It's the same. Let M = 3N, then
((1-1/N)^N)^3 = (1-3/M)^M
Quote:
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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.
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