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    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
    
    
    

       
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