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Basic theorems

events should occur equally often. De Mere reasoned as follows: A occurs in just
six ways (6 : 4 : 1, 6 : 3 : 2, 5 : 5 : 1, 5 : 4 : 2, 5 : 3 : 3, 4 : 4 : 3), and B also occurs in
just six ways (6 : 5 : 1, 6 : 4 : 2, 6 : 3 : 3, 5 : 5 : 2, 5 : 4 : 3, 4 : 4 : 4). Therefore A and B
have the same probability P(A) = P(B).
The fallacy in this argument was found by Pascal, who showed that the outcomes
listed by de Mere are not actually equiprobable. In fact, one must take account
not only of the number of spots showing on the dice, but also of the particular
dice on which the spots appear. For example, numbering the dice and writing the
number of spots in the corresponding order, we find that there are six distinct
outcomes leading to the combination 6 : 4 : 1, namely (6, 4, 1), (6, 1, 4), (4, 1, 6), (1, 6, 4)
and (1, 4, 6), whereas there is only one outcome leading to the combination 4 : 4 : 4,
namely (4, 4, 4). The appropriate equiprobable outcomes are those described by triples
of numbers (a, b, c), where a is the number of spots on the first die, b the number
of spots one second die, and c the number of spots on the third die. It is easy
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to see that there are then precisely n = 6 = 216 equiprobable outcomes. Of these,
m(A) = 27 are favourable to the event A (in which the sum of all the spots equals
11), but only m(B) = 25 are favourable to the event B (in which the sum of all the
spots equals 12). To see this, note that a combination a : b : c occurs in 6 distinct
ways if a, b and c are distinct, in 3 distinct ways if two (and only two) of the
numbers a, b and c are distinct, and if only 1 way if a = b = c. Hence A occurs in
6 + 6 + 3 + 6 + 3 + 3 = 27 ways, while B occurs in 6 + 6 + 3 + 3 + 6 + 1 = 25 ways. This
fact explains the tendency observed by de Mere for 11 spots to appear more often
than 12. ,

Basic theorems

From (2.1) we may deduce the following properties of the probability P(A).
1. For any event A in a sample space S,

0 ≤ P(A) ≤ 1. (2.4)
If P(A) = 1 then A is a certainty; if P(A) = 0 then A is an impossibility.

2. For the entire sample space S we have

n S
P(S) = = 1, (2.5)
n S
which simply states that we are certain to obtain one of the possible outcomes. Besides
P(∅) = 0.
3. The Addition Theorem. The probability of the sum of two events is equal to the sum of their
probabilities minus the probability of their intersection, that is

P(A + B) = P(A) + P(B) − P(AB). (2.6)

PROOF. Let n be the total number of uniquely possible, equiprobable and incompatible outcomes;
m the number of outcomes leading to the occurrence of the event A; l the number of outcomes leading
to the occurrence of the event B and k the number of outcomes leading to the occurrence of both A and
k
B. Then P(A + B) = m+l−k = m + l − , but m = P(A), l = P(B), k = P(AB), and hence
n n n n n n n
P(A + B) = P(A) + P(B) − P(AB).
Observe, that k = 0 if the events A and B are incompatible. 2

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