, fraction
$c_j,\ c_k$ money flow on the branch at the date of the first decision; costs have a minus sign, US$ million
$x_j,\ x_k$ value of the node that the branch leads to; at the end of a path it is zero, US$ million

Units: US$ million at the date of the first decision (any units, the same for the whole tree)

Holds when

the attitude to risk is neutral: a chance node is valued by EMV; how to allow for risk aversion is shown in the advanced part
all the flows are brought to one date at one rate and in one kind of money
the probabilities of each chance node add up to one and take into account everything known on the path to it
in each decision node we choose only by what is already known at that moment

The value of the root is the EMV of the best strategy. It is the average of the results of all the paths, weighted by the probability of each path (the product of the probabilities along it), with the best action chosen in each decision node. It makes no difference whether we put the flows on the branches or add up the result of each path at once: the answer is the same.

The example continues the lessons on geological risk and on EMV. An untested segment of a play has two prospects (a segment is the part of a play that has one common risk). The probability that the play works is 0.6. The local elements of each prospect (the conditions that concern that prospect only) work with a probability of 0.5 (the numbers are assumed and round), so each prospect has $P_g = 0.30$ (common and local risk). Prospect A is the one from the lesson on EMV, and prospect B is smaller. The result of the first well tells us about the play and changes the chances of the second. In the lesson on geological risk, the same $P_g \approx 0.30$ was split as $0.56\cdot0.54$ . Here round numbers are taken, so that prospect A coincides exactly with the prospect of the lesson on EMV. The split matters: it decides how strongly the result of one well changes the chances of its neighbour.

Worked exampleTwo prospects of one play: both wells at once, or one after the other

Given (assumed data)

Play segment: $P_\text{play} = 0.6$ , each prospect has $P_\text{loc} = 0.5$ , so $P_g = 0.30$
Prospect A: the discovery is worth $V_A = 91.03$ million dollars at the date of the decision; the well costs $C = 12$ million dollars
Prospect B: production 20% lower than the base, with the same costs: the NPV of the development is 49.58 million dollars, and at the date of the decision $V_B = 45.07$ million dollars; the well also costs 12 million dollars
Time: the result of the first well is known before the decision on the second; waiting costs nothing

Solution

Chances of the second prospect after the first well. A discovery proves the play: only the local risk is left, 0.5. After a dry well the play works with a probability of $0.6\cdot(1 - 0.5)/(1 - 0.3) = 0.4286$ , and the chance of the neighbour is $0.4286\cdot0.5 = 0.2143$ .
Decision nodes for B after the well on A. After a discovery on A: $0.5\cdot45.07 - 12 = 10.54$ million dollars, so drill. After a dry well: $0.2143\cdot45.07 - 12 = -2.34$ , so do not drill, and the value of the node is 0.
The chance node of well A and the branch “A first”: $0.3\cdot(91.03 + 10.54) + 0.7\cdot0 - 12 = 18.47$ million dollars.
The branch “B first”: after a dry well on B, drilling A still pays: $0.2143\cdot91.03 - 12 = 7.51$ million dollars. The result of B does not change the decision on A, so the branch is worth the same as drilling both at once: $0.3\cdot91.03 + 0.3\cdot45.07 - 24 = 16.83$ million dollars.
Comparison at the root: A first, 18.47; B first or both at once, 16.83; do not drill, 0. The gain from the order is 1.64 million dollars. It is the probability of a dry well on A times the loss on B that we managed to avoid: $0.7\cdot2.34 \approx 1.64$ .

Answer: The best strategy is to drill A, and to drill B only if A finds an accumulation. Its EMV is 18.47 million dollars, against 16.83 when both are drilled at once.

What it means. Taken separately, both prospects pay (EMV 15.31 and 1.52 million dollars), and drilling both at once simply adds their EMV. But B pays only a little, and a dry well on A makes it unprofitable. By waiting for the result of A, the company drills B only when it is worth it. It makes no sense to start with B: its result does not change the decision on A, because A pays whatever happens. So the rule is this. Of the prospects of one play, drill first the one that pays whatever happens to its neighbours. Then use its result to decide the fate of the more doubtful ones. The data in this example are assumed for illustration.

SchematicThe tree of the example after rolling back: the numbers at the nodes are their values, US$ million. The branches that are cut off are crossed out. On the right are the results of the paths (the sum of the flows along the path). The options “both at once or B first” and “do not drill” are shown as single numbers: their EMV. The best strategy is worth $18.47 million.

Here we also see the difference between EMV and risk. Drilling both wells at once and drilling B first give the same EMV. But EMV does not show that, when both wells are drilled at once, both are dry with a probability of 0.55, because their failures are linked by the common risk of the play (dependent prospects). If such an outcome is dangerous for the company, we must compare the strategies also by the spread of the results and by the certainty equivalent (attitude to risk).

The value of perfect information

Imagine a “clairvoyant”: a source that tells us without error, before the drilling decision, whether the trap holds an accumulation. How much would it be worth to listen to it? The answer is the expected value of perfect information (EVPI): the difference between the EMV of the decision taken after such a message and the EMV of the best decision without it.

Expected value of perfect information

\mathrm{EVPI} = \sum_{i} P(\theta_i)\,\max_{k}\,\mathrm{NPV}_{k}(\theta_i) - \max_{k} \sum_{i} P(\theta_i)\,\mathrm{NPV}_{k}(\theta_i)

where:

$\theta_i$ state of nature, for example “there is an accumulation” and “there is none”, —
$P(\theta_i)$ probability of the state before the information, fraction
$\mathrm{NPV}_{k}(\theta_i)$ NPV of action $k$ in state $\theta_i$ at the date of the decision, US$ million
$\max_k$ choice of the best action; among the actions there is always walking away, with an NPV of zero, —

Units: US$ million at the date of the decision

Holds when

the state becomes known before the decision, without errors
the information is free and does not delay the decision; its price and time are counted separately
the attitude to risk is neutral

The first term is the average over the states of nature of the best result when the state is known. The second is the best average result without such knowledge. Choosing after the state has become known is never worse than choosing before, so EVPI is never negative. It is a ceiling: imperfect data about the same states are worth less, and it makes no sense to pay more than EVPI for them.

Expected value of perfect information for a drilling decision

\mathrm{EVPI} = \min\{(1 - P_g)\,C,\; P_g\,(V - C)\}

where:

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

Units: US$ million at the date of the decision

Holds when

two outcomes: a discovery worth $V$ , or a dry well; $V > C$
without information we choose the better of “drill” ( $P_g V - C$ ) and “walk away” (0)

If drilling pays without information, the clairvoyant saves us from the dry wells. The gain is their expected cost $(1 - P_g)\,C$ . If walking away is better, the clairvoyant returns the discoveries that we missed: $P_g\,(V - C)$ . Both numbers are the expected “regrets” of the two actions, that is, the expected differences from the best action when the outcome is known (the regret matrix). EVPI equals the regret of the better action, which is the smaller of the two.

Worked exampleWhat it is worth to know in advance whether there is an accumulation

Given (assumed data)

Prospect: from the lesson on EMV: $P_g = 0.30$ , $V = 91.03$ million dollars, $C = 12$ million dollars
Information: error-free, free and instant (a thought experiment)

Solution

Without information, drilling pays: $0.3\cdot91.03 - 12 = 15.31$ million dollars.
With information, we drill only where there is an accumulation. With a probability of 0.3 we get $91.03 - 12 = 79.03$ million dollars, and with a probability of 0.7 we spend nothing. EMV $= 0.3\cdot79.03 = 23.71$ million dollars.
EVPI $= 23.71 - 15.31 = 8.40$ million dollars. This is exactly the expected cost of the dry wells that the information saves us from: $0.7\cdot12 = 8.40$ .
A check through the regrets. If we drill, in 70% of cases we regret the 12 million dollars, which is $8.40$ . If we walk away, in 30% of cases we regret the missed 79.03 million, which is $23.71$ . The smaller of the regrets is 8.40.

Answer: EVPI ≈ 8.40 million dollars: no data about whether this trap holds an accumulation are worth more.

What it means. EVPI depends on how close the decision is to break-even. At $P_g = 0.6$ , drilling pays with a large margin, dry wells are rare, and EVPI falls to 4.80 million dollars. It is largest, 10.42 million dollars, at the break-even chance ( $P_g^{*} = C/V = 0.132$ ). There, without information, drilling and not drilling are equally good, and any knowledge decides the matter. EVPI is a quick upper estimate before any calculation of reliability: if a survey costs more, there is no need to calculate its value in detail. The data in this example are assumed for illustration.

Imperfect information and Bayes' theorem

There are no clairvoyants. Real data, such as a seismic survey, an analysis of seismic amplitudes, a geochemical survey or an appraisal well, are wrong in both directions. They miss some accumulations and “find” some empty traps. A false indicator can come, for example, from gas of low saturation (gas that fills only a small part of the pore space) or from a thin bed of dense (tight) rock. A real accumulation in a thin or deep layer may show nothing in the record.

Usually we know how the data respond in each state of nature: over what share of the accumulations the indicator appears (the detection rate $P(+\mid H)$ ) and over what share of the empty traps it appears (the false-alarm rate $P(+\mid \bar H)$ ). Here $H$ is the event “there is an accumulation”, $\bar H$ is “there is none”, and the sign $+$ is a positive answer. We get these numbers from past cases (wells drilled on objects with the indicator and without it) and from modelling. But for the decision we need the reverse probability: the chance of an accumulation after we have the answer, $P(H\mid +)$ . Bayes' theorem takes us from one to the other.

Bayes' theorem for a positive answer

P(H\mid +) = \frac{P(+\mid H)\,P_g}{P(+\mid H)\,P_g + P(+\mid \bar H)\,(1 - P_g)}

where:

$P(H\mid +)$ posterior probability of an accumulation, after a positive answer, fraction
$P_g$ prior probability of an accumulation, before the data, $P(H)$ , fraction
$P(+\mid H)$ probability of a positive answer over an accumulation: the detection rate, fraction
$P(+\mid \bar H)$ probability of a positive answer over an empty trap: the false-alarm rate, fraction

Units: dimensionless

Holds when

the events $H$ and $\bar H$ exclude each other and cover all the possibilities
the detection and false-alarm rates refer to objects of the same type as the one being assessed: an indicator that is reliable in one play may mean nothing in another
for a negative answer the formula is the same, with $P(-\mid H) = 1 - P(+\mid H)$ and $P(-\mid \bar H) = 1 - P(+\mid \bar H)$

The denominator is the total probability of a positive answer $P(+)$ . It is made of the right answers over accumulations and the false answers over empty traps. The numerator is the part of it that fell on accumulations. The rarer the accumulations (the smaller $P_g$ ), the larger the share of false answers among the positive ones, even for a good method.

The formula is easiest to understand with “natural frequencies”: instead of probabilities, we imagine many identical prospects and count them.

CalculatedBayes' theorem in frequencies: 1000 prospects with

P_g = 0.30

before the survey (top) and after it (bottom), grouped by the answer. There is an accumulation in 255 of 465 prospects with the indicator, and in 45 of 535 prospects without it.

Worked exampleA survey before drilling: updating the probability and its value

Given (assumed data)

Prospect: $P_g = 0.30$ , $V = 91.03$ million dollars, $C = 12$ million dollars; without data, the EMV of drilling is 15.31 million dollars
Survey: the indicator appears over 85% of accumulations and over 30% of empty traps: $P(+\mid H) = 0.85$ , $P(+\mid \bar H) = 0.30$
Price of the survey: 1.5 million dollars; the result is known before the decision to drill

Solution

Of 1000 such prospects, 300 hold an accumulation: over 255 of them the indicator will appear, and over 45 it will not. Of the 700 empty traps, the indicator will “find” an accumulation over 210 and will stay silent over 490.
The indicator will appear in $255 + 210 = 465$ cases out of 1000, and the accumulation is in 255 of them: $P(H\mid +) = 255/465 = 0.5484$ . Without the indicator there are 535 cases, and the accumulation is in 45 of them: $P(H\mid -) = 45/535 = 0.0841$ .
Decisions after the answer. Indicator present: $0.5484\cdot91.03 - 12 = 37.92$ million dollars, so drill. No indicator: $0.0841\cdot91.03 - 12 = -4.34$ , so do not drill, because the probability is below the break-even chance 0.132.
EMV with the survey, without its price: $0.465\cdot37.92 + 0.535\cdot0 = 17.63$ million dollars. The value of the survey: $17.63 - 15.31 = 2.32$ million dollars.
After the price: $2.32 - 1.5 = 0.82$ million dollars > 0, so the survey is worth ordering.

Answer: The survey raises the EMV of the decision from 15.31 to 17.63 million dollars before its price is deducted. It is worth paying no more than 2.32 million dollars for it. At a price of 1.5 million, the net gain is 0.82 million.

What it means. The value of the survey is 3.6 times smaller than EVPI (8.40 million dollars), because the survey is wrong in both directions. Out of 1000 prospects, a negative answer saves us 490 dry wells, but makes us miss 45 discoveries, and a false indicator still leads to 210 dry wells. Note where the value comes from. It does not come from a positive answer, because we would have drilled without it anyway, but from a negative answer, which changes the decision. The data in this example are assumed for illustration.

The updated probabilities equal the original one on average: $0.465\cdot0.5484 + 0.535\cdot0.0841 = 0.30$ . Data do not raise the chances “on average”: they push the estimate apart in both directions (the same after a dry well). The value of data is not in a higher probability but in better decisions: after a negative answer we can decide not to drill.

When information is worth buying

Here is a general definition. The value of information (VOI) is the difference between the EMV of the decision taken after getting the data and the EMV of the best decision without them. Both are calculated with the right to walk away from the project. The price of the data is not subtracted from the first one: VOI itself is the highest price that is worth paying for them. The difference between VOI and the price is the net gain from the purchase.

Value of information

\mathrm{VOI} = \sum_{r} P(r)\,\max_{k}\,\mathrm{EMV}_k(r) - \max_{k}\,\mathrm{EMV}_k

where:

$r$ a possible result of the data, for example a positive or a negative answer, —
$P(r)$ probability of getting the result $r$ , fraction
$\mathrm{EMV}_k(r)$ EMV of action $k$ with the probabilities updated after the result $r$ , US$ million
$\mathrm{EMV}_k$ EMV of action $k$ without the data; among the actions is walking away, with an EMV of zero, US$ million

Units: US$ million at the date of the decision

Holds when

the price of the data is not subtracted: VOI is compared with the price
the data do not delay the decision; if they do, the first term is discounted by the time of the delay
the attitude to risk is neutral

VOI is never negative, because we may simply not use the data we got, and it does not exceed EVPI. It is zero if the best action stays the same for every result of the data.

Value of a survey before drilling

\mathrm{VOI} = (1 - P_g)(1 - f)\,C - P_g\,(1 - s)\,(V - C)

where:

$s$ detection rate $P(+\mid H)$ , fraction
$f$ false-alarm rate $P(+\mid \bar H)$ , fraction
$P_g,\ V,\ C$ chance of success before the data, value of the discovery and cost of the well, —, US$ million

Units: US$ million at the date of the decision

Holds when

without data, drilling pays; after a positive answer we drill, and after a negative one we do not (if neither answer changes the decision, VOI = 0)
at $s = 1$ and $f = 0$ the formula turns into EVPI $= (1 - P_g)\,C$

The first term is the dry wells that a negative answer saves us from. The second is the discoveries that it makes us miss. In the example: $0.7\cdot0.7\cdot12 - 0.3\cdot0.15\cdot79.03$ $= 5.88 - 3.56 = 2.32$ million dollars. A survey is valuable as long as the first term is larger than the second.

The value of perfect information and of a survey at different prior probabilities, US$ million

$P_g$: 0.05
Without data: do not drill: 0
EVPI: 3.95
Survey: 0.00

$P_g$: 0.10
Without data: do not drill: 0
EVPI: 7.90
Survey: 3.48

$P_g$: 0.20
Without data: drill: 6.21
EVPI: 9.60
Survey: 4.35

$P_g$: 0.30
Without data: drill: 15.31
EVPI: 8.40
Survey: 2.32

$P_g$: 0.40
Without data: drill: 24.41
EVPI: 7.20
Survey: 0.30

$P_g$: 0.50
Without data: drill: 33.52
EVPI: 6.00
Survey: 0.00

$P_g$	Without data	EVPI	Survey
0.05	do not drill: 0	3.95	0.00
0.10	do not drill: 0	7.90	3.48
0.20	drill: 6.21	9.60	4.35
0.30	drill: 15.31	8.40	2.32
0.40	drill: 24.41	7.20	0.30
0.50	drill: 33.52	6.00	0.00

$V = 91.03$ million dollars, $C = 12$ million dollars; survey: $s = 0.85$ , $f = 0.30$ . The break-even chance is $C/V = 0.132$ . The data in this example are assumed for illustration.

CalculatedThe value of perfect information and of the survey against the prior chance of success. Both are largest at the break-even chance of 0.13. The survey has value only for

P_g

from 0.051 to 0.41, and it pays for its price of $1.5 million for

P_g

from 0.07 to 0.34.

The table and the chart show where data are worth money. At $P_g = 0.05$ even a positive answer leaves the probability below break-even, and at $P_g = 0.5$ even a negative one leaves it above. The survey will not change the decision, so its value is zero, although EVPI is noticeable in both cases. The value is largest near the break-even chance, at $P_g \approx 0.13$ , where without data drilling and not drilling are equally good.

The data must be able to change the decision. If you act in the same way whatever their result, their value is zero, even if they reduce the uncertainty. For a drilling decision, the answer must be able to move the probability across the break-even chance $C/V$ .
No more than EVPI. Imperfect data are worth no more than perfect ones. EVPI is a quick upper estimate that we can get before any calculation of reliability.
Compare with the best alternative. Without data we can also walk away from the project. For a prospect below break-even, the value of data is the whole EMV of the option “with data”, not its difference from loss-making drilling.
Time is also a price. While the data are being collected, all the outcomes of the project move later, and the delay is discounted with them. See the example below.
The value is an average. A particular result can disappoint, by lowering the estimate. This does not mean that we overpaid for the data: they saved us from a worse decision.

So far the information has come at once. An appraisal well after a discovery gives information on the size of the accumulation, but the development has to be put off while the well is drilled and tested. In the lesson on EMV, a small discovery was simply not developed. That is, its size was taken as known before the decision on development. The example below shows what such knowledge is worth.

Worked exampleAn appraisal well, or development at once

Given (assumed data)

Discovery: the size is unknown: ×0.5 of the base production with probability 0.3; ×1 with 0.5; ×1.5 with 0.2, as in the lesson on EMV
NPV of the development: −26.40, 100.13 and 226.51 million dollars; the decision on development is taken at the origin of the project's NPV
Appraisal well: it will show the size without error; development is put off by a year or by half a year; we do not count the cost of the well yet

Solution

Develop at once, without knowing the size: $0.3\cdot(-26.40) + 0.5\cdot100.13 + 0.2\cdot226.51$ $= 87.45$ million dollars.
If the size became known at once and for free, the small accumulation would not be developed: $0.5\cdot100.13 + 0.2\cdot226.51 = 95.37$ million dollars. The value of perfect information is 7.92 million: exactly the loss of the small case, $0.3\cdot26.40$ .
The well puts everything off by a year: $95.37/1.1 = 86.70$ million dollars. This is less than 87.45 without the well, even if the well is free.
By half a year: $95.37/\sqrt{1.1} = 90.93$ million dollars. The gain is $90.93 - 87.45 = 3.48$ million: this is what we can pay for the well.

Answer: A year of waiting costs more than knowing the size: an appraisal with such a delay does not pay. An appraisal well that delays the decision by half a year is justified if it costs less than 3.48 million dollars.

What it means. A delay discounts all the outcomes of the project, the good ones too. Here a year costs about 8.7 million dollars, while perfect information returns only 7.92 million, by saving us from the small case. The example is deliberately simplified in two ways. A real appraisal well makes errors, and its value is lower. But in the example one development plan serves all the sizes, while in reality knowing the size lets us choose the number of wells and the capacity of the facility. This adds value to the appraisal (Geological and technical uncertainty of a project). We shorten the delay by running the appraisal in parallel with the design, or by starting the development in stages. The data in this example are assumed for illustration.

Terms and symbols

Decision tree: A diagram of a sequential problem of choice. It is made of decision nodes and chance nodes, with branches that carry money flows and probabilities, and the results of the paths stand at the ends.
Decision node, chance node: A decision node (a square) is a choice made by the decision-maker. A chance node (a circle) is an outcome decided by nature.
Rolling back: Calculation from right to left: EMV in the chance nodes, the best branch in the decision nodes.
Strategy: The chosen actions in all the decision nodes: what to do now and after each possible outcome.
Prior and posterior probability: The probability of an event before the data and after them. They are linked by Bayes' theorem.
Detection rate $s$ , false-alarm rate $f$: The share of accumulations over which the data give a positive answer, and the share of empty traps over which they give it wrongly.
EVPI: The expected value of perfect information: the increase of EMV if the outcome became known in advance without errors. It is the upper limit of the price of data.
VOI: The value of information: the increase of EMV from data received before the decision. It is the highest price that is worth paying for them.

Common mistakes and misconceptions

Wrong

“The probability of a discovery after a positive answer is 85%: this is the reliability of the survey.”

In fact

85% is the probability of a positive answer over an accumulation. The probability of an accumulation after a positive answer also depends on the prior probability and the share of false alarms. In the example of the lesson it is 0.55.

Wrong

“Data are valuable because they raise the probability of success.”

In fact

On average the probability after the data equals the probability before them: in the example 0.465 · 0.5484 + 0.535 · 0.0841 = 0.30. The value of data lies only in the decisions that they change.

Wrong

“The value of a survey is the difference between the EMV with the survey and the EMV of drilling without it.”

In fact

Without a survey we can also not drill. For a prospect below break-even, drilling without data loses money, and we must compare with walking away (EMV = 0). Otherwise the value of the data is overstated by the size of this loss.

Wrong

“We learn the size of the accumulation after the discovery, so in the tree it stands before the decision on development.”

In fact

That is right only if between the discovery and the decision there is an appraisal that will show this size, with its own price and delay. Otherwise the tree gives itself free perfect information and overstates the value of the discovery.

Wrong

“The order of drilling the prospects does not matter: their EMVs simply add up.”

In fact

The EMVs add up for a programme that is drilled as a whole. If the result of the first well can change the decision on the second, the order creates value: in this lesson's example, $18.47 million against $16.83 million.

In depth: advanced detail

Bayes' theorem in the form of odds

It helps to write Bayes' theorem in terms of odds: the ratio of the probability of an event to the probability of the opposite event. A probability of 0.3 has odds of $0.3/0.7 \approx 0.43$ , that is, “3 to 7”.

Bayes' theorem in the form of odds

\frac{P(H\mid r)}{P(\bar H\mid r)} = \frac{P_g}{1 - P_g}\cdot\frac{P(r\mid H)}{P(r\mid \bar H)}

where:

$r$ the result received: a positive or a negative answer, —
$\dfrac{P(r\mid H)}{P(r\mid \bar H)}$ the likelihood ratio (LR): how many times more likely such a result is over an accumulation than over an empty trap, —
$P_g$ prior probability of an accumulation, fraction

Units: dimensionless

Holds when

the same as for Bayes' theorem
for several indicators, the likelihood ratios are multiplied only if the indicators are independent in each state of nature; two attributes of the same seismic record are usually dependent

The odds after the data are the odds before the data times the likelihood ratio. A result with LR above one raises the probability, a result with LR below one lowers it, and a result with LR equal to one tells us nothing. A good survey is one for which the LR of a positive answer is large and the LR of a negative answer is small.

Worked exampleUpdating the odds, and the limits of usefulness of a survey

Given (assumed data)

Prior probability: 0.30
Survey: $P(+\mid H) = 0.85$ , $P(+\mid \bar H) = 0.30$
Break-even chance of drilling: $C/V = 0.131824$

Solution

Prior odds: $0.3/0.7 = 0.4286$ . LR of a positive answer: $0.85/0.30 = 2.833$ . LR of a negative one: $0.15/0.70 = 0.21429$ .
After the indicator: odds $0.4286\cdot2.833 = 1.214$ , probability $1.214/(1 + 1.214) = 0.548$ . Without the indicator: odds $0.4286\cdot0.21429 = 0.0918$ , probability 0.0841.
The limits of usefulness. The break-even chance has odds of 0.15184. A positive answer raises prior odds of $0.15184/2.833 = 0.0536$ to this level, that is $P_g = 0.051$ . A negative answer lowers odds of $0.15184/0.21429 = 0.7086$ to it, that is $P_g = 0.415$ .

Answer: The survey can change the decision, and so has value, only for prospects with

P_g

from 0.051 to 0.415.

What it means. Below the lower limit, even a positive answer does not make drilling profitable. Above the upper limit, even a negative one does not make it loss-making. Two indicators that are independent in each state of nature would multiply the likelihood ratios: two positive answers would give odds of $0.4286\cdot2.833^2 = 3.44$ and a probability of 0.77. The data in this example are assumed for illustration.

The tree and the attitude to risk

Rolling back by EMV assumes a risk-neutral attitude. A risk-averse company values the chance nodes by expected utility or, more conveniently, by the certainty equivalent (the utility function and the certainty equivalent). With an exponential utility function, we can roll the tree back directly in certainty equivalents. Adding a guaranteed sum to all the outcomes adds it to the equivalent as well. So the value of a node can stand in for its whole lottery (its set of risky outcomes) without distortion. For other utility functions, we roll back the expected utility of the results of the paths.

Worked exampleA survey for a small company

Given (assumed data)

Company: risk tolerance $R = 30$ million dollars, exponential utility function
Prospect and survey: as in the example above: $P_g = 0.30$ , $V = 91.03$ , $C = 12$ million dollars; $s = 0.85$ , $f = 0.30$

Solution

Without the survey, the certainty equivalent of drilling, −1.91 million dollars, is below zero: the company will give up the prospect, and the value of the decision is 0.
With the survey, the company drills only after the indicator. The certainty equivalent of drilling after it is 10.14 million dollars > 0.
The outcomes of the strategy “survey, then decision”: a discovery (+79.03 million dollars) with a probability of 0.255, a dry well (−12) with 0.21, and walking away after a negative answer (0) with 0.535. The certainty equivalent is 4.30 million dollars, without the price of the survey.
The price of the survey, 1.5 million dollars, reduces the equivalent by exactly this sum: 2.80 million dollars.

Answer: For a small company, the survey is worth up to 4.30 million dollars. This is more than for a risk-neutral company (2.32 million): the survey turns an unacceptable stake into an acceptable one.

What it means. A risk-neutral company would drill without data, and the survey changes its decision only after a negative answer. A risk-averse company would give up the prospect without data. With data it drills in 46.5% of cases, and it loses 12 million dollars on a dry well much more rarely, so the survey changes its decision after a positive answer too. That is why data are often worth more to small companies than to large ones, and so are partners who share the risk. The data in this example are assumed for illustration.

Large trees in practice

A tree grows fast: ten sequential events with two outcomes each give more than a thousand paths. In practice we keep it manageable with a few methods.

Three branches instead of a distribution. A continuous uncertainty (volume, flow rate, costs) is replaced by three branches, P90, P50 and P10, with probabilities of 0.3, 0.4 and 0.3 (three cases instead of a distribution).
Identical subtrees once. If after different paths the problem is the same (with the same probabilities), we roll it back once and substitute the value.
A model at the ends of the paths. We calculate the result of a path with the cash-flow model of the project (Cash flow model of an oil and gas project), not by hand.
Sensitivity of the strategy. We look for the values of probabilities and costs at which the best strategy changes, such as the limit of 56 million dollars for the value of prospect B in the example of the lesson. If such a limit is close to the accepted estimate, the decision needs more accurate data.
Influence diagrams. An influence diagram is a compact record of the same problem: nodes that are quantities and arrows of dependence, without listing all the branches. We unfold it into a tree for the calculation.

A tree is a way to value flexibility: the possibility to wait, to stop after a failure, to expand the project after a success. The value of the sequential strategy in the example with two prospects is exactly the value of such flexibility. Simulation (simulation of NPV) shows the spread of the result for a given strategy, but it does not find the best strategy by itself. Decisions that depend on intermediate outcomes must be set in each trial by the logic of the tree.

Connections to other topics

Expected monetary value and preference theory — EMV, the utility function and the certainty equivalent: the numbers with which a tree is rolled back.
Geological risk and chance of success — the chance of success, the common risk of the play, and updating the chances of neighbouring prospects after a well.
Geological and technical uncertainty of a project — where the ranges of volumes, flow rates and costs come from, and how an appraisal narrows them.
Sensitivity analysis and scenarios — the payoff matrix and regret, and simulation of NPV.
Probabilistic volumetrics by Monte Carlo simulation — the distribution of volumes from which the P90, P50 and P10 branches are taken.
Risk reduction and diversification — partnership and the portfolio as other ways to reduce risk, besides buying data.

Check yourself

1
A chance node: with a probability of 0.4 a gain of 50 million dollars, with a probability of 0.6 a loss of 15 million. In the same decision node there is a guaranteed option of 8 million dollars. What should we choose by EMV, and what is the value of the decision node?
1. A.The risky option: 20 million dollars
2. B.The risky option: 17.5 million dollars
3. C.The risky option: 11 million dollars
4. D.The guaranteed option: 8 million dollars
Show the answer
Correct answer: C. The risky option: 11 million dollars
The EMV of the risky option is $0.4\cdot50 + 0.6\cdot(-15) = 20 - 9 = 11$ million dollars > 8, so the value of the decision node is 11. The number 17.5 is the plain average of the two outcomes without probabilities, and 20 is the gain without the loss. A risk-averse company might prefer the guaranteed 8 million: this is decided by the certainty equivalent.
2
A prospect has $P_g = 0.2$ , the discovery is worth 70 million dollars, and the well costs 9 million. What is the value of perfect information about whether there is an accumulation?
1. A.7.2 million dollars
2. B.14 million dollars
3. C.5 million dollars
4. D.12.2 million dollars
Show the answer
Correct answer: A. 7.2 million dollars
Without information, drilling pays: $0.2\cdot70 - 9 = 5$ million dollars. With information, we drill only when there is an accumulation: $0.2\cdot(70 - 9) = 12.2$ million. EVPI $= 12.2 - 5 = 7.2$ million dollars, which is the expected cost of the dry wells, $0.8\cdot9$ . The number 5 is the EMV without information, and 12.2 is the EMV with it.
3
The prior probability of an accumulation is 0.2. The indicator appears over 90% of accumulations and over 30% of empty traps. The indicator has appeared. What is the probability of an accumulation now?
1. A.0.43
2. B.0.75
3. C.0.20
4. D.0.90
Show the answer
Correct answer: A. 0.43
Out of 100 prospects, the 20 with an accumulation give 18 positive answers, and the 80 empty ones give 24 false ones. Of the 42 positive answers, 18 are right: $18/42 \approx 0.43$ . This is less than a half, although the indicator finds 90% of the accumulations. The answer 0.90 mixes up the detection rate with the probability of an accumulation after the answer. The answer 0.75 is $0.9/(0.9 + 0.3)$ , a calculation that ignores that there are four times as many empty traps as accumulations. The answer 0.20 is the probability before the answer.
4
A survey costs 1 million dollars. Without it, drilling pays, and after any answer of the survey drilling still pays. What is this survey worth for the drilling decision?
Show the answer
Zero. Information is valuable only because it can change the decision. Here, whatever the answer, the company will act as it would without the survey, and the EMV of the decision will not change. The survey will make the probability more exact, but no decision depends on this accuracy. It is different if its results are needed for another decision, for example where to place the well or how to design the development: then the value is calculated for that decision.
5
In one play there are two prospects: a large one, which pays whatever happens to its neighbour, and a small one, which pays only after a discovery on its neighbour. Which one should be drilled first, and why?
Show the answer
The large one. Its result can change the decision on the small one: after a dry well, the small prospect becomes loss-making, and it is not drilled. The result of the small one does not change the decision on the large one, which pays in any case. So to start with the small one would mean giving up the value of the information. The gain from the right order is compared with the price of waiting: the idle rig and the delay of the project.
6
Why can waiting for the result of an appraisal well reduce the EMV of a project, even if the well is free and error-free?
Show the answer
While the well is drilled and tested, the development is put off, and the delay discounts all the outcomes of the project, the good ones too. Information pays only where it changes the decision, for example by saving us from developing a small accumulation. In the example of the lesson a year of waiting reduces the value by about 8.7 million dollars, and knowing the size returns 7.9 million. So the price of information includes the time needed to get it.

Updated: 7 October 2026