Probability of Geologic Success Associated with an Amplitude Anomaly
By David Swinehart

How does the presence of an amplitude anomaly affect the probability of finding oil and gas? How should the amplitude anomaly affect the probabilities as calculated on BR’s geologic chance worksheet? Is there a consistent way to tie amplitude anomalies to the geologic chance of success? Thomas Bayes, an English clergyman, answered those questions more than two hundred and thirty years ago. In an excellent paper in The Leading Edge1 Richard Houck of Mobil demonstrates the use of Bayes’ Theorem to calculate the chance of success using AVO. However, the methodology is not limited to AVO. It is attempted here to generalize the approach taken by Houck and to present some elementary examples.

We understand that the probability of two independent events happening is the product of each happening. For example, the probability of getting two heads on coin flips is 0.5x0.5 or 0.25. This is what is done on BR’s geologic chance worksheet, when we multiply the probability of source rock times the probability of reservoir rock. It is assumed that the presence of one has no effect on the presence of the other.

What if the two events are not independent? This is the case with amplitude anomalies. The presence of amplitude anomalies affects our opinion as to presence of source rock and reservoir rock. Conversely, the presence of source rock and reservoir rock affects the probability that an amplitude anomaly will be observed.

This is where the Rev. Bayes comes in. Bayes’ Rule represents the probable state of nature after some experiment has been preformed. (Bayes’ Rule is used in BR’s multi-zone evaluation worksheet and the value of information worksheet. See appendix A for a pictorial explanation of Bayes’ Rule.) Bayes’ Rule in exploration words states:

Probability of an amplitude anomaly given gas
New P(gas)=old P(gas)________________________________________________________________________
Total Probability of an amplitude anomaly

In the above, old P(gas) is the probability of finding gas ignoring the amplitude anomaly, and new P(gas) is the probability of finding gas given that we have an amplitude anomaly. The numerator is the probability of seeing an amplitude anomaly, if there is gas. If whenever gas is present you see an amplitude anomaly then the numerator would be 1.0. If sometimes you have gas with no amplitude, the numerator would be less than 1.0. The less than 1.0 case accounts for false negatives. The denominator includes what’s in the numerator and adds to it the probability of observing an amplitude anomaly when there is no gas present. It thereby accounts for the false positive case. (See the appendix B for an "easy" example of Bayes’ rule using dice.)

The above formula can be written somewhat more formally as:

P(a.a.|gas)
P(gas|a.a.) = P(gas) _________________________________
P(gas) P(a.a.|gas) + P(no gas)P(a.a.|no gas)

where P(gas|a.a.) is read as the probability of gas given an amplitude anomaly; P(gas) is the probability of gas without regard to amplitudes (your geologic as opposed the geophysical probability); P(a.a.|gas) is the probability of an amplitude anomaly given gas is present; P(no gas) is the geologic probability of no gas [just 1-P(gas)]; and P(a.a.|no gas) = the probability of having an amplitude anomaly like the one you have, if there were no gas present.

To illustrate the point lets consider a trivial example. Assume the Pg worksheet came up with a probability of finding gas of 0.2, and amplitude anomalies were ignored in deriving that number. Now by definition

P(gas) + P(no gas) = 1.0

therefore

P(no gas) = 1 – 0.2 = 0.8

To make it really easy lets assume that every time gas is present an amplitude anomaly is present, and we never see an amplitude anomaly like ours when gas is not present. What is P(gas|a.a.), the new probability of gas? If you said 1.0; then you’ve followed the discussion up to this point, and it gets easier from here. Although we already know the results, let’s do the numbers.

We’ve assumed,

P(a.a.|gas) = 1.0 and P(a.a.|no gas) = 0.0. [Do you see where these came from?]

then

1.0
P(gas|a.a.) = 0.2 ___________________________ = 1.0
0.2x1.0 + 0.8x0.0

That is to say, if an amplitude anomaly like ours is always present when there is gas and is never present when there isn’t gas then, we can be certain we have gas here.

What if we sometimes get amplitude anomalies like ours even when gas is not present, but we always get amplitudes anomalies when gas is present? Will our P(gas|a.a.) = 1.0? If you said "No", then you get a passing grade for the course. Want to try for a B? Assume

P(a.a.|no gas) = 0.5

That is, half of the time we get amplitude anomalies like ours when gas is not present (for example, fizz water). We are still assuming that when gas is present we always get an amplitude anomaly. So what’s the answer?

1.0
P(gas|a.a.) = 0.2 ________________________ = 0.3333
0.2x1.0 + 0.8x0.5

If you got the above answer, you’ve got at least a B going.

Let’s go for an A. What if we want to account for more than just gas/no gas? Let’s add a condensed section to our model. Assume,

P(gas) = 0.2

P(wet) = 0.7 [Probability of a wet sand regardless of the geophysics.]

P(condensed) = 0.1 [Probability of a condensed section.]

Note that,

P(gas) + P(wet) + P(condensed) = 1.0

That is we assume these are the only possibilities (no salt, no limestone, no coal, etc). P(wet) represents either fizz with an amplitude anomaly or a wet sand with no amplitude anomaly.

We’ll assume that P(a.a.|condensed) = 0.2 and P(a.a.|wet) = 0.5 [as in the above case]. That is, we assume that 20% of the time we’ll get amplitudes like ours when a condensed section is present, while the rest of the time we won’t (for example, in those other cases the polarity is wrong).

1.0
P(gas|a.a.) = 0.2 ____________________________________ = 0.3509
0.2x1.0 + 0.7x0.5 + 0.1x0.2

Note that we added a complication to our model and our Pg improved! However we reduced the probability of a wet sand (from 0.8 to 0.7) with its associated high value of P(a.a.|wet) of 0.5 and substituted it (0.1 of it) with a low P(a.a.|condensed) of 0.2. The net result is a higher Pg. If you understand the above (not necessarily agree with) you get an A.

All that remains is to add more assumptions and use realistic probabilities.

Notes:

1. Houck, Richard T., The Leading Edge, "Estimating uncertainty in interpreting seismic indicators", March, 1999, page 320

Appendix: A

Bayes’ Rule

Bayes’ Rule states that probability of A happening given that B has happened is the probability of both A and B happening divided by the probability of B happening. That is written in mathematics as

P(A Ç B)
P(A|B) = _____________
P(B)

P(A|B) is read, the probability of A happening given that B has happened. The "|" is read as "given that". P(A Ç B) is read, the probability of the intersection of A and B. Intersection here means both A and B happening. This perhaps can be more easily understood by viewing the following diagram.

Where circle A overlaps with circle B (the blue area) is the intersection of A and B. If we divided this area by the area of the B circle, we get the probability of A happening given that B has happened.

With a little bit of mathematics Bayes’ Rule can be expanded to yield the following:

P(A Ç B)               P(A) P(B|A)
P(A|B) = _____________ = ________________________________________________
P(B) P(A)     P(B|A) + P(NOT A ) P(B | NOT A)

where P(NOT A) is the probability of A not occurring [Note that P(A) + P(NOT A) = 1], and P(B| NOT A) is the probability of B happening given that A has not happened.

We’ll leave the proof to the reader. For a numeral example of Bayes’ Rule see appendix B.

Appendix: B

An "easy" example of Bayes’ rule using dice.

What is the probability of rolling a 5 on a fair die? If you answered 1/6, congratulations you know all the mathematics you need to understand Bayesian probability theory. You knew to count the possibilities, 6 of them. You assumed that each was equally likely. You knew to divide the number of ways to get 5 (one way) by the total possibilities 6, getting 1/6. The computation of probability is merely a problem of counting.

What is the probability of rolling a 5 or 6? Here, there are two desired outcomes divided by six possibilities, yielding 2/6 or 1/3. Now what is the probability of rolling a 5 or 6 with two dice? Consider the table below. The dice have been labeled "A" and "B". There are 36 possible pairs. The desired outcomes are colored in red. Counting the red pairs we get 20. Therefore, the probability of rolling a 5 or 6 with two dice is 20/36 or 5/9. In mathematical shorthand, P(5 or 6) = 5/9.

 A B A B A B A B A B A B 1 1 2 1 3 1 4 1 5 1 6 1 1 2 2 2 3 2 4 2 5 2 6 2 1 3 2 3 3 3 4 3 5 3 6 3 1 4 2 4 3 4 4 4 5 4 6 4 1 5 2 5 3 5 4 5 5 5 6 5 1 6 2 6 3 6 4 6 5 6 6 6

Bayesian Theory explains the effect on the probability of an outcome after gathering information (see appendix B). This probability is usually stated thusly, "What is the probability of rolling a 5 or 6 given that die "A" is a 1." This is written in mathematical shorthand as P(5 or 6 | A=1). The "|" is read as "given that".

So what is the probability of rolling a 5 or 6 given that die "A" is a 1? Here we need only consider the pair of results that appear in column 1 of the table above (that is A = 1). The answer of course is 2/6 or 1/3. That is, P(5 or 6 | A=1) = 1/3.

Since we are a bit masochistic and started out trying to learn Bayesian Theory, we will calculate the answer above this time the hard way. Bayesian Theory would state the problem as

P(5 or 6)P(A=1|5 or 6)
P(5 or 6|A=1) = _____________________________________________________
P(5 or 6)P(A=1|5 or 6)+P(no 5 or 6)P(A=1|no 5 or 6).

We already know that his rather impressive formula should yield 1/3. The numerator is the probability of getting a 5 or 6 and the probability of getting an "A" on die 1. The denominator is the probability of die "A" being a 1.

Consider the numerator first. We have already calculated P(5 or 6), the probability of getting a 5 or 6 not knowing what’s on die "A". P(5 or 6) was 5/9. P(A=1 | 5 or 6) is read, the probability of die "A" being a 1 given that there is a 5 or 6 also showing. The ways of getting a 5 or a 6 are highlighted in red below.

 A B A B A B A B A B A B 1 1 2 1 3 1 4 1 5 1 6 1 1 2 2 2 3 2 4 2 5 2 6 2 1 3 2 3 3 3 4 3 5 3 6 3 1 4 2 4 3 4 4 4 5 4 6 4 1 5 2 5 3 5 4 5 5 5 6 5 1 6 2 6 3 6 4 6 5 6 6 6

Only 2 of the 20 red pair have a 1 on die A. Therefore, there are 2 ways [(1,5) and (1,6)] out of 20 possibilities. Hence, the P(A=1 | 5 or 6) = 2/20 or 1/10. P(5 or 6) times P(A=1 | 5 or 6) is 5/9 times 1/10 or 5/90 or 1/18.

To calculate the numerator another way, remember that the numerator is the probability of getting a 5 or 6 and the probability of getting an "A" on die 1. There are 2 ways of doing this, [(1,5) and (1,6)] out 36 total combinations, yielding 2/36 or 1/18. Thus, whichever way we calculate/count it, the numerator is 1/18.

Now let’s consider the denominator. This time we’ll do the global approach first and then recalculate the denominator using the individual parameters. Remember that the denominator is the probability of die "A" being a 1. Let us count the ways. There are 6 ways that A=1. Thus, the denominator is 6/36 or 1/6.

Now lets recalculate the dominator using P(5 or 6) P(A=1 | 5 or 6) + P(no 5 or 6) P(A=1 | no 5 or 6). We already know P(5 or 6) and P(A=1 | 5 or 6), [5/9 and 1/10, respectively]. P(no 5 or 6), the probability that neither of the dice having a 5 or 6, are the pairs in black below divide by 36, thus 16/36 or 4/9.

 A B A B A B A B A B A B 1 1 2 1 3 1 4 1 5 1 6 1 1 2 2 2 3 2 4 2 5 2 6 2 1 3 2 3 3 3 4 3 5 3 6 3 1 4 2 4 3 4 4 4 5 4 6 4 1 5 2 5 3 5 4 5 5 5 6 5 1 6 2 6 3 6 4 6 5 6 6 6

Note that P(5 or 6) + P(No 5 or 6) = 5/9 + 4/9 = 1.0. Hopefully that doesn’t come as a shock. The last parameter we need is P(A=1 | no 5 or 6), the probability of die A=1 given that a 5 or 6 was not rolled. That will be the black pairs in column one in the chart above divided by the total number of black pairs, thus 4/16 or 1/4. Substituting into t