in reply to Spooky math problem
Actually, you do. In order to generate a random number, you have to know the probability distribution. Therefore your original assertion was right, and my analysis was right. I didn't get it 100% last time. In the problem the probability distribution is the 'trivial' uniform, infinite distribution and with a mean of zero you can actually beat better than 5050 odds with only the ability to generate a random number  you don't actually have to do it. If you are randomly given a second number without knowledge of the distribution, the result tells you something about the distribution again giving you better than 5050 odds. I'm done!
RE (tilly) 2 (benford): Spooky math problem by tilly (Archbishop) on Nov 04, 2000 at 20:32 UTC 
Please look at Spooky math  with Perl. You will see that the numbers
I have are parameters to the experiment, the guesser uses
no knowledge about numbers, and the trick is that the
guesser is sometimes guaranteed of being right.
The only limits to the guessing rule in that program are
internal to how computers select pseudorandom numbers and
the floating point math that Perl uses.
Other incidental notes. Your "trivial" uniform, infinite
distribution actually does not and cannot exist. Its not
existing has deep consequences. The details of why not are
covered in real analysis. In the US and Canada this would
traditionally be taken either by an advanced 4'th year
math student or a beginning graduate student.
And random trivia. Not only is an infinite uniform
distribution impossible, but attempts to look for really
random numbers invariably turn up patterns that don't fit
with a uniform distribution. For instance Benford's law
states that the first digit obeys a logarithmic
distribution. It isn't really a theorem, but other than
that detail the
following
is a good introduction for the general public. Knuth tries
to explain it in his series, but does not manage IMNSHO to
show why his abstract model has anything to do with reality.
Just thought I would throw that out there...  [reply] 
