February 27, 2015 No.3

February 27, 2015

https://www.facebook.com/mehtarahulc/posts/10152623327721922

A proposed "un-fixable" procedure to select a large number of Jurors randomly
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Say we have to select 1500 Jurors from a list of N = about 4 crore voters between age of 18 years and 120 years .
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Order then in list by their assembly constituency number , booth number, serial number in section (as it is done today)
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So they are all numbered from 1 to N = about 4 crore. 
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Say following procedure is used to select 1500 Jurors between age of 30 years to 55 years.
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m = a prime number at between 0 and N selected at random in public using multiple dice-pair throw 
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k = a prime number at randomly between 1 lakh and 2 lakh selected at random in public using multiple dice-pair throw 
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select m
select m + k 
select m + 2k 
....
(if number is larger than N, then divide N and take remainder i.e. modulo).
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If person selected is below 30 years or above 55 years , then select next person. Repeat the process till 2000 are selected. Send summons to all 2000. At least 1900 will show up (100 may be dead of transferred) . And select first 1500 from these 1900.
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===
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Here is how dice pair throw can decide a number between 0 and 4 crore.
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The number has 8 digits . 
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Two dices are thrown. One is white, another is black. And translation is done as
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(W = 1 , B = 1) , digit = 0
(W = 1 , B = 2) , digit = 1
(W = 1 , B = 3) , digit = 2
(W = 1 , B = 4) , digit = 3
(W = 1 , B = 5) , digit = 4
(W = 1 , B = 6) , digit = 5
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(W = 2 , B = 1) , digit = 6
(W = 2 , B = 2) , digit = 7
(W = 2 , B = 3) , digit = 8
(W = 2 , B = 4) , digit = 9
(W = 2 , B = 5) , digit = 0
(W = 2 , B = 6) , digit = 1
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(W = 3 , B = 1) , digit = 2
(W = 3 , B = 2) , digit = 3
(W = 3 , B = 3) , digit = 4
(W = 3 , B = 4) , digit = 5
(W = 3 , B = 5) , digit = 6
(W = 3 , B = 6) , digit = 7
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(W = 4 , B = 1) , digit = 8
(W = 4 , B = 2) , digit = 9
(W = 4 , B = 3) , digit = 0
(W = 4 , B = 4) , digit = 1
(W = 4 , B = 5) , digit = 2
(W = 4 , B = 6) , digit = 3
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(W = 5 , B = 1) , digit = 4
(W = 5 , B = 2) , digit = 5
(W = 5 , B = 3) , digit = 6
(W = 5 , B = 4) , digit = 7
(W = 5 , B = 5) , digit = 8
(W = 5 , B = 6) , digit = 9
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(W = 6 , B = 1) , reject
(W = 6 , B = 2) , reject
(W = 6 , B = 3) , reject
(W = 6 , B = 5) , reject
(W = 6 , B = 6) , reject
(W = 6 , B = 7) , reject
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So each digit between 0 and 9 both inclusive has equal chances of getting selected.
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So if 8 digits are to be selected, the dices have to be thrown 8 times (or may be 9 , 10 times if rejects show up)
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If the number selected is larger then N, then dice throws have to be repeated.
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