

Finding 2013 in Pi - wmat
http://www.johndcook.com/blog/2013/01/01/finding-2013-in-pi/?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+TheEndeavour+%28The+Endeavour%29

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omonra
<http://www.angio.net/pi/piquery>

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omra
Assuming the digits of pi are randomly distributed, any finite digit sequence
can be found in pi.

The probability any sequence of length d is found in N digits of pi is 1 -
1/exp(N*0.1^d) (Poisson distribution for approximating the binomial). Then the
limit as N approaches infinity is 1 for any finite d.

~~~
hypersoar
The probability of an event being 1 is _not_ the same thing as that event
being completely certain. For example, if you pick a random real number
between 0 and 1, the probability of getting something rational is zero. It's
clearly not impossible, though.

~~~
gizmo686
Do you know of any proof for that? My math intuition is telling me that
randomly picking a real number is guarenteed to be irrational, based on the
fact that there is an uncountable infinity real numbers, but only a countable
infinity of rational numbers. But, without assuming a probability of 0 means
impossible, I do not know how to go about proving/disproving this.

~~~
hypersoar
We can't really pick "random real numbers" in any practical sense, so this is
pretty much a theoretical distinction. It's essentially a matter of
definition. The rational numbers have probability zero of being drawn, but
they still lie in the sample space.

One way to see it is this: The probability of picking any particular point in
the interval is 0. But that doesn't mean that picking that point is
impossible. _Some_ point has to show up when you pick one at random.

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jey
He buried the lede!

    
    
       In fact, no one has proved that any particular number is
       normal. However, we do know that almost all numbers are 
       normal. That is, the set of non-normal numbers has 
       Lebesgue measure zero.

