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Dear George,

Sorry, but now I think you've got it wrong because you're playing John Karl's game of actually measuring three specific LOPs. In such a case you have to calculate the probability contours, and you will not get exactly 25% probability contained within the triangle. If the lines almost intersect the probability will approach zero even though this appears to be a good outcome. If the lines miss by a wide margin the probability of including the location will be large.

Your 25% probability of inclusion applies to an average over all possible LOP triplets. In this game you measure N LOP triplets where N is a very large number. The larger the number N of measurements taken, the closer you will get to the ideal 25% figure.

This is what I meant about both of you having the right answer to different problems.

Regards,
George

From: george---me.uk
Date: 11 Dec 2010 00:05

I think John Karl and I may be at cross-purposes, obtaining different
answers because we are calculating different things, as George Brandenburg
has suggested.

Let me explain to John Karl exactly what is the event for which I claim the
probability to be 75%.

A vessel is at some position. It does not matter whether the exact position
is known to those aboard the vessel, or whether it isn't known.
Nevertheless, the vessel has a certain position.

Three position lines are measured to three stars. There is no systematic
error, only a scatter which makes a towards error, with respect to the
vessel's position, just as likely as an away error. It does not matter
whether the distribution is Gaussian or otherwise. Nor does it matter how
wide that distribution is. From arguments that are by now familiar, if we
construct a triangle from those three position lines, the probability that
the triangle will not embrace the position will be exactly 75%..

And if we repeat the exercise, from that same position or from any other
position, the probability will be 75% each time.

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