# [RC5] Winning key statsistics stuff

Skip Huffman SHuffman at Atl.Carreker.Com
Fri Apr 3 08:00:57 EST 1998

```On Wed, 01 Apr 1998 21:36:47 -0500, Brice D. Fleckenstein wrote:

>> Seriously, the original poster is confusing two different levels.  If we
>> solved a whole bunch of these contests and graphed where in the keyspace
>> the "winning" key was, then it would form a bell curve
>
> Not it wouldn't. The plot would be FLAT give or take standard
>deviation, if enough contests were run to be statistically significant.
>
> You're not using multiple "dice" for key generation - you're using a
>SINGLE die of a heck of a lot of sides.

Did my message below ever make it to the list?

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On Mon, 30 Mar 1998 23:27:40 -0500 (EST), Phil Gregory wrote:

>If we
>solved a whole bunch of these contests and graphed where in the keyspace
>the "winning" key was, then it would form a bell curve

Dead wrong.  The curve will be flat if we measure where in the
keyspace the key was.  If I take a die and roll it 1000 times, I am
going to get just about as many ones and sixes as I am threes and
fours.

The key is a randomly chosen number between 1 and 2^64.  A large
number of randomly generated numbers is going to be evenly
distributed within its limits.

Now if you generate two random numbers and add them, the sum will
tend towards a bell curve.  Look at two dice.  There are six
combinations that could result in a six,seven or eight: (1,6) (2,5)
(3,4) (4,3) (5,2) (6,1); (6,2) (5,3) (4,4) (4,4) (3,5) (2,6); (1,5)
(2,4) (3,3) (3,3) (4,2) (5,1).  Four that result in a four, five,
nine, or ten: (1,4) (2,3) (3,2) (4,1); (3,6)(4,5) (5,4) (6,3); (1,3)
(2,2) (2,2) (3,1) ;(4,6) (5,5) (5,5) (6,4).  Two that result in two,
three, eleven or twelve: (1,1) (1,1); (1,2) (2,1); (5,6) (6,5); (6,6)
(6,6).  This does form a nice bell curve.

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If you consider that pairs are really the same roll the curve is
even prettier

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(By the way, use a fixed font.)

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