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Old 03-15-2005   #48 (permalink)
noise_gaining
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Join Date: Mar 2005
Posts: 1
half dozen is six

It's the same. Let M = 3N, then

((1-1/N)^N)^3 = (1-3/M)^M




Quote:
Originally Posted by Madory
This makes sense, thanks.

Perhaps it is a case of 6 and one-half-dozen. I got my explanation from "Attacks On RC4 and WEP" by FMS:

"The probability that three locations will not be pointed to by a pseudo random index during the
remaining N - 1 - x rounds is better than ((1-1/N)^N)^3 ~ e^-3 ~ 5%."

((1-1/N)^N)^3
can be reduced to
(e^-1)^3
and finally
e^-3

-OR-

(1-3/N)^N
reduced directly to
e^-3

Anyway, thanks for the general formula - crystal clear now.

Last edited by noise_gaining : 03-15-2005 at 12:01 PM.
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