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The Law of Large Numbers
The nut does not fall far from the tree
 but in fact, how far away does it fall?
Suppose I roll a die 100 times, and find the percent of times I rolled a 4.
I expect about
of the rolls to be 4s.
1
6
Of course, in any particular series of 100 rolls, I don’t expect the percent to be
exactly
.
(In fact, it can’t be, because
of 100 rolls is not a whole number.)
But
1
6
1
6
how far away will my percent be?
To answer this question we calculate the standard
deviation of the percent.
But first, some technical details:
The jargon:
Repeating some action such as rolling a die and counting the number of times
you get a particular outcome is called “Bernouilli trials.”
(The trials must be
independent.)
The outcome you are counting is called a “success”  even if you are
counting failures!
The number of successes is a random variable called the “Binomial
random variable.”
The notation:
the number of trials (rolls) is n
the chance of a “success” on any one trial (roll) is p
the actual percent of “successes” in an actual series of trials (rolls) is
$
p
Then
is a random variable (It will come out differently for different actual series.)
$
p
( 29
μ
σ
$
$
p
p
p
p
p
n
=
=

1
The formulas for
and
have been found using algebra.
If you are interested, I
$
p
$
p
can show you the details.
Notice that the number of trials (n) appears in the denominator of the formula for the
standard deviation.
That means that as the number of trials increases, the actual
percent of successes is likely to get closer to the longrun percent, p.
That is the Law
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 Spring '08
 Staff
 Statistics, Law Of Large Numbers, Probability

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