lecture 2b more prob

8/13/2019 Lecture 2B More Prob

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Geostatistics for ReservoirCharacterization

Lecture 2b More probability

Geostatistics for ReservoirCharacterization

Lecture 2b More probabilityIn this section we will continue covering the

following topics ...

Combination of Events

Basic Operations

Conditional Probabilities


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Definitions . . .

Experiment - using equipment to measure asample under given conditions (T)

Hassler cell for gas perm; density log; drill-stem test

Sample space - all possible values that the

measurement might give 0 < kplug


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Example 1

Net Pay and Core Data

0.01

0.1

1

10

100

0 5 10 15 20

Per

meability,mD

Porosity, %

86 pts

86 experiments, = {0 < kplug 1 md is an event E1


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Example 2


0.01

0.1

1

10

100

0 5 10 15 20

Per

meability,mD

Porosity, %

86 pts

86 experiments, = {0 10% is an event E2


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Combining Events . .

Intersection

E1 E2 x x E1and x E2

the symbol means is an element of

E1

E2


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Example 3


0.01

0.1

1

10

100

0 5 10 15 20

Per

meability,mD

Porosity, %

86 pts

86 experiments, = {0 < kplug 10% is an event E2

E1

E2


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Example 2 from textIntersection & Logging Tool

Responses

E1 E3 Facies 3

Facies 1 Facies 2

Facies 3

Porosity

GR

GR

GR

GR1

2

3

4

1 2 4 3

E

E

E E

1

2

3 4

-

-

-

-


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Basic Operations . .Union

E1 E2x x E1 or x E2}

E1

E2


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Example 2, textUnion & Logging Tool Responses

E1

E3

Facies 1 and Facies 3

Facies 1 Facies 2

Facies 3

Porosity

GR

GR

GR

GR1

2

3

4

1 2 4 3

E

E

E E

1

2

3 4

-

-

-

-


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Basic Operations . .Complement

AE

Ec

= {x: x E}


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Basic Operations . .Null Set

E1 E2 E1and E2are disjoint or mutuallyexclusive

AB

E1E2


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Probability . . .

Define an event E

Make n measurements (T)

Suppose m measurements satisfy E

We define the probability of E to be:

Prob(E) = limn (m/n)


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Probability - Examples Procedure consists of two parts

An identifiable event E, e.g.,

E = Facies A at a particular location

E = 100 < k < 300 md for a core plug

A number p expressing event likelihood e.g.,

Of 300m gross pay in a well, 240m is productive

E = One meter of productive or net payp = Prob(E) = 240/300 = 0.80

Eighteen of thirty well tests had k > 100 md

E = well test with k > 100 mdp = Prob(E) = 18/30 = 0.6

Interpret probability as a frequency


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Probability Example

Assessing Net Pay from Core Data

0.01

0.1

1

10

100

0 5 10 15 20

Permeability,mD

Porosity, %

86 pts

Let E1 = {k | k > 1 md}

Prob(E1) = 47/86 = 0.55


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Probability Example


0.01

0.1

1

10

100

0 5 10 15 20

Permeability,mD

Porosity, %

86 pts

Let E2 = { | > 10%}

Prob(E2

) = 38/86 = 0.44


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Key Point . . .

If areas on Venn Diagram are

proportional to m(E1) and m(E2),they represent probabilities, e.g.

Prob(E1c) = 1- Prob( E1)

Leads to fundamental algebraof probabilities


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Fundamental Law of Addition . . .

Prob(E1 E2 (and)

Prob(E1)+ Prob(E

2)-Prob(E1E2Prob(E1 E2 Prob(E1)+ Prob(E2)

for E1 ,E2 mutually exclusive(probabilities add)


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Conditional Probability . . .

Prob(E1E2) = Conditionalprobability (probability of E1 given

than E2 has occurred)

Pr ob(E1

| E2

)Pr ob(E1 E2 )

Pr ob(E2 )


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Schematic of Conditional Probability . . .

E1 E2

Prob(E 1| E 2) =


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In terms of Prob definition . . .

E2

E1 E2

Prob(E 1 | E2) =P(E1 E2)

P(E2)


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Probability Example 6, text


0.01

0.1

1

10

100

0 5 10 15 20

Per

meability,mD

Porosity, %

86 pts

Recall E1 = {k | k > 1 md} and E2 = { | > 10%}

Prob(E1) = 0.55 and Prob(E2) = 0.44

Prob(E1 E2) = 36/86 = 0.42 so


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Probability Example


0.01

0.1

1

10

100

0 5 10 15 20

Per

meability,mD

Porosity, %

86 pts

Prob(E1 | E2) = Prob(E1 E2)/Prob(E2) = 0.42/0.44 = 0.96so IF > 10%, we have a 96% chance of having net pay

On the other hand...


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Probability Example


In summary, based on core data given,

> 10% is reliable indicator of net pay (96% correct)

< 10% is less reliable non-pay (1 - 0.23 = 77% correct)

0.01

0.1

1

10

100

0 5 10 15 20

Permeability,mD

Porosity, %


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Cond. Prob. . . . Special cases

E2

E1

Prob(E1E

2)

Prob(E1)

Prob(E2)


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Cond. Prob. . . . Special cases

E1

E2

Prob(E1|E2 ) = 1


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Cond. Prob. . . . Special casesNull Event (ME events)

E1 E2

Prob(E1E2) = 0Prob(E1cE2) = 1


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Multiplication Law

Conditional probability definitionmore commonly written as . . .

Prob(E1E2) = Prob(E1E2) Prob(E2)


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An Important Special Case. . .

For E1 and E2 independent

Prob(E1 | E2) = Prob(E1) . . . then

Prob(E1 E2 ProbE1 ProbE2(or)

(probabilities multiply)


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Example . . . Probability of Rolling a Seven

with two dice

Prob(16) = (1/6)(1/6) = 1/36Multiplication Law . . . Die independent

Prob(16)+ Prob(25) + Prob(34) + . . . =1/36 + 1/36 + 1/36 + . . .= 1/6

Addition Law . . . Each outcome mutually exclusive


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Conditional Probabilities Uses . . .

Evaluating independence Probability trees

Bayes theorem


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Evaluating Independence . . .

Example 7, Text

Let A, B, C be facies in 5 vertical wells

Posi ti on Well 1 Well 2 Well 3 Well 4 Well 5

Top C B A B AMiddle A A B A B

Bottom B C C C C


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Evaluating Independence . . .

Let E1 = A is middle facies

E2 = C is bottom facies

Are E1 and E2 independent?

Prob(E1) = 3/5

Prob(E2) = 4/5 , and

Prob(E1E2 . . . henceProb(E1) Prob(E2) = 12/25 Prob(E1E2 E1 and E2 are not independent (i.e. there is a

characteristic sequence in the facies data)


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Probability Tree for Exploration . . .

F4 F4F4F4 F4 F4

F4 F4 S4S4S4 S4

S4S4S4S4

Prob( 2 successes)X10 4: 144 108 72 64164872 128 96 64-

0.20.40.6 0.80.80.8 0.80.9 0.20.2 0.20.4 0.40.60.60.1

F3F3F3F3 S3 S3 S3 S3

0.2 0.20.80.8 0.6 0.40.10.9

F2F 2 S2S2

0.9 0.1 0.8 0.2

F1S1

Start

162

0.10.9

Want to drill four wells with initial 0.1 prob of successSuccess prob doubles with previous success

What are probabilities of various outcomes?


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Bayes Theorem . . .Can express Prob(E1E2) two ways Prob(E1E2) = Prob(E1E2) Prob(E2)

or

Prob(E1E2) = Prob(E2E1) Prob(E1) Equating above gives cond prob formula

Prob(E 1|E 2) =Prob(E 2|E 1) Prob( E 1)

Prob( E 2)

E l E i i N P i h


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Example - Estimating Net Pay with

GRLet

E1 = k > 1 mD (ie net)

E2 = GR < 30 API

-The field NTG is 0.7

-Core GR showsGR < 30 for 60% of the time when k > 1

GR > 30 for 90% of time when k < 1

-New well shows GR < 30 for 80% of interval

-What is best NTG estimate for the new well?


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Example - Estimating Net Pay with GR E1 = k > 1 mD (ie net) E2 = GR < 30 API

-Probabilities BEFORE new well is measured

Field NTG is 0.7: Prob(E1) = 0.7

Core GR < 30 for 60% when k > 1: Prob(E2

|E1

) = 0.6

Core GR > 30 for 90% of time when k < 1: Prob(E2c|E1

c) = 0.9

-Probabilities AFTER new well is measured

GR < 30 for 80% of interval: Prob(E2) = 0.8


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Key Points . . .

Definitions:

Null set, complement, mutually

exclusive, independent, union,

intersection, a priori, a posteriori Fundamental laws: addition and

multiplication

Conditional probabilities

Bayes theorem

lecture 2b more prob

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