[EMAIL PROTECTED] wrote:
>
> I was watching the "random" drawing of letters to determine ballot
> order in the California primary. They had a cylindrical bingo style
> cage with a door in the middle. They inserted each letter of the
> alphabet, beginning with "A", etc into little film tubes. As they did
> this, the later letters naturally forced the earlier letters toward
> the ends of the cylinder. They then rotated the cylinder 4 or 5 times
> and proceded to draw the letters. Of course, the film tubes mostly
> rolled in place, occasionally tumbling over each other, but really not
> being disturbed much. The "drawer" then reached down in and pulled out
> a letter. The cylinder was rotated once more, another letter drawn,
> etc.
>
> I said to myself, 'the chances of the first few letters coming from
> the beginning of the alphabet are pretty slim. <p> The letters came
> out as follows.
>
> R, W, Q, O, J, M, V, A, H, B, S, G, Z, X, N, T, C, I, E, K, U, P, D,
> Y, F and L
>
> 4 of the first thirteen letters drawn were from the first half of the
> alphabet. All of the first five letters drawn were from the last
> half.
No, they weren't. J is the tenth letter.
I very much doubt whether anything can be shown. Even if it is not
properly randomized, you would need a very strong pattern to appear to
allow for the huge Bonferroni correction you would have to make.
That is: there are dozens of _a_priori_ plausible tests you could make
(early letters in the first few, late letters in the first few, early
letters in the middle, middle letters at the end...) and you would have
to multiply any p-value you got by the number of tests you's high-graded
it out from among. So nothing much weaker than (say) all of the first
ten letters being in the first half of the alphabet (p=.001) would be
much use. And you haven't got that. All of the first four, six of the
first seven, nine of the first 13: none give very small p-values
individually. (Six of the first seven, at a bit over 5%, is the best,
and even that won't stand up to Bonferroni.)
Everything *else* about this process is mad enough. (Consider: if 45%
support the governor, 35% support the Terminator, and 20% support Mickey
Mouse, the Terminator wins. That's unles the governor's supporters were
smart enough to vote _en_masse_ to remove him and then make him his own
replacement, in which case he'd win. And presumably if Arnie wins the
Democrats can imediately start a recall against *him* which may well go
through too.)
-Robert Dawson
.
.
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