Suppose we have a bagatelle:
                 \    //
                  \  o/
                   \ /
                   / \
                  /   \
                 /  @  \
                /       \
               /  @   @  \
              /           \
             /  @   @   @  \
            /               \
           /  @   @   @   @  \
          /                   \
          |                   |
          |   |   |   |   |   |
          |   |   |   |   |   |

            A   B   C   D   E
When we drop the ball, it is going to end up in one of A, B, C, D or E, let's say, with probability a, b, c, d and e, respectively.

Because the ball must land in one of the slots, we know that

a + b + c + d + e = 1
(because for probabilities, 1 represents certainty.)

Suppose we take bets on where the ball will land... I bet on A, and you bet on D.

  • The odds of my winning are a and the odds of your winning are d.
  • The odds of at least one of us winning are just a + d.
  • The odds of both of us winning are 0 (just one ball - it can't end up in both slots - so probability zero.)
Now suppose I bet on 'A or B' and you bet on 'D or B'
  • My odds are now a + b and yours are b + d
  • The odds that one of us will win are a + b + d
  • The odds that both of us will win are just b, since we only both win if the ball ends up in b.
  • The odds that neither of us will win are just the sum of the probabilites of the outcomes where neither of us win, namely c + e, or alternatively put: 1 - (a + b + d)
There is a pattern here, to do with sets. Call our set of outcomes S, that is: S = {A,B,C,D,E}; we have just been considering the probabilites of subsets of S - when I bet on 'either A or B', I am betting on the subset {A,B}.

Bear with me while I define some terms..

In set theory, we have the operations: union, intersection and complement. For example the union of the one-element sets {A} and {D}, is just the two-element set {A,B}.

The intersection of {A,B} and {B,D} is {B}, the set of the common elements.

The intersection of {A} and {D}? That's the empty set: {}, because there are no elements in common.

The complement of {A,B,D} is the set of the other elements of S, just {C,E}. The complement of {C,E} is {A,B,D}. The complement of S (our original set) is {} and the complement of {} is S.

So looking back at the odds of the various bets, above, we can see that, for two subsets of S, X and Y:

  • The odds of 'X or Y' are equal to the odds of the union of X and Y.
  • The odds of 'X and Y' are the same as the odds of their intersection
  • The odds of 'not X' are equal to the odds of the complement of X.
If you know Boolean algebra, this may be beginning to look a bit familiar... and in fact, in this way we can represent any Boolean algebra as a set algebra.

(There are many Boolean algebras, it's Boolean logic that only takes values of 0 and 1)

Set algebra is a bit of a scary term, but in this case it just means the collection of S, and its subsets, together with our probability map A->a,B->b,C->c,D->d,E->e, with a,b,c,d,e adding up to 1, and the union, intersection, and complement operators. In other words, just what is discussed above.

Conversely, we can treat normal probability theory (which is what our 'set algebra' amounts to) as a Boolean algebra, with the Boolean 'or' 'and' and 'not' defined, as above, as the union, intersection and complement operations on subsets of the set of possible outcomes. (This Boolean algebra can take values anywhere between 0 and 1, inclusive, rather than just 0 and 1.)

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