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 notu 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
righthand 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:


A 


A! 



B 

B! 
B 

B! 

c 
a 


b 
c 

d 
c! 
e 

f 
g 

h 

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 righthand 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 righthand 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:


A 


A! 


B 

B! 
B 

B! 
c 
.126 

.044 
.154 

.176 
c! 
.054 

.076 
.066 

.304 
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.
