Geological risk and chance of success

Quick refresher

An accumulation needs all the elements at once, so the chance of success is the product of the probabilities of the elements. We assess each one given that the elements before it have worked. Multiplying independent estimates is a special case.
The product is no greater than the weakest factor, and it falls with every new factor. In the example of the lesson, five elements with probabilities from 0.7 to 0.9 give 0.30. So we define the elements so that no uncertainty enters the product twice.
$P_g = P_\text{play}\cdot P_\text{loc}$ : the common play risk is the same for all prospects in one segment. If the play does not work, all of them will be dry, so the prospects of one play depend on one another.
The expected number of discoveries equals the sum of the probabilities, however the prospects depend on one another. But the probability that all the wells will be dry is greater for prospects of one play. A dry well lowers the chances of its neighbours in the play, and a discovery raises them.
Geological success is not the same as commercial success: $P_c = P_g\cdot P(V \ge V_\text{min})$ . The P90–P10 estimates (a low case that is exceeded with 90% probability and a high case that is exceeded with 10% probability) describe the volume given success. The risked volume is the mean volume given success, multiplied by $P_g$ .
$P_g$ measures our confidence. It is not a property of the rock: a trap either has oil or it does not. We can check the estimates only on many wells: the share of discoveries should be close to the mean probability that we assigned.

Key relationships

P_g = P_1\cdot P_{2|1}\cdot P_{3|1,\,2}\cdots

each factor is the probability of an element given that the preceding ones have worked

P_g = P_\text{play}\cdot P_\text{loc}

common play risk and local prospect risk

P_0 = (1 - P_\text{play}) + P_\text{play}\,(1 - P_\text{loc})^{n}

all $n$ prospects of one segment will be dry

P_c = P_g\cdot P(V \ge V_\text{min})

commercial chance of success

\bar k = \sum_i P_{g,i}

expected number of discoveries of a drilling programme

Common pitfalls

Mixing up the chance of success with risk. The geological risk, as the probability of failure, is $1 - P_g$ . The expressions “common play risk” and “local risk” name groups of uncertainties. In numbers, we express them by the chances of success $P_\text{play}$ and $P_\text{loc}$ .
Multiplying the estimates of linked elements as if they were independent, or counting one uncertainty in two elements.
Treating the prospects of one unexplored play as independent chances.
Mixing up the geological and the commercial chance of success, and taking the P90 of the volume given success for a cautious risked estimate.
Judging an estimate from one well: a dry well at $P_g = 0.6$ does not prove that the estimate was wrong.

After this lesson you can

state, for each element of a play (a group of similar prospects that share the same source rock, reservoir and seal), exactly which event the probability is about, and link the assessment to data
calculate the geological chance of success as the product of the probabilities of the elements, and explain when you can simply multiply and when you need conditional probabilities (probabilities that depend on another event)
split the chance of success into the common play risk and the local prospect risk, and read maps of common risk segments
calculate the probability of at least one discovery among several prospects, both independent ones and ones linked by a common play risk, and explain how the result of the first well changes the chances of its neighbours
move from geological success to commercial success, and to risked volume (the volume multiplied by the chance of success)
explain what the chance of success means and what it does not mean, and check the calibration of the estimates (whether the stated chances match what happens) against drilling results

Risk and uncertainty: is there an accumulation, and how much is in it

Before drilling an exploration well, we ask two different questions about a prospect. The first is whether there is an accumulation in the trap at all. The answer is “yes” or “no”, and before drilling we can name only the probability of “yes”. The second is how much oil and gas the accumulation holds, if it exists. Here the answer is a number from a range, and we describe it with a probability distribution (Probabilistic volumetrics by Monte Carlo simulation).

In exploration, risk usually means the first question: the possibility that there is no accumulation, so the money for the well is spent without a discovery. Uncertainty means the second: the spread of possible volumes, given that an accumulation exists. We assess them in different ways, and one cannot stand in for the other. A P90–P10 curve says nothing about the chance of an empty trap, and a chance of success says nothing about the size of the discovery.

Two questions about a prospect

: Answer
Is there an accumulation: yes or no
How much oil and gas is in it: a number from a range

: Measure
Is there an accumulation: geological chance of success $P_g$
How much oil and gas is in it: volume distribution: P90, P50, P10, mean

: What it is made of
Is there an accumulation: probabilities of the elements: charge, reservoir, trap, seal, timing
How much oil and gas is in it: distributions of the parameters of the volumetric formula

: What reduces it
Is there an accumulation: data on the weakest element; the first well in the play
How much oil and gas is in it: seismic and appraisal wells (drilled after a discovery to measure its size)

	Is there an accumulation	How much oil and gas is in it
Answer	yes or no	a number from a range
Measure	geological chance of success $P_g$	volume distribution: P90, P50, P10, mean
What it is made of	probabilities of the elements: charge, reservoir, trap, seal, timing	distributions of the parameters of the volumetric formula
What reduces it	data on the weakest element; the first well in the play	seismic and appraisal wells (drilled after a discovery to measure its size)

Before we assign probabilities, we agree what counts as success. Usually geological success means a well that has found moveable hydrocarbons (hydrocarbons that can flow): a flow on test, or reliable signs from well logs and fluid samples. Some companies add a threshold, such as a minimum accumulation height or flow rate. The definition must be the same for the chance of success and for the volume distribution. The distribution describes volumes starting from the smallest one that still counts as success. Otherwise some outcomes are either lost or counted twice.

The elements of risk: what each probability is about

We break the chance of success down into the elements of the play (geological chance of success). For each element, we state the event precisely, so that data can answer the question. The vague question “is there a reservoir?” turns into a sharper one: “is there, within the closure, rock with porosity and permeability above the cutoffs (the lowest values that count as reservoir), and thick enough for flow?”

Charge: In the area that feeds the trap there is a mature source rock, and enough hydrocarbons have reached the trap. Oil and gas shows in neighbouring wells (signs of oil or gas in the rock), the chemistry of the oils, and heating models raise our confidence (Source rocks, maturation and migration). A typical failure: the rock did not mature, or the hydrocarbons went another way.
Reservoir: In the trap there is rock with porosity and permeability above the cutoffs. Confidence comes from analogue wells (nearby wells in similar rock), core and seismic facies (rock types recognised on seismic data). A typical failure: shale replaces the sandstone sideways, or cement closes the pores.
Trap: The closure exists, even if it is smaller than mapped. Its size already belongs to the volume uncertainty. Confidence comes from dense seismic data, often three-dimensional (3D), and a reliable conversion of times to depths (seismic data record time, and a velocity model turns it into depth; Traps and timing). A typical failure: an error in the velocity model, so the “crest” turns out to be a monocline (a bed that dips one way, with no closure).
Seal: The top seal and the lateral (side) seals hold the hydrocarbons in the trap from the time it was filled to the present day. Confidence comes from the thickness and rock type seen in wells, entry pressure (the pressure that hydrocarbons need to push into the pores of the seal) and the assessment of fault seal (Reservoir and seal). A typical failure: a fault breaks the seal. If the seal holds only a short column, that mainly reduces the volume of the accumulation, and we take it into account in the volume distribution.
Timing and preservation: The trap formed before the main migration and was not destroyed after it. Confidence comes from burial history curves (plots of how deep a rock was buried through time) and the history of deformation. A typical failure: the trap was late, or the accumulation was later redistributed.

We set the number from data, not from a feeling. A common scale is convenient: it links the probability to how well the element is proven. It must be one scale for all of a company's prospects, otherwise they cannot be compared.

Example of a scale for the probability of an element

Probability: 1.0
What is known about the element: proven in the same play nearby: works in discovered accumulations

Probability: 0.8–0.9
What is known about the element: the data are favourable, there are reliable analogues

Probability: 0.6–0.7
What is known about the element: there are few data, but they lean in favour

Probability: 0.5
What is known about the element: no data for or against

Probability: 0.3–0.4
What is known about the element: the data lean against

Probability: 0.1–0.2
What is known about the element: the data are against, the element is unlikely

Probability	What is known about the element
1.0	proven in the same play nearby: works in discovered accumulations
0.8–0.9	the data are favourable, there are reliable analogues
0.6–0.7	there are few data, but they lean in favour
0.5	no data for or against
0.3–0.4	the data lean against
0.1–0.2	the data are against, the element is unlikely

The scale is only an example: every company has its own. A probability of 0.5 means “I don't know”, not cautious optimism. We give 1 only to an element that is proven in the same play.

The product of probabilities and its assumptions

Success means that all the elements work at once: the charge arrived, and there is a reservoir, and the trap is closed, and the seal holds, and the timing is right. The probability that several events all happen equals the probability of the first, times the probability of the second given the first, and so on. This rule always holds. Conditional probabilities turn into ordinary ones only when the elements are independent (one does not affect another).

Geological chance of success through conditional probabilities

P_g = P_1\cdot P_{2|1}\cdot P_{3|1,\,2}\cdots P_{k|1,\ldots,k-1}

where:

$P_g$ geological chance of success: all the elements have worked, fraction
$P_1$ probability that the first element is present and works, fraction
$P_{2|1}$ probability of the second element given that the first has worked; likewise $P_{3|1,\,2}$ and the following ones, fraction
$k$ number of elements, —

Units: dimensionless

Holds when

The elements are defined so that success happens when all of them have worked, and only then.
Each uncertainty is counted in exactly one factor.
The order of the elements does not matter, but we assess each following probability given that the elements before it have worked.
For independent elements $P_{2|1} = P_2$ and so on, and the formula becomes the simple product $P_g = \prod_i P_i$ .

The product is no greater than the smallest factor: a prospect is no more reliable than its weakest element. Every new factor smaller than one reduces the result. So if we split the elements into ever smaller ones without moving to conditional probabilities, we understate the chances.

Worked exampleChance of success of a prospect, element by element

Given (assumed data)

Charge (common to the play segment): 0.8
Regional seal (common to the segment): 0.7
Trap: 0.75
Reservoir within the trap: 0.8
Timing: 0.9
How the probabilities were assessed: each given that the elements above it in the list have worked

Solution

Charge and seal together: $0.8\cdot0.7 = 0.56$ . This is the common play risk $P_\text{play}$ , which is the same for all prospects of the segment.
Add the trap: $0.56\cdot0.75 = 0.42$ . Add the reservoir: $0.42\cdot0.8 = 0.336$ . Add the timing: $0.336\cdot0.9 = 0.302$ .
Local risk: $P_\text{loc} = 0.75\cdot0.8\cdot0.9 = 0.54$ . Check: $P_\text{play}\cdot P_\text{loc} = 0.56\cdot0.54 = 0.302$ .
The probability of failure, which is the geological risk, is $1 - P_g = 0.698$ .

Answer:

P_g \approx 0.30

: about three chances in ten, although no element is assessed below 0.7.

What it means. Taken separately, each element looks likely, but together they give less than a third. The weakest factor is the regional seal. If new data raise its probability to 0.9, $P_g$ rises to 0.39. The common play risk ( $P_\text{play} = 0.56$ ) counts here almost as much as the local one ( $P_\text{loc} = 0.54$ ). One successful well in the segment would remove it at once for all neighbouring traps. The data in this example are assumed for illustration.

Independence is a strong assumption. Elements are linked when one geological process created them. The sand body and the shales above it were laid down by one environment. The same fault decides both the trap and the lateral seal at the fault. One burial history decides both the charge and the timing. Then the probability of the second element assessed “in general” and its probability given the first are different numbers, and the second one must go into the product.

Worked exampleA reef and its seal

Given (assumed data)

Carbonate buildup (a reef: a mound of limestone built by organisms) — the reservoir: exists with a probability of 0.6
Regional seal over the whole area: present over 70% of the area: 0.7
Seal over the buildup: the buildups grew in a marine basin and were buried by marine shales: 0.9

Solution

If the elements are treated as independent: $0.6\cdot0.7 = 0.42$ .
With the conditional probability: $P_R\cdot P_{S|R} = 0.6\cdot0.9 = 0.54$ .
Check that the estimates agree. The seal probability averaged over the area is made up of the parts with a buildup and the parts without one: $0.7 = 0.9\cdot0.6 + P_{S|\bar R}\cdot0.4$ . From this, where there is no buildup, the seal is present with probability $P_{S|\bar R} = 0.40$ .

Answer: Multiplying as if the elements were independent gives 0.42 instead of 0.54, which is 22% less than the correct value.

What it means. The error was not in the arithmetic but in the question. We assessed the seal “in general”, but success needs a seal exactly where the reservoir is. The link can also work the other way round. A thick sand body is a good reservoir, but it may itself cut away (erode) the shales that would serve as its lateral seal in a stratigraphic trap. Then multiplying as if the elements were independent overstates the chances. The data in this example are assumed for illustration.

Common play risk and common risk segments

Some elements are the same for all traps over a large area: whether the kitchen (the area where the source rock makes hydrocarbons) works, whether there is a regional seal, and whether the reservoir sequence is widespread. We assess them for the play as a whole, not for each trap, and show them on maps. For each common element, we divide the area of the play into zones by degree of confidence. Overlaying these maps divides the play into common risk segments (CRS): areas with the same probability that the play works, $P_\text{play}$ .

A prospect inherits the $P_\text{play}$ of its segment. The second factor, the local risk $P_\text{loc}$ , is the probability that the elements of the trap itself work: closure, reservoir within it, and timing (common and local risk). We decide separately for each play which elements are common and which are local. Sometimes we split one element in two. For example, the extent of the reservoir sequence goes to the play, and its quality at a given point goes to the prospect. In a segment where discoveries have already proven the play, $P_\text{play} = 1$ , and only the local risk decides the chances of the prospects.

The same prospect in three segments of a play

Segment: A
Charge: 1.0
Seal: 1.0
$P_\text{play}$: 1.00
$P_g$: 0.54

Segment: B
Charge: 0.8
Seal: 0.7
$P_\text{play}$: 0.56
$P_g$: 0.30

Segment: C
Charge: 0.5
Seal: 0.4
$P_\text{play}$: 0.20
$P_g$: 0.11

Segment	Charge	Seal	$P_\text{play}$	$P_g$
A	1.0	1.0	1.00	0.54
B	0.8	0.7	0.56	0.30
C	0.5	0.4	0.20	0.11

The local risk is the same in all three cases: $P_\text{loc} = 0.54$ . Segment A: the play is proven by discoveries. Segment B: the segment of the prospect from the example above. Segment C: the edge of the basin, far from the kitchen, where the regional seal thins out. The data in this example are assumed for illustration.

We need segment maps even before prospects appear. They show in which parts of the basin it is worth looking for traps. They also show where any prospect carries a heavy common risk from the start, and where the first well will tell us most about the play.

Dependent and independent prospects

Prospects of one segment are linked by the common factor $P_\text{play}$ . If the play does not work, they all fail together. If it works, each trap tests only its own local elements. Prospects of different plays have no common elements, so their outcomes are independent.

Probability that all prospects are dry

P_0^{\text{ind}} = \prod_{i=1}^{n} \left(1 - P_{g,i}\right), \qquad P_0^{\text{play}} = \left(1 - P_\text{play}\right) + P_\text{play}\left(1 - P_\text{loc}\right)^{n}

where:

$P_0^{\text{ind}}$ probability that all $n$ independent prospects are dry, fraction
$P_0^{\text{play}}$ the same for $n$ prospects of one segment with the same local risk, fraction
$P_{g,i}$ geological chance of success of the $i$ -th prospect, fraction
$P_\text{play},\ P_\text{loc}$ common play (segment) risk and the local risk of each prospect, fraction
$n$ number of prospects, —

Units: dimensionless

Holds when

In the first formula the outcomes of the prospects are independent.
In the second the prospects are linked only by the common play risk, and their local elements are independent.
The probability of at least one discovery equals $1 - P_0$ .

The second formula says: either the play does not work (and then all the prospects are dry), or it works but all the traps have failed for their own, local reasons. If each prospect has the same $P_g$ , the probability that all are dry is higher for prospects of one play, because their failures come together.

Worked exampleThree prospects: in one play and in different ones

Given (assumed data)

Three prospects of one segment: $P_\text{play} = 0.56$ , each has $P_\text{loc} = 0.54$ , that is, $P_g = 0.3024$
Three prospects of different plays: the outcomes are independent, each has the same $P_g = 0.3024$

Solution

The expected number of discoveries is the same in both cases: $3\cdot0.3024 = 0.9072$ . The mean adds up whatever the dependence.
Different plays: all three are dry with probability $(1 - 0.3024)^3 = 0.34$ . At least one discovery has probability 0.66.
One play: all three are dry with probability $0.44 + 0.56\cdot0.46^3 = 0.49$ . At least one discovery has probability 0.51.
But three discoveries out of three are more likely in one play: 0.088 against 0.028 for independent prospects.

Answer: At least one discovery: 66% for prospects of different plays and 51% for prospects of one play.

What it means. Dependence does not change the mean number of discoveries. But it stretches the distribution towards the ends: “all three” and “none” both become more likely. For a company that needs at least one discovery, three prospects in different plays are more reliable than three in one, although the chance of each one taken separately is the same. The data in this example are assumed for illustration.

CalculatedDistribution of the number of discoveries among three prospects with the same

P_g \approx 0.30

: independent prospects of different plays, and prospects of one segment with a common play risk. The mean number of discoveries is the same: about 0.91.

The link between prospects also works over time: the result of the first well in a segment tells us something about the whole play.

Worked exampleHow the result of the first well changes the chances of a neighbour

Given (assumed data)

Segment: $P_\text{play} = 0.56$ ; each prospect has $P_\text{loc} = 0.54$
First well: a discovery or a dry hole; the cause of failure is not established

Solution

A discovery proves the play. The neighbouring prospect is left with only the local risk: $P_g = P_\text{loc} = 0.54$ instead of 0.30.
A dry hole. Imagine 1000 identical segments. In 560 of them the play works. There, the first well finds an accumulation in 302 cases and is dry for local reasons in 258 cases. In the remaining 440 segments all the wells are dry.
In all, there are 698 dry first wells, and the play works in only 258 of them. So after a dry hole the play works with probability 0.37.
The chance of the neighbouring prospect after a dry hole: $0.37\cdot0.54 = 0.20$ .

Answer: After a discovery the neighbour's chance rises from 0.30 to 0.54, and after a dry hole it falls to 0.20.

What it means. So the first well in a play is worth twice as much: it tests its own trap and the whole play. The order in which we drill the prospects of one play is also a decision. Often we drill first the prospect that tests the common risk most cheaply. The topic Decision trees and the value of information covers how much such information is worth and how to choose the order. The data in this example are assumed for illustration.

From geological to commercial success

Geological success does not yet mean that the discovery will pay for its development. A small accumulation, viscous (thick) oil, low flow rates or a long distance to pipelines can make it uneconomic. So we introduce the minimum economic field size $V_\text{min}$ (MEFS). It is the smallest recoverable volume whose development still pays back at the assumed price, costs and discount rate (the rate that brings future money back to today's value; Net present value (NPV)). We also introduce the commercial chance of success $P_c$ .

Commercial chance of success

P_c = P_g\cdot P(V \ge V_\text{min})

where:

$P_c$ probability of discovering an accumulation no smaller than the minimum economic field size, fraction
$P_g$ geological chance of success, fraction
$V$ recoverable volume of the discovery, if there is an accumulation, million m³ or million bbl
$V_\text{min}$ minimum economic field size, same units
$P(V \ge V_\text{min})$ the share of the volume distribution (given success) that is at least the threshold; we read it from the exceedance probability curve, fraction

Units: dimensionless; $V$ and $V_\text{min}$ in the same units

Holds when

The volume distribution is built given geological success and starts at the smallest volume that counts as success.
$V_\text{min}$ depends on price, costs, taxes and infrastructure: for the same trap it differs between companies and between years.
The chance of success and the volume distribution relate to the same accumulation under one definition of success.

$P_c$ is no greater than $P_g$ , and the gap grows with the threshold. If $V_\text{min}$ is below the smallest possible volume, $P_c = P_g$ . If it equals the P50 of the volume given success, $P_c = P_g/2$ . If it is above the largest possible volume, $P_c = 0$ .

Risked volume

\bar V_\text{risked} = P_g\,\bar V

where:

$\bar V_\text{risked}$ mean volume over all outcomes, including a dry hole (risked volume), million m³ or million bbl
$\bar V$ mean (not median) recoverable volume given success, same units
$P_g$ geological chance of success, fraction

Units: volume units

Holds when

If the well fails, the volume is zero, so the mean over all outcomes is $P_g\bar V + (1 - P_g)\cdot 0$ .
We use the mean, not the P50, because the mean volumes of prospects add up, but the P50 values do not.

Risked volume is the mean over all outcomes. We use it to compare prospects and to add up the volumes of a portfolio (a set of prospects). The topic Expected monetary value and preference theory covers how probabilities and volumes are turned into money.

Worked exampleCommercial success and risked volume

Given (assumed data)

Geological chance of success: 0.3024 — the prospect from the example above
Recoverable volume given success: the distribution from the lesson Probabilistic volumetrics by Monte Carlo simulation: P90 = 3.3, P50 = 4.1, P10 = 5.1 million m³, mean 4.152 million m³
Minimum economic field size: 3.5 million m³

Solution

The share of trials (simulation runs) in which the volume is at least 3.5 million m³: 0.820.
Commercial chance of success: $P_c = 0.3024\cdot0.820 = 0.248$ .
Risked volume: $0.3024\cdot4.152 = 1.26$ million m³, or 7.9 million barrels (there are 6.2898 barrels in a cubic metre).

Answer: The commercial chance of success is about 25%, against 30% for the geological one. The risked volume is 1.26 million m³.

What it means. The risked volume is not a possible outcome. It is the mean over all outcomes: an accumulation of 1.26 million m³ will never be found here. We will find either nothing or, in the simulation trials, between 2.3 and 7.2 million m³. This number is convenient for comparing prospects, but not for designing development. We design development for the volume given success. The volume distribution is borrowed from another lesson to keep the example clear. The data in this example are assumed for illustration.

CalculatedThe probability that the recoverable volume is at least a given value, shown given success (the curve starts at 100%) and risked (it starts at

P_g \approx 30

%). The minimum economic field size, 3.5 million m³, is marked.

What the chance of success means — and what it does not

A trap either has oil or it does not: nature has already decided. A probability of 0.3 is not about the rock. It is about our knowledge: how confidently we can expect an accumulation, given the data we have. So two teams with the same data can arrive at different numbers, and new data change the estimate even though the trap has stayed the same.

Such a probability is useful because of its frequency meaning: of many prospects that were assigned $P_g \approx 0.3$ , about three in ten should turn out to have an accumulation. This property is called calibration. We cannot check it from one well: a dry hole at $P_g = 0.6$ is an ordinary outcome, and it happens in four cases out of ten. We check calibration on the results of many wells.

Expected number of discoveries and its scatter

\bar k = \sum_{i=1}^{n} P_{g,i}, \qquad \sigma_k = \sqrt{\sum_{i=1}^{n} P_{g,i}\left(1 - P_{g,i}\right)}

where:

$\bar k$ expected (mean) number of discoveries, —
$\sigma_k$ standard deviation of the number of discoveries, calculated from the assigned probabilities (not from a sample), —
$P_{g,i}$ chance of success assigned to the $i$ -th prospect before drilling, fraction
$n$ number of prospects drilled, —

Units: dimensionless (a number of wells)

Holds when

$\bar k$ holds whatever the dependence between the prospects.
The formula for $\sigma_k$ is for independent prospects. A common play risk increases the scatter.
The probabilities were recorded before drilling and were not corrected afterwards.

If the number of discoveries keeps falling more than two $\sigma_k$ away from $\bar k$ on the same side, the estimates are poorly calibrated. Most often they are too optimistic.

Worked exampleChecking the estimates against drilling results

Given (assumed data)

Exploration wells drilled: 24
Probabilities assigned before drilling: 8 wells — 0.5; 10 — 0.3; 6 — 0.15
Discoveries made: 3

Solution

Expected number of discoveries: $8\cdot0.5 + 10\cdot0.3 + 6\cdot0.15 = 7.9$ .
The sum of $P(1 - P)$ is $8\cdot0.25 + 10\cdot0.21 + 6\cdot0.1275 = 4.865$ , so $\sigma_k = 2.21$ .
Three discoveries is 2.2 standard deviations below the expected number. An exact calculation for independent wells gives this: if the estimates are correct, we get three discoveries or fewer with a probability of about 1.8%.

Answer: The drilling result agrees poorly with the assigned probabilities: most likely they were overstated.

What it means. We state the conclusion with caution, for two reasons. First, if some of the wells were drilled in the same plays, the outcomes depend on one another and the scatter is larger than we calculated, so three discoveries are less surprising. Second, 24 wells is a small sample. But the direction of the error is typical: a team that prepares a prospect for drilling tends to be optimistic. So companies compare the assigned probabilities with drilling results, and they send risk assessments to an independent review. The data in this example are assumed for illustration.

What the chance of success does not mean:

It is not the commercial chance of success. What is found may not be enough to pay for development: $P_c \le P_g$ .
It is not a measure of value. A prospect with $P_g = 0.5$ and a small volume may be worth less than one with $P_g = 0.15$ and a large volume. We compare them by expected value (Expected monetary value and preference theory), not by chances.
It is not a threshold for drilling. There is no universal probability below which we do not drill. With a large possible gain and a cheap well, even a low chance is worth it. With a small gain, even a high chance is not.
It is not a property of the rock. The number changes with the data and with the definition of success. We can compare the probabilities of different prospects only if they use one definition and one scale.
The probabilities of prospects do not add up. The sum of $P_g$ over several prospects is the expected number of discoveries, not the probability of at least one discovery. That probability is no greater than the sum, and it is usually noticeably smaller.

Terms and symbols

Geological risk: The probability that an exploration well will not find an accumulation: $1 - P_g$ .
Geological chance of success $P_g$: The probability that all the elements of the play have worked and the well will find moveable hydrocarbons.
Conditional probability $P_{2|1}$: The probability of the second element given that the first has worked.
Common play risk $P_\text{play}$: The probability that the common elements of the segment work. It is the same for all its prospects.
Common risk segment: An area of the play with one assessment of the common elements (CRS).
Minimum economic field size $V_\text{min}$: The smallest recoverable volume whose development pays back (MEFS).
Commercial chance of success $P_c$: The probability of discovering an accumulation no smaller than the minimum economic field size.
Risked volume: The mean volume over all outcomes, including failure: $P_g\bar V$ .
Calibration: How well the assigned probabilities match the share of discoveries among many wells.

Common mistakes and misconceptions

Wrong

“Every element is assessed at 0.7 or higher, so the prospect is reliable.”

In fact

Probabilities multiply. In the example of the lesson, five elements with probabilities from 0.7 to 0.9 gave 0.30. A prospect is no more reliable than the product, and the product is smaller than the weakest factor.

Wrong

“The seal is present over 70% of the area and the reservoir with a probability of 0.6, so together 0.42.”

In fact

We need the probability of a seal exactly above the reservoir. If one process created both, this probability is different: in the reef example it is 0.9, and together they give 0.54.

Wrong

“Three prospects in one new play are three independent chances, so at least one discovery is almost guaranteed.”

In fact

They share the common play risk. In the example of the lesson, the probability of at least one discovery among three prospects of one play is about 51%, but for three independent prospects with the same chances it is 66%.

Wrong

“A dry hole at $P_g = 0.6$ proves that the geologists were wrong.”

In fact

A failure at such an estimate happens in four cases out of ten. Only a systematic gap between the share of discoveries and the assigned probabilities, over many wells, shows an error.

Wrong

“The P90 of the volume of a prospect is a cautious estimate that can be relied on.”

In fact

The P90 refers to the volume given that an accumulation exists. With probability $1 - P_g$ the prospect will give nothing, so always state $P_g$ beside the P90.

In depth: advanced detail

Updating the chances after a well

We can write the calculation with a thousand segments from the example of the lesson as a single formula. A dry hole in a segment means either a play that does not work, or a working play whose local elements failed. The share of the second case among all dry holes is the new probability that the play works.

Probability that the play works after a dry hole

P_\text{play}' = \frac{P_\text{play}\left(1 - P_\text{loc}\right)}{1 - P_\text{play}\,P_\text{loc}}

where:

$P_\text{play}'$ probability that the play works after a dry hole in the segment, fraction
$P_\text{play}$ the same probability before drilling, fraction
$P_\text{loc}$ local risk of the prospect drilled, fraction

Units: dimensionless

Holds when

The well was drilled on a prospect of the same segment, and the cause of the failure has not been established.
The local elements of different prospects are independent.
If the well data show that exactly a local element failed (for example, there is no closure), the common risk hardly changes. The formula does not suit that case.

The denominator is the probability of a dry hole. The numerator is the part of it that falls on a working play. The more reliable the local elements are, the harder a dry hole hits the play: if everything in the trap is almost certainly in order, nothing else can explain the failure. The general rule for updating probabilities with new data is Bayes' theorem. The topic Decision trees and the value of information covers it.

That is why it is so important to find out which element failed. A well that did not meet a reservoir says almost nothing about the charge. A well that drilled through a good reservoir under a reliable seal in a proven closure, but found no traces of hydrocarbons, says a great deal about the charge.

Worked exampleData do not raise the chances “on average”

Given (assumed data)

A neighbour in the segment before the first well is drilled: $P_g = 0.302$
After the first well: 0.54 for a discovery (probability 0.302), 0.199 for a dry hole (probability 0.698)

Solution

The mean over the two possible outcomes is $0.302\cdot0.54 + 0.698\cdot0.199 = 0.302$ .

Answer: The mean probability after drilling equals the probability before drilling: 0.302.

What it means. This is a general property of consistent estimates: if we expected new data to raise the probability, we should raise it now. Data push the estimate towards zero or one, and so they help us decide whether to drill the neighbour or give up. The value of information lies in better decisions, not in “raising the chances” (Decision trees and the value of information). The data in this example are assumed for illustration.

Connections to other topics

Petroleum system and the play concept — the elements of the petroleum system, the play and the prospect, common and local risk.
Source rocks, maturation and migration — how the charge is assessed: source rock maturity, generation, migration.
Reservoir and seal — data on the reservoir, and on what size of accumulation the seal will hold.
Traps and timing — the closure of a trap and its age relative to the migration.
Probabilistic volumetrics by Monte Carlo simulation — the volume distribution given success and the exceedance probability curve.
Expected monetary value and preference theory — how probabilities and volumes are turned into a monetary estimate and a drilling decision.
Decision trees and the value of information — the value of new data and the order of drilling dependent prospects.

Check yourself

1
Probabilities of the elements of a prospect (independent): charge 0.9, reservoir 0.8, trap 0.6, seal 0.7, timing 0.95. What is the geological chance of success?
1. A.0.60 — from the weakest element
2. B.0.79 — the mean of the elements
3. C.0.29
4. D.0.12
Show the answer
Correct answer: C. 0.29
Success needs all the elements at once, so we multiply the probabilities: $0.9\cdot0.8\cdot0.6\cdot0.7\cdot0.95 \approx 0.29$ . The weakest element sets only an upper limit (0.6), and the mean of the elements is not a chance of success at all.
2
Two prospects in one unexplored segment: $P_\text{play} = 0.5$ , each has $P_\text{loc} = 0.6$ . With what probability will both be dry?
1. A.0.70
2. B.0.16
3. C.0.49
4. D.0.58
Show the answer
Correct answer: D. 0.58
Either the play does not work (0.5), or it works but both traps fail for their own reasons: $0.5\cdot0.4^2 = 0.08$ . So the total is $0.5 + 0.08 = 0.58$ . If the prospects were independent with the same $P_g = 0.3$ , the result would be $0.7^2 = 0.49$ : the common risk makes a double failure more likely.
3
Why does a dry hole in an unexplored play noticeably reduce the chances of a neighbouring prospect, while a dry hole in a play proven by discoveries hardly does?
Show the answer
In a proven play the common risk has already been removed ( $P_\text{play} = 1$ ). Only local causes can explain a failure, and the neighbour keeps its own $P_\text{loc}$ . In an unexplored play a dry hole is evidence that the play itself does not work, and that factor is common to all its prospects. So after the dry hole, the probability that the play works falls, and so do the chances of the neighbours.
4
A prospect has $P_g = 0.25$ ; the volume given success: P90 = 2, P50 = 4, P10 = 8 million t, mean 4.6 million t. What is the risked volume?
1. A.4.0 million t — the P50
2. B.1.15 million t — $0.25\cdot$ mean
3. C.1.0 million t — $0.25\cdot$ P50
4. D.2.0 million t — the P90
Show the answer
Correct answer: B. 1.15 million t — $0.25\cdot$ mean
The risked volume is the mean over all outcomes: $0.25\cdot4.6 + 0.75\cdot0 = 1.15$ million t. We use the mean, not the P50. For a right-skewed distribution (one with a long tail of large values) they differ, and only the means add up across prospects.
5
Over five years a company drilled 30 exploration wells, each assigned a chance of success of 0.4 before drilling, and made 6 discoveries. What does this say about the estimates?
Show the answer
We expected 12 discoveries ( $30\cdot0.4 = 12$ ). The standard deviation for independent wells is $\sqrt{30\cdot0.4\cdot0.6} \approx 2.7$ . Six discoveries is 2.2 standard deviations fewer, so the estimates were most likely optimistic. We state the conclusion with caution, because with common play risks the scatter is wider, and the sample is small. But a systematic gap is a reason to revise the probability scale.
6
What does $P_g = 0.3$ mean?
1. A.The volume of the accumulation is not less than the P50 with a probability of 30%
2. B.The company risks losing 30% of its investment
3. C.The trap is filled with oil to 30% of its volume
4. D.Of many similar prospects, about three in ten will turn out to have an accumulation
Show the answer
Correct answer: D. Of many similar prospects, about three in ten will turn out to have an accumulation
The chance of success is a measure of confidence that an accumulation exists, and we check it by frequency: among many prospects with such an estimate, about 30% should turn out to be discoveries. It says nothing about how full the trap is or about the volume.

Updated: 5 October 2026