If this is like most toy problems, it shouldn't be a 15-variable problem. Assuming that pigs within the same class have same probability of winning, then there's only one variable involved - the bias in favor of the strong pigs vs the weak pigs.

The problem can be simplified by classifying the pigs into {Strong,Weak} x {Odd,Even}. And the top three positions as {Even,Odd} x {Even,Odd} x {Even, Odd} - pick only the outcomes that result in an even sum.

Then the rest of the problem can be solved with a probability tree. Okay, that would be how i'd do it coz i'm stupid - but it will work!

Darkwall, hope you don't mind if I tumpang your thread.

A friend asked me earlier. Let's say you have these parents. The parents have two children. One of them is a girl. What is the probability that the other child is a boy?

Assume there's 1 girl to every 1 boy in the world (it's close). When calculating the the probability of the other child's gender, do we consider the population as a whole (6 billion, so the probability that it will be a boy is a tiny bit higher than 1/2), or do we consider this family in itself (so, 2/3 that it will be a boy because they already have a girl).