Sorry, all. New to the list. I accidentally posted my
previous under a nonsense user id :-)

> > Unless I'm mistaken, in the 5e6-from-1e9 sampling,
> the probability
> > that a sampling contains exactly one number from a
> given 200-numbers
> > interval is 200.0*(1/200)*(199.0/200)**199 =
> 0.3688. The probability
> > that this happens for 20 such 200-numbers
> intervals is
> > 0.3688**20 = 2.1e-09.
> 
> I think you are wrong. A rough estimates of this
> probability gives me
> 1e-2171438.
> Actually, it should be a bit smaller, but at least
> the order of
> magnitude of the order of magnitude is right :)
> Quite impressive...
> 
> Paolo
> 
> 
I think he is wrong too (sorry. Missed the previous
post, so I don't know who it was) but I think I think
it's wrong for a different reason.

I got the probability of getting exactly one number in
a specific interval to be as follows:
Probability of one *specific* number in the interval
1/20
Probability of all the other 19 OUT of that interval
(19/20)**19
But we have 20 specific numbers to consider, so:
20 * 1/20 * (19/20)**19
or
(19.0/20)**19
which gives a remarkably similar 0.3773536 (don't know
if this is a co-incidence. My brain hurts :-)

Where I think this goes wrong is in assuming that
expanding this to 20 intervals just needs a **20. 

After we get exactly one number in the first interval,
we then are looking for exactly one number in the
second interval, which is only one interval out of 19.
In other words, the probability should
be(18.0/19)**18. and so forth. The overall probability
therefore is
(19.0/20)**19 * (18.0/19)**18 * ... * (2.0/1)**1
which my RubyShell informs me is 0.377353602535307e-8,
or thereabouts :-)

Graeme

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