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Both the encryption algorithm and the hypothetical algorithms used by
the adversary involve random events. We will need a theorem that says
that, if an event with probability is tried times, the chance that
the number of successes is not close to is small. The paper uses^{7}(p. 293)

**Lemma 24**
*Let be the number of successes in tries.
For any *
*
*
**Proof:** is a random variable, which is the sum of
independent random variables, each having value 0 or 1. Let be the
variance of . Each of the 0-1 variables has variance ,
so

Lemma 24 provides a very rough estimate of the probability.
An improvement requiring much more work is:

For comparison, if , , the probability that there are
successes is .1087. Lemma 24 gives^{8}an upper limit of .3125,
while Lemma 25 gives .1498. (these figures courtesy of
Mathematica)

One reason the paper does not
use Lemma 25 is that it does not give a simple formula for
how large would have to be in terms of the other quantities. We will
not use this result later, and you should skip to section 7.3
unless you like to manipulate formulas.

**Proof:** We will assume is integer.
From the binomial theorem:

implies and

which gives the second factor of . We
use Stirling's formula on the binomial coefficient and group it with the
powers of and to obtain:

The first factor of is the first factor of . We
obtain upper bounds on the rest of , using

(the lower bound on
involves a geometric series)

Adding these and using gives the remaining factor of .

#### Footnotes

- ... uses
^{7}
- The usual central limit theorem cannot be used because it does
not tell you how large must be for the normal distribution to give
a good estimate.
- ... gives
^{8}
- We divide by 2 to eliminate the probability of .

** Next:** The magic of sampling
** Up:** Probabilistic Encryption
** Previous:** The Goldwasser-Micali encryption system
** Contents**
*Translated from LaTeX by Scott Sutherland *

2002-12-14