There are some difficulties to overcome in order to convince persons that the subject of this book is worth knowing something about. George Boole's theory of probability has had an extremely bad press for more than 100 years. It has almost universally been considered to be too complicated to understand, too difficult to calculate, and wrong in addition. As evidence that this is true, try and find a book on probability in, say, the 20th century which has any material on Boole's theory of probability. This is quite remarkable treatment for a person as distinguished as George Boole. In fact, his theory is not wrong, it is quite easy to make the calculations after a few simple emendations, it can solve a number of important problems which cannot be handled by the orthodox theory of probability, and there are quite a few areas in which these problems arise. In the following panes we will try and convince skeptics that there are good reasons for believing that a book on the subject is justified.

Boole gives a summary of the probability principles that have been applied to the solution of questions of probability. With regard to the extent and relative sufficiency of these principles he writes, "It is always possible, by the due combination of these principles, to express the probability of a compound event, dependent in any manner upon independent saimple events whose distinct probabilities are given." (Laws of Thought; pp. 248-9) An even more striking statement is also made: "It may, therefore, be fully conceded that all problems having for their data the probabilities of independent simple events fall within the scope of received methods." (Laws of Thought; p. 250)

It is beyond the bounds of rational opinion to imagine that the person who wrote these summary statements would then proceed to publish examples of solutions to probability problems which were not in accord with orthodox theory, granted that the requirements of the orthodox theory were available.

Hailperin in the two editions of his very excellent book, Boole's Logic and Probability, considered, and solved, in a different sense, Boole's general problem of probability. Everyone with an interest in Boole should surely read this fine book. However, Professor Hailperin fully accepts the contention of the orthodox critics that Boole's method introduced equations that were not part of the data and that his "solutions", therefore, were of a different problem than the actual problem with a lesser number of equations. As the capstone of his very fine book, Professor Hailperin offers his own analysis of how to solve Boole's general probability problem. He writes, "In this section we are going to show that Boole's general problem can be solved without additional assumptions, solved not in his sense but in the sense of giving best possible upper and lower bounds on the probability of the objective event, these bounds being determined by the given data." Hailperin shows that obtaining such bounds is a reasonably straightforward linear programming problem. (2nd ed.; p. 338)

Therefore, even granted the fact that Professor Hailperin's two editions have answered a large number of questions that have been debated about Boole's theory of probability, it remains the case that no book known to the present author deals with Boole's theory in terms of its claims or its relative usefulness. In fact, the treatment that history has accorded Boole's theory of probability is really quite extraordinary. Here is a person who is held almost universally in high esteem and who claims to have a theory of probability that enables one to solve very many problems that cannot be solved with the usual theory. Yet in the vast number of books dealing with probability theory or applications written in the past sixty years one looks in vain for a reference to Boole. This is a remarkably long time to ignore a theory produced by a person of such intellectual stature as George Boole.

I believe that it is possible to state three specific reasons why Boole's method appears to be so difficult. First, Boole never recognized the enormous advantages of representing the information known about a specific probability problem in the form of a logical table. As an illustration, suppose that some, or all, information about the probabilities of three events (A, B, and C) is known. Then the associated logical table would be:

where the small letters represent the probabilities of the specific combination of events. Thus, f = probability that A occurs without B or C occurring and c = probability that Both B and C occur and A does not. Our goal is always to reconstruct this whole logical table for a given probability problem. This was never Boole's goal. This difference has two major implications. First, Boole is always trying to solve for some one of the eight small letters and this is generally much more difficult than solving for all of them. Second, when we have reconstructed as much of the logical table as is possible with the information given then we know everything about the problem that it is possible to know. Any other information required is immediately available from the logical table. For Boole it is quite otherwise and he frequently has to repeat an entire onerous algebraic exercise in order to find the value of a second small letter.

Second, suppose there are a number of random events, say x, y, z, . . . Some of these events may be simple and some may be compound but Boole assumes that this distinction is not an important one. If some of the events are compound, Boole intends to designate each of them by a single letter and then to treat them as if they were simple. We are therefore entitled to to assume that the probability problem involves some number of simple events x, y, z, . . . Our desideratum is to find the probability of some event, w, that is defined in terms of one or more of the simple events. We will know the probabilities of some, perhaps all, of these simple events. The first step in Boole's procedure is to use his algebra of logic to eliminate all of the variables representing simple events for which the probabilities are not known and then to express the event w in terms of simple events for which the probabilities are known. The procedure is straightforward but it quickly becomes on! erous when the number of simple and compound events increases. Everyone, including Boole, agrees that his method, as he presented it, requires heavy algebraic manipulations. There are two main reasons for this and the step now being discussed is one of them. Generally in Boole's own examples the algebra becomes disconcertingly heavy. However, using a different approach we can accomplish Boole's purpose with very much less algebra. In fact, the method we will propose makes this whole step very easy, even for pencil-and-paper calculations.

The third reason for the difficulty of Boole's procedure is his final step, involving the solution of a quite complicated algebraic equation for a quantity that Boole designates as V. The argument that leads to V is lengthy and elaborate so it is not an appropriate subject for here. It is, of course, presented in the book which is here under report. It is quite remarkably difficult to solve the equation and Boole, who reports his solutions to many of the examples in his book, never gives any hints as to how he does it. Therefore, it is extremely important to emphasize that we have developed a method which permits the solution of the equation for V. In the great majority of cases it is very straightforward to solve the equation, using the approach presented in the book. It is always relatively easy as compared to Boole's method.

In the Ad Example tab we will present a typical example of a Boolean analysis of a probability problem. In that section several of the topics mentioned above will be illustrated. We intend to show Boole's own procedure and then to contrast that with our emended procedure. It is important to emphasize that the emended procedure is entirely based on Boole's arguments. I have no doubt that had Boole been granted a few more years of life he would have found these emendations himself. Or, if the new field that Boole founded had followed a pattern of normal development instead of being killed almost at birth by the developments of mathematical logic then the emendations would have been early improvements resulting from that normal development.

One area which generates an extremely large number of probability problems is that of advertising decision making. The goal of the advertiser is like the goals of most decision makers in business: to achieve the best value possible for the money that is to be expended. In advertising this often reduces to the question: What selection of media will be the optimal one for our product? Often, the choice ranges over many different media: network television, local television, radio, FM radio, magazines, newspapers, billboards, and so forth. We can confine our attention here to the selection from among a small number of magazines since that decision problem will be typical of the more general ones.

The basic problem is that generally only partial information is available and the decision maker needs to achieve estimates of the missing information in order to make rational decisions. Suppose, then, that two magazines, A and B, have already been selected and another magazine, C, is under consideration as to whether it should be included in the advertising program. To make a more or less typical situation, suppose that the target audience is sophisticated home computer users. This would be "typical" in the sense that probably only partial information about this specific target audience would be available. Suppose, then, that it is known that the proportion of members of the target audience who read A (called the "reach" of A in the industry) is p(A) = p, it is also known that the reach of B is p(B) = q. The reach of C is unknown. However, from an industry study it is known that the proportion of the target audience that reads both A and C is p(AC) = r (called the "duplication" in the industry) and the proportion that reads both B and C is p(BC) = s.

Analyzed using Boole's Method

It is very important to recognize that Boole's method does NOT in any way start from the logical table since Boole seems never to have noticed the very great advantages of using the logical table. Instead, Boole would start with a summary of the information available:

p(A) = a + b + e + f = p;
p(B) = a + c + e + g = q;
p(AC) = a + b = r;
p(BC) = a + c = s.

For Boole's method we must specify the desideratum of our analysis. It is very often the case that the decision maker wants to know the triplication to be expected, meaning here the proportion, w, of the target audience that would read all three magazines, ABC. We will assume that this is the desideratum. We now see that we have three compound events, including the desideratum, and following Boole's procedure we need to assign a single letter to each compound event since we are going to treat the compound events exactly as we treat the simple events. A and B are simple events and we now assign u = AC, v = BC, and w = ABC. We know that p(u) = r and p(v) = s, and p(w) = t is to be determined. Following Boole, we want to convert each of these equations defining three variables into an equation = 0. This is a standard, and easy, operation in Boolean logic so we immediately obtain:

u!AC + uA! + uC! = 0 ;
v!BC + vB! + vC! = 0;

and

w!ABC + wA! + wB! + wC! = 0

Here u! means not-u and has the same meaning as 1 - u in Boolean algebra and similarly for the other letters. Boole's argument shows that we are entitled to set the sum of these three equations = 0, giving

u!AC + uA! + uC! + v!BC + vB! + vC! + w!ABC + wA! + wB! + wC! = 0.

There is a major difference among these variables. We know the probabilities of A, B, u, and v and this knowledge means that these four variables are Boolean variables. w and w! are a special case since we are going to solve for these probabilities. C, however, is different since we do not know the probability of C and we are not going to solve for it. Therefore, we want to eliminate C from the above equation (and we would similarly eliminate any other variable for which we did not know the probability). The procedure for so doing is a standard part of Boolean algebra. There are four terms in the above equation which do not include C or C! and we will put these aside until we have eliminated C. These four terms are uA! + vB! + wA! + wB! = 0. We set the remaining six terms = f(C):

f(C) = u!AC + uC! + v!BC + vC! + w!ABC + wC! = 0

Following the standard procedure, we want f(1), in which C = 1 and C! = 0, and we want f(0), in which C = 0 and C! = 1. We find:

f(0) = u + v + w

and

f(1) = u!A + v!B + w!AB

Therefore, the product of these plus the four terms set aside above gives

uv!B + uw!AB + u!vA + vw!AB + u!wA + v!wB + uA! + vB! + wA! + wB! = 0 .

This is the desired equation. We have succeeded in expressing w in terms of Boolean variables, each of which has a known probability. We intend now to solve this equation for w. In doing so we must remember to write each occurrence of w! as 1 - w. We obtain:

w = (uAB + vAB + u!v!B + u!vA + uA! + vB!)/(uAB + vAB - u!A - v!B - A! - B!)

and it will be seen that we have written an equation for w which has on the right-hand side only variables for which the probabilities are known.

Boole now shows that we can express w as a developed function of A, B, u, and v. There are two steps involved. First, we must recognize all the possible constituents of such a development and, second, we must calculate the coefficients of each of the constituents because in Boole's system the coefficients will tell us the role played in the development by the corresponding constituent. It is easy to list all the possible constituents. There are four variables and each variable can assume a value of either 1 or 0 so there are 16 constituents. Boole shows that that the coefficients can only be 1, 0, 0/0, or any other number, which Boole conventionally writes as 1/0. Boole shows that the use of the various constituents depends on the coefficient in the following ways: 1. If the coefficient is 1 then the whole of the class represented by that term is to be taken; 2. If the coefficient is 0 then none of the class represented by that term is to be taken; If the coefficient is ! 0/0 then an indefinite amount (none, some, all) of the class represented by that erm is to be taken; and 4. If the coefficient is any other number (conventionally, Boole writes it as 1/0) then the term associated with the coefficient must be equated to zero. Boole writes the general result of these manipulations in the form

w = k + 0l + (0/0)m

where I am using small letters merely to avoid any conflict with other uses for the capital letters. Boole always writes the equation with capitals but we do not have much need to discuss this formula of Boole's in the present document so we will use small letters. Following Boole, designate the four classes of constituents as a, b, c, and d, respectively, including among them all the possible constituents that can be formed from the four Boolean variables. This means that one can write a perfectly arbitrary constituent - say uv!A!B! - and this, like every other constituent formed from the given four Boolean variables, will be in a, or b, or c, or d. How can one determine which? Evaluate separately the numerator and the denominator of the expression for w using this rule: if the constituent involves x as a factor, change x into 1 and x! into 0, but if it involves x! as a factor change x into 0 and x! into 1. Proceed similarly with each of the Boolean variables in the expre! ssion for w. Thus, for our example we will make u = 1 and u! = 0, v = 0 and v! = 1, A = 0 and A! = 1, and B = 0 and B! = 1. So doing, we find that the numerator = 1 and the denominator = -2. Therefore, w is the ratio or w = 1/(-2). Boole shows us that "a" includes all constituents for which w = 1; "b" includes all constituents for which w = 0; "c" includes any, and all, constituents for which w = 0/0; and "d" includes all constituents for which w = any other value than the three just specified, so our example is in "d".

The difficulty is that we must make this calculation for 24 = 16 constituents. If there are n Boolean variables then there will be 2n constituents and the calculations required go up quite sharply as n increases. Further, we need to know the class into which each constituent falls because Boole's answer to the desideratum depends on this analysis. Specifically, Boole shows that any constituent = 1, hence in "a", means that the whole of the class represented by that constituent is to be taken for w; any constituent = 0, hence in "b", means that none of the class represented by that constituent is to be taken for w; any constituent = 0/0, hence in "c", means that some (none, all, or some) of the class is to be taken for w; and, finally, any constituent with 1/0, hence in "d", means that the constituent must be set = 0. Note that the symbol 1/0 is a purely conventional symbol selected by Boole as the symbol for any constituent which does not = 1, 0, or 0/0.

Boole shows that a + b + c + d = 1 and he introduces V = a + b + c = 1, since d = 0. He shows that the conclusion from w = a + 0 b + 0/0 c + 1/0 d is that

w = a + qc
V = 1

where q is Boole's symbol for an indefinite number from a class. The probability of w, the desideratum, is given by

Prob. w = (a + kc)/V

where k is the probability of the indefinite event q. So, if we have found the values for each of the possible constituents then we know a and c and Boole shows how to find k, which we do not need for our example here. In fact, for our example we find that only uvAB has coefficient = 1 so the only constituent in "a" is uvAB. We find six constituents with coefficients = 0 so there are six constituents in "b". There are no constituents with coefficient = 0/0 so all of the remaining constituents have coefficient 1/0. Hence, there are nine constituents in "d".

For our example there are no constituents in "c" and, therefore, Prob. w = a/V and a = uvAB. So, all we need in order to find the desideratum is V. Unfortunately, this is the very difficult algebraic equation we have mentioned several times. Actually, we can write two different equations for V. The first one comes from Boole's definition of V:

V = a + b + c
= uvAB + uv!AB + u!vA!B + u!v!AB + u!v!AB! + u!v!A!B + u!v!A!B!

where the last six of the entries on the right are the constituents which each had coefficient 0. The second equation for V can be found in the book under discussion here.

All we need is to solve one of these equations for A, B, u, and v and the analysis can be completed and we will have found the desideratum. But either equation seems to be very hard to solve. I have never found an adequate general method and I am sure that Boole did not have one either. Yet, the equation can be solved. The solution for an example similar to this one is given on p. 55 of the book here under discussion. The mere form of the solution suggests that it is not easy to find. It seems to require an exorbitant amount of calculation (determining the coefficients of the constituents) and a lot of algebra (to solve the equation). Further, it is annoying because we really do not need the values of the Boolean variables. As we will now see, we can solve the whole problem quite easily without ever knowing the values of A, B, u, and v in our present example. While we cannot discuss it here, it is a fact that we can calculate those values quite easily once we have solved the problem.

Analyzed using an Emendation of Boole's Method

We are going to use the same example so that it is easy to make comparisons between Boole's original method and the emendations proposed in the book under discussion here. To state the example again, suppose that two magazines, A and B, have already been selected and another magazine, C, is under consideration as to whether it should be included in the advertising program. To make a more or less typical situation, suppose that the target audience is sophisticated home computer users. This would be "typical" in the sense that probably only partial information about this specific target audience would be available. Suppose, then, that it is known that the proportion of members of the target audience who read A (called the "reach" of A in the industry) is p(A) = p, it is also known that the reach of B is p(B) = q. The reach of C is unknown. However, from an industry study it is known that the proportion of the target audience that reads both A and C is p(AC) = r (called the! "duplication" in the industry) and the proportion that reads both B and C is p(BC) = s. We will mention another possible piece of information in order to illustrate the other kind of information that Boole can incorporate in his analyses. This is information about "conditioned" variables. By "conditioned" Boole means that some variables may be restricted in terms of the values they can assume. If this is so then that information must be incorporated in an appropriate equation. Thus, suppose that C is a markedly more advanced treatment of computers and this is reflected by the fact that essentially any reader of C will also be a reader of either A or B or both. This means that d = 0 in our logical table since we are told that anyone who reads C also reads at least one of A and B. This is a simple illustration of Boole's meaning of conditioned, which was so egregiously misunderstood by Keynes. We will NOT assume that in our present analysis we have d = 0.

We will use our logical table in summarizing the information we have:

where the small letters represent the probabilities of the specific combinations of events. Thus, f = probability that A occurs without B or C occurring and c = probability that both B and C occur and A does not. Our goal is always to reconstruct this whole logical table for a given probability problem. This was never Boole's goal. This difference has two major implications. First, Boole is always trying to solve for some one of the eight small letters and this is generally much more difficult than solving for all of them. Second, when we have reconstructed as much of the logical table as is possible with the information given then we know everything there is to be known about the underlying probabilities. Any other information required is immediately available from the logical table. For Boole it is quite otherwise and he frequently has to repeat an entire onerous algebraic exercise in order to find the value of a second small letter.

We can see immediately that the information we already know can be easily expressed in terms of the small letters. Thus, the total proportion of the target audience that reads A is a + b + e + f and we have been told that this sum is p; the total proportion of the target audience that reads B is a + e + c + g = q; the total proportion reading C is a + b + c + d; the total proportion reading both A and C is a + b = r; and the total proportion reading both B and C is a + c = s. Since the whole table represents the total target audience we know that the sum of all of the small letters = 1. Suppose that for a specific example we know that p = .3, q = .4, the proportion reading C is .5, r = .18, and s = .28. The sum of all eight small letters is 1. We have, then, six equations involving eight unknowns. Hence, we know that we can write our knowledge about the likelihoods of occurrence of the eight unknowns in the form of six equations involving two unknowns. I picked ! "b" and "e" as the unknowns and by straightforward solution and substitution I get the equations:

a = .18 - e
c = .10 + e
d = .22 - b
f = .12 - b
g = .12 - e
h = .26 + b.

With Boole's own method it is necessary to have a specified desideratum. We used uvAB above and this meant that there were three compound events to which we had to assign single letters. Here we have only two compound events and we will assign s = AB, hence s! = A! + B!, and t = BC, hence t! = B! + C!. We will find what possibilities exist by direct multiplication of the Boolean variables. We can do it most easily in two steps. First, we have

st = ABC
st! = ABC!
s!t = A!BC
s!t! = A!B! + A!C! + B!C! + C!

Here we have used nothing more profound than that for any Boolean x we always have x(x!) = 0. Now we multiply these four products by AB, A!B, AB!, and A!B!, remembering that, for example, uv!(AB) = 0 since B(B!) = 0. We obtain the following as being the only possibilities:

stABC = ABC = a
st!A!BC = A!BC = c
s!tABC! = ABC! = e
s!t!A!B!C= A!B!C = d
s!t!A!B!C! = A!B!C! = h
s!t!AB!C = AB!C = b
s!t!AB!C! = AB!C = f
s!t!A!BC! = A!BC! = g

If you will make the comparison you will find that these are exactly the same constituents that had coefficients of 1 (a single constituent) or 0 (six other constituents). as found by Boole's method, above. Here we have found them by simple Boolean multiplication and we have identified, on the right-hand side, the specific cell of the logical table which is represented by each of the constituents.

The argument by means of which Boole developed the second equation for V, mentioned above, is a central part of his theory and cannot be presented in this here. However, we can call attention to the fact that the basic structure of the fractions required for the second equation for V is that the numerator is the sum of the constituents (or simple events) that produce some compound event with probability equal to the denominator of the fraction. Thus, by Boole's method there are four ways to obtain A and the first fraction in the equation for V would have had denominator p = p(A). Since we had four such compound events (u, v, A, B) in our previous application we had four such fractions, each = V. Our present point is that we now have eight such events with known probabilities: a, b, c, d, e, f, g, h. Therefore, we can now write eight fractions, each equal to V. Thus, we have:

stABC/a = st!ABC!/e = s!tA!BC/c = s!t!A!B!C/d = . . . = s!t!AB!C!/f = s!t!A!BC!/g = V .

This is already an enormous improvement since it is very much easier to solve eight such equations than it is to solve only four but we can do very much better than this improvement. As we already noted, we have no particular use for the values of the Boolean variables (u, v, A, and B). Obviously, if we were told these values then we could calculate the quantities we really want (a, b, c, d, e, f, g, h) but we surely have no desire to solve for the Boolean variables if we could, instead, solve directly for the quantities we really want.

This is, in fact, quite easy to do. Now, if we multiply the first fraction by the eighth fraction we get the same numerator as when we multiply the second fraction by the third fraction:

(stABC/a)*(s!t!A!BC!/g) = (st!ABC!/e)*(s!tA!BC/c) = (constant/ag)*(constant/ce) = V^2

whence we have ag = ce. Substituting our original equations for a, g, and c we have a quadratic for e which gives e = .054. Similarly, the product of the fourth and seventh fractions equals the product of the fifth and sixth, whence we have df = bh. Replacing these by their equations we have a quadratic for b and we find b = .044. We now have the values of the two variables we used on the right-hand side of our initial equations. Therefore, we can calculate the values of each of the entries (a, b, . . ., h) in our original table of probabilities. We find:

You can verify that all of the requirements of the original data are fulfilled. Thus, the proportion reading A is .3, the proportion reading B is .4, the proportion reading C is .5, and so forth.

Notice that these estimates are not available using standard probability theory. It should also be noticed that the calculations we have undertaken do not involve any hidden assumptions about the meaning of "independence", a frequent criticism of Boole's theory. In fact, Boole's argument is coherent, straightforward, and extremely useful. It seems, to the present author at least, that it is shameful that Boole's extraordinary extension of orthodox probability theory has received such a negative reception for so long.