probability estimates of field areas and trapped oil volumes

P1: MRM/FYX P2: MRM/RKP QC: MRM

Natural Resources Research (NRR) PP202-312125 July 4, 2001 11:53 Style file version Nov. 07, 2000

Natural Resources Research, Vol. 10, No. 2, 2001

Probability Estimates of Field Areas and TrappedOil Volumes

Malcolm Anderson1 and Robert S. Gullco2

Received 3 December 2000; accepted 30 March 2001

A portion of a sedimentary basin is subdivided conceptually into hexagons of equal area. Thearea of each hexagon is equal to the minimum area an oil field should have to be commercial.Hexagons can be ‘full’ of oil or ‘empty.’ A field size 1 consists of a cell with oil surrounded bysix empty cells; a field size 2 consists of two adjacent cells with oil surrounded by eight emptycells, etc. Principles of Percolation Theory are used to determine the probability distributionof the areas of the oil fields existing in this portion of the basin. The only piece of informationnecessary to determine this probability distribution is the Success Ratio (number of successfulexploration wells/total number of exploration wells drilled in this portion of the basin). Thisapproach has several practical applications.

A probabilistic model is introduced to predict to which extent potential oil traps are filledwith oil. The model assumes that the probability that an oil unit will end up in a particulartrap, is proportional to the surface area of the trap. The model predicts that independently ofthe distribution of the trap volumes, there will be a critical trap volume. All the traps havinga volume less than this critical volume, will be filled to spill point. An equation is deduced topredict, for all traps having a volume greater than the critical, the volume of oil that can beencountered in the trap, provided the volume of the trap is known.

KEY WORDS: Hexagons; triangular lattice; Percolation Theory; Success Ratio; critical volume; spillpoint.

INTRODUCTION

If an oil company wants to drill 5 explorationwells in a certain basin, they would be interested toknow what is the probability that at least one of thewells will be successful. Or, if they have discovered astructure capable of holding 100 million barrels, theywould like to know what is the probability that thestructure is filled to spill point or eventually, what isthe probability that the structure is empty.

A stockbroker also might be interested in know-ing these probabilities, in order to advise his/herclients whether to buy shares of that oil company.

1 Department of Mathematics, University of Brunei, Negara BruneiDarussalam.

2 Paradigm Geophysical, 1 William St., Perth, Western Australia,Australia; e-mail: [email protected]

Finally, in an even less conventional situation, aninsurance company that intends to insure the oil com-pany against ‘geological risk,’ also would be keen todetermine these probabilities.

It is clear that knowing the probability of occur-rence of certain events has great economic implica-tions. In this paper, we will be concerned only withthe probability of occurrence of some events, such asthe probability that a trap is filled to spill point or theprobability of success of a well. However, the prob-abilities calculated here should be the ‘raw material’for further economic calculations.

This paper has two sections. Although the twosections are concerned with the broad subject of prob-ability, they are conceptually different.

In the first section, we attempt to obtain infor-mation regarding the oil potential of a basin whenthe only piece of information is the drilling history

1491520-7439/01/0600-0149$19.50/1 C© 2001 International Association for Mathematical Geology



150 Anderson and Gullco

in the basin (number of wells, number of successes,and evolution of the success ratio through time). Thissort of information is easy to obtain, because it canbe provided by several companies at a relatively lowcost. It is true that a detailed ‘post mortem’ of all theexploration holes drilled in the basin (successful ornot) would provide the basis for a refined statisticalanalysis, from which the probability of occurrence ofcertain events could be rigorously derived. However,in such a situation, an enormous amount of informa-tion (if available) would have to be digested and thetime and money spent in this exercise would be sub-stantial. This is perhaps the only justification for theapproach taken here: fast and inexpensive. The ba-sic idea here is to divide conceptually the portion ofthe basin under study into a set of cells of equal areawhich may contain oil. With this model, several pre-dictions can be made, even though the only piece ofinformation available to us is the success ratio.

In the second section, we propose a probabilisticmodel to predict to which extent potential oil trapswill be actually filled with oil. The model has not beentested with real data because a certain amount of in-formation, which is in possession of oil companies, andnot readily accessible by the general public, has to bedigested. If the model proves to be correct, it wouldhave important practical applications. However, someoil fields have to exist in the basin and the original oilin place and the volume of the trap for each of thesefields, have to be known to be able to use the model.

A MODEL TO PREDICT THE DISTRIBUTIONOF FIELD AREAS AND ITS CONNECTIONWITH THE ‘SUCCESS RATIO’

Let us imagine a sedimentary basin (or a portionof it), of known area, which we subdivide conceptuallyinto a set of hexagons of unit area. The magnitude ofthe ‘unit area’ will depend upon which is the minimumarea that an oil field should have in order to be com-mercial. Onshore, for shallow targets, this minimumarea could be 0.5 km2. Offshore, it could be perhaps2 km2, depending on the seafloor and target depths.For the purpose of the following discussion, we willconsider the unit area to be 1 km2.

Each of the unit area hexagons can contain oilor not. The number of hexagons containing oil, di-vided by the total number of hexagons will be a num-ber (‘p’), which, as will be shown later, is connectedclosely with the ‘success ratio.’ The ‘success ratio’ canbe defined as the number of successful exploration

Figure 1. Portion of basin divided into hexagons, showing fieldsize 1 and field size 3.

holes drilled in a basin (or a portion of it), divided bythe total number of exploration holes.

A field is size one (i.e., has an area of 1 km2) ifit consists of a single cell with oil surrounded by sixempty cells. A field is size two (with an area equalto 2 km2), if it consists of two adjacent cells with oil,sharing a side, surrounded by eight empty cells, etc.Figure 1 represents a portion of a basin divided intohexagons, where a field size one and a field size three,are shown. We would be interested in determining theaverage field size and the distribution of field sizes.Our only piece of information is ‘p’, which at thisstage we will consider to be known exactly.

The geometrical model that we have selected torepresent the distribution of “full” and “empty” cells,can be regarded as a triangular lattice, where the cen-ter of each one of the hexagons coincides with eachone of the points of the lattice. In a triangular lattice,each point is surrounded by six equidistant points. Theselection of this lattice is of course, arbitrary. In thefollowing discussion we will assume that the spatialdistribution of “empty” and “full” cells, is completelyrandom and spatially independent.

The fact that the model can be regarded as a lat-tice of points which can belong to two mutually exclu-sive sets, allows the use of Percolation Theory (Shanteand Kirkpatrick, 1971) to predict characteristics of themodel. There is an abundant literature on this subjectand most of the problems in which we are interestedhave already been dealt with.



Probability Estimates of Field Areas and Trapped Oil Volumes 151

If we regard our lattice as a porous medium,where the cells (or points) with oil are considered as‘pores’ and the barren cells as ‘solids,’ there will be acritical value of p, below which all clusters of ‘pores’will be of finite size. However, when p is greaterthan this critical value, a cluster of infinite size or a‘percolation channel’ (Shante and Kirkpatrick, 1971),spreads through the medium. In other words, the av-erage size of the pore clusters becomes infinite. Thiscritical value is 0.5 for triangular lattices (Shante andKirkpatrick, 1971). This indicates that the quotient:cells with oil to the total number of cells, cannot begreater than 0.5. Otherwise, we would have oil fieldsof infinite size. It should be stressed that this resultholds if the spatial distribution of “empty” and “full”cells, is completely random.

Let us now define two discrete random variables,Y and P:

Yk = (Number of cells with oil belonging to fieldssize k)/(Total number of cells with oil)

Pk = (Number of fields size k)/(Total number offields)

It can be shown readily that the two variables arelinked by the equation:

Pk =Ykk∑∞

j=1Yj

j

(1)

It also can be shown readily the mean value of P,µ (i.e. the average value of the field sizes), is given by:

µ = 1∑∞j=1

Yj

j

(2)

The reason why we have introduced the random vari-able Y, is that in principle, it can be calculated by in-spection. Assume, for instance, that we pick at randoma cell with oil. To constitute a field size 1, this cell withoil must be surrounded by six empty cells. The proba-bility that the six cells will be empty is given by (1−p)6,where p is the number of cells with oil over the totalnumber of cells. The result comes directly from the bi-nomial law: we have zero successes in six trials (i.e. wehave no cell with oil amongst the six cells surroundingthe initial cell we picked). Hence, we can state.

Y1 = (1− p)6

Suppose now that we pick at random a cell withoil. To be part of a field size 2, there has to be justone cell with oil in the ring of six cells surrounding theoriginal cell. Furthermore, this second cell shares sideswith three additional cells. These three cells have tobe empty, so that the field remains size two. The prob-

ability of these two events occurring simultaneouslyis given by:

Y2 = 6p(1− p)8

Again, this expression results from the direct applica-tion of the binomial law.

We proceed in a similar fashion with fields size 3,getting the expression:

Y3 = p2(1− p)9[27(1− p)+ 6]

It is apparent that these expressions become increas-ingly complex as the field sizes or areas becomelarger. Furthermore, it is clear that there is no sim-ple analytical expression to describe this distribution.Fortunately, work done by Sykes and Glen (1976)on Percolation Theory, can be helpful. These au-thors have calculated the values of a function whichthey call Dk(q), where k is the cluster size (k = 1,2, . . . , n) and q = (1− p), up to k = 14. This func-tion is related closely to Yk/k, the exact relationshipbeing:

Dk pk−1 = Ykk−1 (3)

The following are some values of Dk, for triangularlattices, taken from Sykes and Glen (1976), being q =1− p. These formulae are valid for p < pcritical whichis 0.5 for triangular lattices.

For k= 1 D1 = q6

For k= 2 D2 = 3q8

For k= 3 D3 = 2q9 + 9q10

For k= 4 D4 = 3q10 + 12q11 + 29q12

As stated, these authors have calculated the func-tion Dk to k= 14. However, we would like to know thebehavior of Dk for bigger values of k and eventually,for k tending to infinity. For doing so, we have fittedthe 14 data points provided by Sykes and Glen with acontinuous function, for values of q ranging between0.99 and 0.55 (that is, for p ranging between 0.01 and0.45).

The following empirical formula has been usedto fit the data:

Dk = ak θλk (4)

In this expression, a, θ , and λ are parameters whichdepend on q and their values have been tabulated inTable 1, for q ranging between 0.99 and 0.55 (or pranging between 0.01 and 0.45).

Combining expressions (2), (3), and (4), we getthe average field size as:

µ = 1∑∞j=1 aj θλ j p j−1

(5)




Table 1. Values for a, θ , λ That Depend on q

p a(p) θ(p) λ(p) µ(p)

0.01 0.19069 −0.81581 5.01512 1.015620.02 0.18236 −0.82164 4.93446 1.048280.03 0.17430 −0.82762 4.85451 1.082540.04 0.16651 −0.83373 4.77528 1.118510.05 0.15899 −0.83999 4.69677 1.156300.06 0.15173 −0.84640 4.61898 1.196050.07 0.14471 −0.85297 4.54190 1.237900.08 0.13795 −0.85969 4.46554 1.282010.09 0.13142 −0.86659 4.38990 1.328550.10 0.12513 −0.87366 4.31498 1.377710.11 0.11907 −0.88090 4.24077 1.429700.12 0.11324 −0.88834 4.16729 1.484760.13 0.10762 −0.89596 4.09452 1.543140.14 0.10221 −0.90378 4.02247 1.605120.15 0.09701 −0.91181 3.95113 1.671040.16 0.09201 −0.92005 3.88052 1.741240.17 0.08721 −0.92852 3.81062 1.816130.18 0.08260 −0.93721 3.74145 1.896150.19 0.07818 −0.94615 3.67299 1.981820.20 0.07393 −0.95533 3.60525 2.073710.21 0.06987 −0.96476 3.53823 2.172480.22 0.06597 −0.97447 3.47193 2.278880.23 0.06224 −0.98445 3.40634 2.393760.24 0.05867 −0.99472 3.34148 2.518120.25 0.05526 −1.00528 3.27733 2.653110.26 0.05200 −1.01617 3.21391 2.800040.27 0.04888 −1.02737 3.15120 2.960500.28 0.04591 −1.03891 3.08922 3.136300.29 0.04308 −1.05081 3.02795 3.329630.30 0.04038 −1.06307 2.96740 3.543070.31 0.03781 −1.07571 2.90757 3.779730.32 0.03537 −1.08875 2.84846 4.043340.33 0.03305 −1.10221 2.79007 4.338440.34 0.03084 −1.11609 2.73240 4.670570.35 0.02875 −1.13044 2.67545 5.046510.36 0.02677 −1.14525 2.61921 5.474580.37 0.02489 −1.16055 2.56369 5.965040.38 0.02311 −1.17637 2.50890 6.530500.39 0.02143 −1.19273 2.45482 7.186390.40 0.01985 −1.20965 2.40145 7.951500.41 0.01836 −1.22716 2.34881 8.848460.42 0.01695 −1.24529 2.29688 9.904230.43 0.01563 −1.26407 2.24567 11.150570.44 0.01438 −1.28352 2.19517 12.624540.45 0.01321 −1.30368 2.14539 14.36916

The corresponding values of µ have been in-cluded in Table 1. It should be pointed out that theseries in the denominator has been calculated nu-merically. It converges rapidly for small values of p,but for values of p greater than 0.40, a great num-ber of terms have to be taken into account. For theparticular example in which θ ∼= −1, correspond-ing to a value of p of about 0.245, the series canbe expressed in a closed form, and the theoretical

and numerical results agree up to the fifth decimalpoint.

At this point it would be worthwhile to mentionthe problems brought about by expressing Sykes andGlen polynomials as a continuous function. The aver-age field size should be infinite for p = 0.50 and thiscould be accomplished if the parameter ‘a’ in Equa-tion (5) were equal to zero for p = 0.50. An inspec-tion of Table 1 shows that indeed the parameter ‘a’decreases as p increases, but calculations show thatalthough small, it is not zero for p = 0.50. For thisreason, values of the parameters have not been cal-culated for p> 0.45. However, for values of p< 0.45,there has been a good agreement between the use ofthe empirical formulae developed here and simulatedexamples. In summary, these simulations are the onlyindication that these empirical formulae are usefulwhenever p< 0.45.

The distribution of field areas now can be ex-pressed as:

Pk = aµk θ pk−1λk (6)

The four parameters in this equation (a, µ, θ , andλ) are functions of p and are given in Table 1. ‘p’ isthe only variable which has to be known to use theformula. Because we will never know ‘p’ exactly (wewill have only an estimate from the success ratio, asshown in the next section), the tabulated values of theparameters, corresponding to values of p with just twodecimals, will be enough for any practical application.

Equation (6) actually is independent of the realarea of the individual hexagons. The number k (k =1, 2, 3, . . . , n) is the ‘size’ of the field. So, a field size1 has a real area of 1.0 km2, if we have selected thisnumber as the area of each individual hexagon. For kequal 2, the area will be 2 km2, etc.

Note that the distribution is discrete, that is, thepossible areas are 1 km2 (for k = 1), 2 km2, (for k =2), etc. If we had taken the unit area as 2 km2, thepossible areas would be 2 km2 (for k= 1), 4 km2 (fork= 2), 6 km2 (for k= 3), etc.

Examples of the practical use of Equation (6) willbe given after discussing the connection of p with thesuccess ratio.

RELATIONSHIP BETWEEN p AND THESUCCESS RATIO

If a great number of wells were drilled in a sed-imentary basin of finite size, and all the wells weredrilled at random, the quotient ‘successful wells/total




number of wells’ would tend to a limit equal to thequotient ‘sum of the areas of all the fields existing inthe basin/total area of the basin.’

The limit is numerically equal to the probabilityof drilling a successful exploration well in just onetrial. The situation is completely analogous to the tossof a coin. If we toss a coin once, the probability of headis 1/2. If we toss a coin a great number of times, thequotient ‘number of heads/number of trials’ will tendto 1/2.

The quotient ‘number of successful explorationwells/total number of exploration wells’ generally istermed the ‘success ratio.’ The observed success ra-tio is related, as shown previously, to an importantgeological quantity, the sum of the areas of all thefields/total area. However, the success ratio is in facta sample and would coincide with the quotient of ar-eas only if infinite wells were drilled. We may toss acoin ten times and get just one head, even though theprobability of one head in one trial is 1/2. In an explo-ration context we do not know what is the true valueof the quotient (which we term the ‘mineralized areafraction’) and we will have to estimate it from a sam-ple, the success ratio. In our triangular lattice model,the ‘mineralized area fraction’ is equal to the num-ber of hexagons filled with oil to the total number ofhexagons (p). We also know that this number has tobe less than 0.5 (otherwise we would have fields ofinfinite size).

In the following discussion it will be assumedthat:

(a) The exploration wells have been drilled atrandom. This is somewhat difficult to justify,because a lot of time and effort has been in-vested in locating the ‘right spot’ to drill thewell. However, at an initial stage in the explo-ration of a basin, when seismic lines are per-haps 3 or 5 km apart, there is little well controland not much is known of the regional geol-ogy, it can be argued that the wells are drilled‘at random.’ Once there are plenty of seismicand well data, the probability of drilling a suc-cessful well at least in theory, should, increase.At any rate, an increase of the success ratiowith time, should be apparent and the hypoth-esis could be tested in several ways. We willassume here, that the success ratio, on aver-age, is independent of time.

(b) Only exploration wells are considered. Ajudgement has to be made to separate explo-

ration from appraisal wells, because the latter,of course, are not independent of the former.

(c) A well is considered successful only if it lo-cates a commercial accumulation of oil.

(d) We will assume that we are sampling ‘withreplacement,’ even though the portion of thebasin we are concerned with, has a finite sizeand there is a finite number of commercial oilaccumulations. It can be shown that this is areasonable assumption.

As stated, although the mineralized area fractionhas just one value for a particular area, we will benever able to know it exactly. For this reason, we willconsider it to be a random variable, which we willterm X. We will try to estimate the characteristics ofthis random variable through the success ratio, theonly piece of information available to us. The problemwill be dealt with differently, depending on the totalnumber of exploration holes.

Case 1: Thirty or more exploration holes havebeen drilled in the area of interest.

In this case, the theory of the sampling distri-bution of proportions for big samples, is applicable.The resulting distribution of X is approximately nor-mal, with a mean equal to R (the success ratio) and avariance equal to R (1− R)/N, where N is the totalnumber of wells drilled in the area.

Knowing the distribution of X, allows us to calcu-late the probability that X> 0.5. According to our tri-angular lattice model, X (that is p, the number of cellswith oil to the total number of cells), cannot be greaterthan 0.5. If the probability that X> 0.5 is greater thansome critical value (perhaps 20% or 30%), we shouldhave to interrupt the analysis, because that would in-dicate that there is a certain probability that the wellswere not drilled ‘at random.’ However, in order tohave this situation, the success ratio itself should beclose to 0.50. So, p, the number we are looking for, isequal to the mean of X, which is the success ratio.

As an example, let us assume that 30 wells havebeen drilled in the area, of which six have been suc-cessful (R = 0.20). Then the mean of X is 0.20 andits standard deviation is 0.073. There are then 68%chances that the true value of X will lie between 0.127and 0.273. In other words, if the area of the basin is10,000 km2, the sum of the areas of all commercial oilaccumulations in the basin will lie (with 68% chances)between 1270 and 2730 km2.

Case 2: Less than thirty exploration holes havebeen drilled in the basin.




In this case, we can consider that the random vari-able X has the Beta distribution. This can be derivedby direct application of Bayes’ formula. The mean andvariance of the beta distribution are given by:

µ = (k+ 1)/(N+ 2);

σ 2 = (k+ 1)(N− k+ 1)/[(N+ 2)2(N+ 3)],

where N is the total number of exploration wells andk is the number of successful wells.

In this example, the mean of X, and not the suc-cess ratio, is our estimate of p. From the propertiesof the Beta distribution, it is possible to check thechances of being p greater than 0.50. Again, p shouldnot be accepted as an estimate of the mineralized areafor total area, if the probability of being greater than0.50 exceeds certain critical value.

APPLICATIONS OF THE TRIANGULARLATTICE MODEL

A usual situation encountered in oil explorationis that a structure has been detected by seismic and theoperator wants to know whether the structure is filledto spill point. If statistical information regarding thedegree of filling of the structures is unavailable, butat least there is an estimate of the mineralized areafraction (via the success ratio, a piece of informationmore easy to get), the triangular lattice model can beused to predict whether the structure is filled to spillpoint.

Consider the following example:A structure has been detected through seismic,

which has an estimated rock volume of 100,000,000cubic meters to spill point. Assuming a porosity of0.20 and an irreducible water saturation of 0.30, thestructure could contain 88 million barrels of oil if filledto spill point. We do not know whether the prospectwill be commercial, but if it so, its minimum area hasto be 1.0 km2. We can ask then the question: if the fieldis commercial, what is the probability that it is filledwith oil to spill point? Certainly, it makes a differencein attracting investors for drilling the well, to knowwhether is likely to have 88 million barrels or 8 millionbarrels of oil.

For developing the example, let us consider thefollowing geometry for the trap: the structure is aspherical cap, having an elevation of 50 meters be-tween the culmination and the spill point level. Thereservoir sand is supposed to be at least 50 meters

thick. The area of the trap, if filled to spill point,would be 4 km2. The radius of the sphere to whichthe gasket belongs (calculated from the seismic data)is 12,750 km. There is a direct relationship betweenthe area of the field and the volume of oil that it con-tains, given by:

V = (π/6)φ(1− Sw)[R− (R2 − A/π)0.5]

×{3R+ [R− (R2 − A/π)0.5]2}The symbols are:

V: oil volumeA: Area of the fieldR: Radius of the spherical cap (derived from

seismic)φ: Porosity

Sw: Water saturation

In a real situation, the equivalent of this equationwould have to be integrated numerically. Althoughthe porosity and water saturation are unknowns, theirranges are limited. The greatest uncertainty is the areaof the field, which is a function of the oil/water contactlevel. However, we know the probability distributionof the field areas from the triangular lattice model andhence, we can use the Monte Carlo method to deter-mine the probability distribution of the oil volume.

Table 2 shows the results of a Monte Carlo sim-ulation. A success ratio of 0.40 was used (in fact theonly piece of statistical information that we have) andthe unit area was 1 km2.

The probability of the structure being filled tospill point (that is, containing 88.17 Mmbbls, corre-sponding to an area of 4 km2, the maximum possible)is low (just 10.8%). In fact it is more likely that thestructure will contain about 5.5 Mmbbls, correspond-ing to a field area of 1 km2. Because fields size 1 arethe most frequent, the results are what one would ex-pect to be, according to the triangular lattice model.It should be remarked, however, that this approachmakes sense if the success ratio is the only piece ofstatistical information available to the operator.

Table 2. Results of Monte Carlo Simulation

Oil volume CumulativeField area (km2) (MMbbls) Frequency frequency

1 5.5 0.534 0.5432 22.01 0.192 0.7263 49.57 0.166 0.8924 88.17 0.108 1.000




A PROBABILISTIC MODEL TO EXPLAINTHE FILLING OF OIL TRAPS AND ITSPRACTICAL APPLICATIONS

Let us imagine the following simple experiment:we have 10 balls and we have two boxes, which we willlabel A and B, capable of containing 5 and 10 balls, re-spectively. Assume that we fill the boxes with 10 ballsat random. We may ask what is the probability thatbox A will contain no balls or what is the probabilitythat box A will be filled to capacity.

The possible configurations of the system are:

(1) Zero ball in box A and 10 balls in box B.(2) One ball in box A and 9 balls in box B.(3) Two balls in box A and 8 balls in box B.(4) Three balls in box A and 7 balls in box B.(5) Four balls in box A and 6 balls in box B.(6) Five balls in box A and 5 balls in box B.

Let us now label the balls 1 to 10 and see forinstance, in how many ways configuration (2) can beachieved. Box A may contain ball number 1, or ballnumber 2, or ball number 3 . . . or ball number 10. Sothis configuration can be achieved in 10 different ways.

In general, the number of ways in which any con-figuration can be achieved is given by N!/[r ! (N− r)!],where in this example, N is equal to 10 (total numberof balls) and r is the number of balls that are in boxA. Results are summarized in Table 3.

The probability of each configuration is given bythe number of ways in which each configuration canbe achieved divided by the total number of ways.

So, the probability that the small box is empty isalmost nil. Furthermore, the most likely configurationis the one in which the small box is full or ‘filled to spillpoint.’

These ideas are of application to the way in whichpotential oil traps are filled with oil in any sedimentarybasin. It is impossible to pinpoint where a particularoil droplet has been generated and the extremely tor-tuous path that the droplet follows to arrive in a trap.

Table 3. Results of Configuration of Balls in a Box

Configuration Balls in box A Balls in box B Frequency

1 0 10 0.0012 1 9 0.0103 2 8 0.0534 3 7 0.1035 4 6 0.1946 5 5 0.639

The process of filling the traps with oil droplets in alegitimate way, can be considered to be a random pro-cess. As in thermodynamics, we cannot predict the be-havior of an individual molecule of gas, but we shouldbe able to predict the average properties of the systemand at least, the probability that certain events mayoccur.

So, instead of boxes, we can talk about potentialoil traps, and, instead of balls, of ‘oil units.’ We want tolocate a certain number of ‘oil units’ in a certain num-ber of potential traps. How will the oil be distributed?

In the previous example with the boxes and theballs, the probability of an individual ball ending upin any of the boxes, was independent of the size ofthe boxes. However, in real life, we would expectthat there should be some sort of relationship be-tween the probability of an individual oil unit end-ing up in a particular trap and some geometric char-acteristic of the trap, such as its volume or surfacearea.

In the following discussion we will assume thatall the potential oil traps in the basin have a similarshape, and that the probability that an oil unit willend up in a specific trap is proportional to vk, where‘v’ is the total capacity of the trap and and k is a realnumber. For instance, in the example of the boxes, kwould be zero. If we postulate that the probability isproportional to the surface area of the trap, k wouldbe 2/3.

Consider now a trap of total volume or capacityequal to v, which contains a volume of oil equal to vo.The amount of oil in the trap will be proportional tovk. Then, the filled fraction, p of a field with capacityv will be:

p= Cvk−1, where C is some constant. However,if k is less than 1, then p→∞ as v→ 0. So all fieldswith v less than some critical capacity vc, will be filledto spill point and therefore:

p = 1 for v <= vc

p = Cvk−1 for v > vc

Because p has to be continuous, we have:

C = vk−1c

So, the volume of oil in a given field of capacity v is:

vo = v for v <= vc

vo = vck−1 vk for v > vc

Then, if the capacity of the traps has a certain proba-bility distribution (such as exponential or lognormal,




etc), we have:

Vo

V= r =

∫ Vc0 v f (v)dv + Vc1−k

∫∞Vc vk f (v)dv∫∞

0 v f (v)dv(7)

In this equation, the variables are:

Vo = Total volume of oil in the basin.V = Sum of the volumes of all the traps in the

basin.VC = Critical trap volume. All the traps with this

volume or less, are filled to spill point.f (v) = Density probability function for the trap vol-

umes.

As an example of the use Equation (7), assume thatthe volume of the potential traps existing in the basin,have a uniform distribution. From now on, we willtake k= 2/3, indicating that we assume that the prob-ability of an oil unit going to a trap of volume v, isproportional to its surface area.

If f (v) = 1/Vm, (uniform distribution) where Vm

is the capacity of the trap of maximum volume, aftermaking the appropriate substitutions in Equation (7),we have:

VoV= r = 0.6Vc

13 V

53

m − 0.1Vc2

0.5V2m

(8)

We can simulate now the filling of a certain number oftraps whose volumes are distributed uniformly, with acertain number of oil units and check the accuracy ofEquation (8). To make such a simulation we proceedas follows:

(1) Take a sample size N from a uniform distri-bution. The trap volumes will range between0 and Vmax, a given value.

(2) Sort the sample from the minimum to themaximum volume and calculate the cumula-tive volume of the N traps.

(3) Enter a value for ‘r ,’ which is the quotienttotal volume of oil/total volume of the traps.The total volume of oil, Voil, will be equal tor times the total volume of the traps. Let usmake Voil an integer, Noil, which representsthe number of ‘oil units.’ The volume of allthe traps are also transformed into integers.

(4) For each trap volume, Vtrap, calculateVtrap2/3. Calculate cumulative distribution ofVtrap2/3 and divide each number by the total(i.e., resulting values will be in ascending or-der from a small number to 1).

(5) Take a random number between 0 and 1. Ifthe random number is less or equal than thecumulative value of Vtrap2/3, an oil unit isassigned to this particular trap.

(6) The process is repeated Noil times, until allthe oil units are assigned to the different traps.

(7) If a particular oil unit is assigned to a trapwhich is already full, (i.e, filled to ‘spill point’),new random numbers are generated until thisoil unit locates a trap which is not full.

An example of such a simulation is shown inFigure 2. The abscissa is the logarithm of the trap vol-ume and the ordinate is the logarithm of the volumeof oil contained in the trap. The outstanding featuresof the plot are the following:

(1) Two straightlines can be seen on the plot. Onehas a slope equal to 1 and passes through theorigin. This line represents the traps whichare filled to spill point. The other line has aslope of approximately 2/3 and intersects theother line at the ‘critical volume,’ Vc. All thetraps having a volume greater than Vc cannotbe filled to spill point.

(2) The approximate value of the critical volumecan be determined, then, by inspection.

These basic features are independent of the actualprobability distribution of the trap volumes.

For the particular case of Figure 2, the simula-tion was performed using a value of Vm (volume ofthe maximum trap) equal to 1000 units and a valueor r (total volume of oil to total volume of the traps)equal to 0.5. By inspection, the approximate valueof the logarithm of the critical volume is about 1.9.The predicted value of the critical volume, accord-ing to Equation (8) is 73 units and its logarithm is1.86, close enough to the value of 1.9 observed in thesimulation.

The ideas outlined here could be checked in anymature basin, where there are a great number of oilfields and there are reasonable accurate estimates ofthe volume of the original oil in place and the volumeof the trap for each one of the fields. It would be idealto work with fields where the gas cap volume, at reser-voir conditions, is negligible as compared with the oilleg volume, to keep things simple.

The methodology to follow would be to plot thelogarithm of the trap volumes on abscissa and the log-arithm of the oil volume on ordinate, and determine iftwo approximately straightlines occur. For small trap




Figure 2. Simulation results. Trap volumes are uniformly distributed.

volumes, the line should have unit slope. For large trapvolumes, the line should have a slope of about 2/3.There should be more dispersion of the data than inthe simulated examples, because the shape of the trapswould be variable and equality of volumes would notimply equality of surface areas. Of course, it mighthappen that the oil fraction, ‘r ’, were small and thatalso would make the critical volume small. In such asituation, there would be only one line with a slope ofabout 2/3 and none of the traps would be filled to spillpoint. Conversely, for a large value of the oil fraction,there might be just one line with unit slope and all thetraps would be filled to spill point.

If these ideas were proved reasonable, theywould have several practical applications, providedthat there is already a small sample of fields of knownvolume in the basin.

(1) If the log-log plot suggests the presence oftwo straightlines, then the critical volume can be cal-culated from inspection. If a company is going to drilla prospect where the trap volume is known, then itcan be predicted that the trap will be filled to spillpoint if the volume of the trap is less than the criticalvolume. If the volume of the trap is greater than thecritical volume, then, the volume of oil in the trap canbe calculated approximately by:

vo = VC1/3V2/3, where vo is the oil volume, VC the

critical volume and v the trap volume. The value ofthe oil volume should be regarded as an ‘expectedvalue.’

If the log-log plot shows just one line of slope2/3, then the critical volume can be calculated fromthe data and hence predict whether the prospectwill be filled to spill point. If not, the above equationis used again to calculate the expected volumeof oil.

If the log-log plot shows just one line of unit slope,and the volume of the prospect is less than the max-imum observed trap volume, the prospect would beexpected to be filled to spill point. If the volume ofthe prospect is greater than the maximum observedtrap volume, a lower limit can be assigned to the oilvolume, assuming that the maximum observed trapvolume corresponds to the critical volume. The upperlimit, of course, is being filled to spill point.

(2) An important consequence of the model isthat, provided there is some oil in the basin, all thetraps would contain a certain amount of oil. If the oilfraction is small, the resulting critical volume would besmall and, according to the equation vo = VC

1/3v2/3,the oil volume in the traps also would be small andhence uncommercial.




However, in a basin where Vc is seen on the log-log plot and, if any four-way closure prospect, havinga trap volume less than the critical, proves to be drywhen drilled, one would be entitled to think that asa result of uncertainty in seismic velocities or poorseismic data, the structure actually does not exist asmapped. This explanation would be more likely thanthe assumption that the structure was never chargedor that there was a ‘leak,’ because, if the proposedmodel is reasonable, all the potential traps should con-tain oil.

Some oil companies estimate the probability thata well will be successful multiplying four probabilities:the probability that the structure, as perceived seismi-cally, exists, times the probability that there is a seal,times the probability that there is reservoir rock, timesthe probability that the trap contains oil. Althoughthis approach is not rigorous, it gives at least an ideaof the ‘goodness’ of a specific prospect. However, oneshould be able to eliminate the ‘probability that thetrap contains oil,’ because that would be a certaintyunder the proposed model, provided there is enoughempirical data which could be analyzed by a log-logplot.

CONCLUSIONS

In the first part of this paper, a portion of asedimentary basin is subdivided conceptually intohexagons of equal area. These hexagons have an areaequal to the minimum area an oil field should have tobe commercial. Hexagons can be “full” or “empty.” Afield of size 1 consists of one empty cell surrounded bysix empty cells; a field of size 2 consists of two adjacent“full” cells surrounded by eight empty cells, etc.

The probability distribution of the areas of thefields, can be estimated by applying Percolation The-ory principles. This probability distribution can be ex-pressed as a function of a single parameter, p, which

is the quotient of the number of cells with oil to thetotal number of cells. From Percolation Theory, it canbe shown that the number ‘p’ has to be less than 0.50.There is a close relationship between ‘p’ and the suc-cess ratio, and according to the proposed model, thesuccess ratio has to be less than 0.50, provided it isaccepted that the wells are drilled ‘at random.’

In the second part of this paper, a probabilisticmodel has been introduced to predict the extent towhich potential oil traps are filled with oil. The modelassumes that the probability that an oil unit will endup in a particular trap, is proportional to the surfacearea of the trap.

The model predicts that independently of the dis-tribution of the trap volumes, there will be a criticaltrap volume. All the traps having a volume less thanthis critical volume, will be filled to spill point. Thetraps having a volume greater than the critical willcontain oil according to the equation vo = VC

1/3 v2/3,where vo is the oil volume, VC, the critical volume andv the trap volume.

The critical volume could be, in real situations,determined by making log-log plots of oil volume vs.trap volume for known fields.

If the model proved to be realistic, it would haveimportant practical applications.

ACKNOWLEDGMENT

The authors wish to thank Daniel Tetzlaff for hisvaluable comments and suggestions.

REFERENCES

Shante, V. K. S., and Kirkpatrick, S., 1971, An introduction to per-colation theory, in Advances in Physics: Taylor & Francis Ltd.,London, v. 20, no. 85, p. 325–357.

Sykes, M. F., and Glen, M., 1976, Percolation processes in two di-mensions I. Low density series expansions: Jour. Physics. A,Math. Gen., v. 9, no. 1, p. 87–95.

probability estimates of field areas and trapped oil volumes

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