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So place the BIG rocks inside the hat for a 75% chance of missing them!
Thomas A. Sult, MD
3rd Opinion
1415 First First St. South #5
Willmar, MN 56201
320 235 2101 Office
www.3rdOpinion.us
tsult@mac.com






On Dec 9, 2010, at 1:31 PM, George Huxtable wrote:

> John Karl writes "I think where George�s argument goes wrong is not
> thinking about integrating a variable probability/area over some  
> specified
> area (like inside the hat)."
>
> That is correct. I am instead applying reasoning that depends on pure
> logic. If Karl wishes to knock it down, he will have to do better than
> integrating a variable probability. He will have to show a flaw in the
> logic.
>
> The argument applies to any combination of azimuths of the three  
> bodies,
> but it's easy to think of in a particular case when they are  
> separated by
> 120�. It does not depend on the nature of the distribution, so it  
> applies
> to the gaussian as well as to other shapes, All that it requires is  
> that
> there is no systematic bias, but only random error, so that the  
> deviation
> of any position line from the true position of the vessel is just as  
> likely
> to be towards the body as it is away. And to keep things simple, we
> consider that the probability of exactly zero deviation is  
> sufficiently
> small as to be negligible.
>
> Then, if we put the three bodies in order, it's clear that there are 8
> possible combinations, where T is toward, and A is away-
> TTT
> TTA
> TAT
> TAA
> AAA
> AAT
> ATA
> ATT
>
> There is no reason for any one of these combinations to be favoured  
> above
> the others, so the probability of each is 12.5%
> But of those combinations, only two, AAA and TTT, create a triangle  
> that
> embraces the body. In the other six cases, 75% of the total, the  
> body is
> outside the triangle. QED.
>
> It makes no difference whether the operator is skilled or unskilled,  
> or if
> the weather is rough or smooth. Those situations will produce cocked- 
> hats
> of various sizes, but the probability of being inside or outside that
> triangle remains exactly the same.
>
> Years ago, I checked it out, to remove all doubt in my mind, by  
> setting up
> a computer simulation. It was not of exactly this same problem, but  
> of the
> analogous problem of compass bearings to three landmarks, perturbed  
> by the
> sea-state, in which the likelihood of the compass bearing being in  
> error to
> the lefy, and to the right, was exactly the same.
>
> I collected sufficient statistics to show that the probability of  
> being
> outside the triangle was in the range 74% to 76%, which was convincing
> enough for me.
>
> If John Karl's conclusion differs from 75%, I will suspect a flaw in  
> his
> mathematics (or in his definitions) unless be can demolish that  
> logical
> argument.
>
> George.
>
> contact George Huxtable, at  george@hux.me.uk
> or at +44 1865 820222 (from UK, 01865 820222)
> or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
> ----- Original Message -----
> From: "John Karl" 
> To: 
> Sent: Thursday, December 09, 2010 5:53 PM
> Subject: [NavList] Re: That darned old cocked hat
>
>
> Yes, George once had me convinced that the probability of being  
> outside the
> hat is 75% for all cases.  But after giving it more thought, I don�t
> believe it.  My logic goes like this:
>
> We�re considering a normal distribution of random errors (no other  
> type of
> errors) that is known a priori from many tests.  The probability of  
> the
> true fix being at any given point is the product of these  
> distributions
> (see the attached PDF file).  This Probability P(x,y) is the  
> probability
> per unit area.  And it can never exceed 1.0.  The probability of the
> correct fix being inside any given area is the integral of P(x,y)  
> over that
> area.  As the area of the cocked hat gets smaller, and tends to zero,
> P(x,y) does not get arbitrarily large.  In fact it can never exceed  
> 1.0.
> Therefore as the hat�s area get smaller, the probability that the  
> true fix
> is inside the hat goes to zero.  If you get a tiny cocked hat, in all
> likelihood you�re not there.
>
> The probabilities quoted in the second attachment below are  
> approximate
> because, for the integration, I approximated the hat triangle with the
> closest contour (inside and outside the triangle to get a good  
> estimate).
> Whatever the accuracy of these numerical integrations are, they don�t
> change the above argument.
>
> I think where George�s argument goes wrong is not thinking about
> integrating a variable probability/area over some specified area (like
> inside the hat).
>
> The attached file also shows that all the best estimates that I have  
> seen
> (Fermat's point, the center of gravity, the bisecting angle point,  
> and the
> Steiner point) can not be the most probable location of the true  
> fix.  (An
> observation that's of no practical use.)
>
> JK
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