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Common knowledge is knowledge of a group that is known to all members, known by all members to be known to all members, and also (a recursive definition) known by all members of the group to be common knowledge.
Logical formulation
To formulate a logical definition of common knowledge, we let <math>\phi<math> represent a sentence (or statement) and let the modal operator E represent the statement that everyone knows a statement, or that it is general knowledge. Usually, "everyone" is defined as some set of agents G, and the operator EG is used. Common knowledge among this group would be described using the operator CG. E2 is used to indicate 2-ply general knolwedge (everyone knows the fact, and everyone knows that everyone knows it.) En, in general, indicates n-ply general knowledge.
The recursive definition of common knowledge is formulated as follows:
<math>C \phi \Leftrightarrow E (\phi \wedge C \phi)<math>
which is equivalent to defining common knowledge in a more mathematically tractible form:
<math>C \phi \Leftrightarrow \bigwedge_{i = 0}^n E^n \phi<math>,
where an E0 operator simply entails that a fact is true.
Strictly speaking, then, something is in the common knowledge if it is known in arbitrarily many ply worth of general knowledge.
Practical applications
Multi-ply general knowledge is important in many practical situations. To provide an example, assume all drivers are legally licensed, thus know how to interpret traffic signals, but do not know that the other drivers understand them. (In other words, there is 1-ply but not 2-ply general knowledge of how to interpret the signals.) In this case, they will not feel safe driving. It is not enough for the knowledge to be in the general knowledge, at least 2 ply are required. More complex arguments can be made where 3- and higher ply general knowledge is necessary.
Textbook example
A common example, in logic textbooks, which illustrates the relevance of common knowledge, is the following logic puzzle (or some variant): On an island, there are n people, of whom k > 1 have blue eyes, and the rest have green. If a person ever knows him- or herself to have blue eyes, he or she must leave the island at dawn the next day. Each person knows every other's eye color, there are no mirrors, and there is no discussion of eye color. At some point, an outsider comes to the island and says, "at least one of you has blue eyes". The problem: Assuming all persons on the island are truthful and completely logical, what is the eventual outcome?
The answer is that, on the kth dawn, all k blue-eyed people will leave the island.
If k = 1, the person will recognize that he or she has blue eyes (by seeing only green eyes in the others) and leave at the first dawn. If k = 2, no one will leave at the first dawn. The blue-eyed people, recognizing that only one other pair of blue eyes are among the others, and that no one left on the 1st dawn, will leave. So on, it can be reasoned that no one will leave at the first k-1 dawns if and only if there are at least k blue-eyed people. Those with blue eyes, seeing k-1 blue-eyed people among the others and knowing there must be at least k, will reason that they have blue eyes and leave.
What's most interesting about this scenario is that, for k > 1, the outsider is only telling the island citizens what they already know: that there are blue-eyed people among them. However, before this fact is announced, the fact is not common knowledge; it is merely k-1 ply general knowledge. The notion of common knowledge therefore has a palpable effect; when a fact already known to all (for k > 1) is put into the common knowledge, the blue-eyed people on this island eventually deduce their status, and leave.
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