There is such a thing as a partition of a positive integer; it's a way to write that positive integer as a sum of positive integers, where order is irrelevant. So, for example, the partitions of 7 are
7, 6+1, 5+2, 5+1+1, 4+3, 4+2+1, 4+1+1+1, 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1, 2+2+2+1, 2+2+1+1+1, 2+1+1+1+1+1, 1+1+1+1+1+1+1
and there are fifteen of them. We can define a function p, where p(n) is the number of partitions of n; thus p(7) = 15.
So how might one pick a partition of n "at random"? There are at least three ways I know of. The first is to pick one uniformly, so each one is chosen with probability 1/15.
The second is as follows: begin with the partition 1. Then generate a partition of 2 by augmenting some part of the partition by 1; we can either augment the first part (giving 2) or the second part (there is no second part, so this corresponds to adding a new part 1, giving 1+1). Then do the same thing to obtain a partition of 3. If you had 2 at the second step, then you get 3 or 2+1 at the third step with equal probability; if you had 1+1 at the second step, then you get 2+1 or 1+1+1 with equal probability. But you can't have 1+2 -- we make the rule that partitions have to be written in decreasing order. So at the third step, you have 3 with probability 1/4, 2+1 with probability 1/2, and 1+1+1 with probability 1/4. At the fourth step, 3 becomes 4 or 3+1 with equal probability; 2+1 becomes 3+1, 2+2, or 2+1+1 with equal probability, and 1+1+1 becomes 2+1+1 or 1+1+1+1 with equal probability. Thus you have the partitions 4 or 1+1+1+1 with probability 1/8 each; the partitions 3+1 or 2+1+1 with probability 7/24 each; the partition 2+2 with probability 1/4.
As you can see, this process doesn't generate partitions uniformly at random. But in a way it's natural, because at each step of the process one is doing something uniform. There's a natural way to draw a picture of a partition as a Ferrers diagram, where a partition of n will have a diagram with n dots; this corresponds to putting down n dots "at random" subject to the constraint that at each step the diagram you obtain is a valid Ferrers diagram. It turns out that the Ferrers diagram obtained in this way tend, up to scaling, to a certain shape as they grow large. As do partitions chosen uniformly at random -- but it's a different shape.
And there's a third way of choosing partitions at random. This is called the "Plancherel measure". We pick a partition λ of n with probability (dim λ2)/n!, where dim λ is the "dimension" of a partition. The dimension of a partition is just the number of ways in which it can be obtained as the nth term of a sequence of partitions, starting with the empty partition and at each step adding one to some part. For example, the partition 5 = 3+2 is the last element of
0, 1, 2, 3, 3+1, 3+2
0, 1, 2, 2+1, (3+1 or 2+2), 3+2
0, 1, 1+1, 2+1, (3+1 or 2+2), 3+2
so there are five such sequences, and dim (3+2) = 5. Thus in the Plancherel measure we choose the partition 3+2 with probability 52 / 5! = 25/120. The dimension of a partition is also the number of standard Young tableaux corresponding to it.
This just seems so artificial, though -- if I wanted to actually pick a partition according to this distribution, how would I do it?
There are n! permutations of n objects. Picking one of these uniformly at random seems like a natural thing to do. (For example, it has a physical instantiation in the shuffling of a deck of cards.) It turns out that there is a correspondence called the Robinson-Schensted correspondence. (Sometimes Knuth's name is in here as well.) This correspondence associates with a permutation of [n] a pair of standard Young tableaux, each containing n squares, and of the same shape. The shape of those Young tableaux gives a partition, then; if we started by picking a permutation uniformly then we've now obtained a partition chosen with the distribution induced by the Plancherel measure.
Suddenly this distribution seems natural. I'm much more comfortable with the idea of picking something from a random distribution if I can imagine how I would actually go and choose it if I had to.