MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 added 71 characters in body

Here is a guide to the intuition. I will not swear that the numerics are exact, but I will bet that the numerical truth is not far off.

Look at the diagonal for the multiplication table of a (labeled) groupoid on $n>3$ elements. Of the n^n possibilities, only one of them is idempotent, so with one exception aa=b will happen for some a and some b different from a. Now all we need for associativity to fail in this case is that ab and ba are different, which will happen for all but n of the n^2 possibilities. So we are already looking at associativity happening only on the order of n^(-n-1) a small fraction of all (non-idempotent) tables, especially as there are often several candidates for a, and only one is needed.

Even for idempotent groupoids, one finds a,b,c distinct and needs to consider only d=ab, g=bc, and the ways in which dc and ag can fail to be equal. Again in rough terms we are talking about n^(-2), and this is just by fixing a,b, and c in advance, and that for the 1 out of n^n tables that are idempotent.

I'll let someone else tighten up the numerics. For strengthening Joseph's intuition, I hope this should will suffice.

Look at the diagonal for the multiplication table of a (labeled) groupoid on $n>3$ elements. Of the n^n possibilities, only one of them is idempotent, so with one exception aa=b will happen for some a and some b different from a. Now all we need for associativity to fail in this case is that ab and ba are different, which will happen for all but n of the n^2 possibilities. So we are already looking at associativity happening only on the order of n^(-n-1) fraction of all (non-idempotent) tables.