# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Still on LOP's**

**From:**Martin Gardner

**Date:**2002 Apr 18, 07:25 -0700

George Huxtable wrote - > Well, let's say we are determining our position by bearings on three > distant landmarks, 1, 2, and 3. > > There is an equal chance that, due to errors in taking the bearing from > landmark 1, that bearing will lie to the left of the true position as to > the right of it. If we take the possibility that the bearing can be exactly > on the line of the true position to be zero, the probability of it being on > the left (L) is 0.5, the same as it being on the right (R). We can say the > same about landmarks 2 and 3. > > There are eight possible combinations, if we list the three bearings, taken > in the order 1, 2, 3, as follows- > > LLL, LLR, LRL, LRR, RLL, RLR, RRL, RRR. > > For each such combination, because it combines 3 terms each with a > probability of 0.5, its probability is (0.5) cubed, or 0.125. There are 8 > such combinations, each with a probability if 0.125, so that looks right, > doesn't it? George, it sure does. But when I draws it out, the plane is divided into only seven areas: the inside of the triangle, the areas outside each side, and the areas outside each vertex. One of the combinations cannot occur on a real plane (which one depends on how you draw and number your lines). > > However, of those 8 combinations, there are only two which place the true > position inside the cocked hat. These are LLL and RRR. This can be seen if > a drawing is made showing all the 8 options. The other combinations put the > true position outside a side or outside a corner. So the probability of the > true position being inside the cocked hat is exactly 0.125 x 2, or 0.25, > which is what we set out to show. When I drew this, RRR and LLL lay outside my triangle. I was able to redraw it so that RRR lay inside, but then there is no area LLL. I think something went awry in your reasoning: for any given three lines intersecting to form a triangle, there is only one area 'inside'. If two different patterns of R and L both described some of the 'inside' then the 'inside' would have to be partitioned, which it is not. > > What seems at first so surprising is that this result is quite independent > of the skill of the navigator. The reason for this is that the better > navigator will produce, on average, a smaller cocked hat, But the > probability of it embracing the true position will remain at 1 in 4. I don't have an opinion on whether the probability is 1 in 4 (or 1 in 7, or one in 8); I have encountered this line of reasoning before, and I think there's something amiss with it, though I can't exactly quantify what. My worry is that I don't think the three bearings are actually independent. It's not as if we were dropping three long straws at random on a chart - we're in some sense measuring the same thing all three times. Martin Gardner Venice CA