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Me culpa. Trevor (and George) is right here. The triangle is the larger
one, and 25% is exacty correct for this case, which is not a more
plausible case than my first one.

On Sat, 20 Apr 2002 19:49:12 -0300, Trevor J. Kenchington wrote:

> >One possibility is that they are just wide enough to make all three
>observations possible. That results in the traditronal center point of
> >the triangle being the only non-zero probability.
> >
> >Another possibility is to make the rectangles just cover all the other
>observations. If the cocked hat happens to be equilateral, the
> >non-zero area expands to fill the triangle, and is zero everywhere
>outside.
> >
>
>Not as I plot it out. I have a non-zero area covering a large
>equilateral triangle, comprising the cocked hat and three identical
>triangles, each abutting one side of the 'hat. In short, that model fits
>George's 25% probability that the true position is inside the 'hat.
>
>
Rodney Myrvaagnes                                                        J36 Gjo/a


"Nuke the gay whales for Jesus" -- anon T-shirt

   

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