NavList:

I believe the question, "What are the odds of being inside the cocked hat?" is ill-posed, comparable to the "Where's the missing dollar?" (see https://en.wikipedia.org/wiki/Missing_dollar_riddle).  In other words, intuition is leading us astray.

Consider the case where the three Lines of Positions (LOPs) cross at a single point. There is no "cocked hat", and no probability of being inside it. But of course, we would expect our position to be near the point of intersection.

By the way, three LOPs are better than two... it's more information.

The proper question, after plotting three LOPs, is, "What region do I have 50% (or 99%) probability of being in?".  With some hand-waving, we can argue each LOP's Probability Density Function can be represented by an elongated (long and thin) ellipse, and we can find the joint probability function of the three ellipses. From that, we outline the smallest region that encompasses, for example, 99% of the probability.

It is inconvenient to use a region as our fix from which to start our dead-reckoning.  Thus we pick some point as representative of our most likely position, and I would argue that in practice, that point could be e.g. the centroid, incenter or symmedian of the cocked hat.  

A reason for choosing the symmedian point is that it is consistent with the "least-squares" approach described in the section "Sight Reduction Procedures", in the back of the Nautical Almanac.  The symmedian point minimizes the sum of the squares of the distances to the sides of the triangle, and thus is a graphical solution (for three LOPs) of the least-squares approach; though the NA's numerical algorithm extends to more than three LOPs.

But in all cases, the plotted fix represents a region where we are (most likely) located.

If you insist on asking what is the probability of being in the cocked hat, one can imagine collecting thousands of three LOP fixes and comparing to a GPS fix.  We can get the answer that way, but it probably would not be illuminating, in terms of navigation.  But perhaps a mathematician would find it an interesting question.