add conditional probability

[btbonval]

Should be fairly easy.

Let the distribution a1d6 (in code) / X (in math) be a pmf for a six-sided die.

p(X = 0 | X > 3) is obviously 0, but p(X > 5 | X > 3) might be interesting and is definitely easy to solve. It's generally p(X > 5 N X > 3) / p(X > 3).

If __and__ is overloaded to make contextual decisions a la the suggestion in #1, then p(X > 5 N X > 3) could be written a1d6 > 5 and a1d6 > 3, which would just resolve to a1d6 > 5. Divide by a1d6 > 3.

If there is an inequality object (which seems necessary for #1), then overloading the | operator (if possible) would be handy although maybe confusing (due to or connotation) resulting in a1d6 > 5 | a1d6 > 3.