Re: Cantorian pseudomathematics

Han de Bruijn wrote:
> cbrown@xxxxxxxxxxxxxxxxx wrote:
>
> > In that case, can you define what /you/ mean by "a randomly selected
> > member of a countably infinite set", potential or otherwise?
> >
> > It might help if you first define what /you/ mean by "a randomly
> > selected member of {1,2,3,..., n}"?
> >
> > Just to anticipate - if I have an n-sided die which I claim is
> > "random", what would you accept as evidence that my claim is true? It
> > doesn't seem sufficient to only require that in m rolls I find "nearly"
> > m/n of them come up each of 1,2,3,.. etc.; for if I rolled the
> > sequence:
> >
> > (1,2,3,...,n,1,2,3,..,n,1,2,3,...n,...)
> >
> > this would hardly be evidence of a "random" die to me.
>
> Maybe it helps if I quote someone else than myself.

Not paricularly. We already know that the standard definitions, which
Stephen is using, result in the conclusion "it is not possible to
select a natural randomly with a uniform distribution". You disagree
with this result, claiming that you use other definitions; so I want to
know what /you/ think "select a natural randomly with a uniform
distribution" means.

> Here is what Stephen
> writes in a recent thread called "Continuum hypothesis", about the same
> subject:
>
> > If you have a finite set (1..n), the probability of choosing
> > a number divisible by a is exactly floor(n/a)/n. If you
> > want to make this close to 1/a, just choose a multiple of
> > a for n, and floor(n/a)/n=1/a. There is no reason to make
> > n particularly large, and no reason to consider limits.
>
> I want to establish a reference point first. Can we all agree on this?
>

Unless we know what "the probability of choosing" actually means, how
do we know that Stephen's numerical result is correct? Does it follow
/only from the above/ that the probability that we choose a number
which is either 0 or 1 mod a is 2/n, or is it (n+1)/n^2? Is the
probability that we /don't/ choose a number divisible by a /also/ 1/n?
Without a definition, we have to guess.

The above is not a definition; it is an /application/ of a definition.
Yes, it /follows/, provably, from the usual definition of "the
probability of choosing 1 from n elements in a uniform distribution"
(as do the answers to the other questions I posed above); but that
won't help us in our current discussion.

What does it mean to say that we choose some element x "at random" from
some set X? When we assign a number such as 1/n as "the probability of
choosing x from X", what does that mean?

Some people, such as yourself, are asserting that "choose at random"
makes sense if X is the naturals; others assert it is not possible. The
means of resolving this dispute is not to find the limit of n/an as n
-> oo; no one disagrees as to the value of that limit. It is to clarify
what is meant by "choose at random" and "probability".

If you can't even define what these terms mean in the finite case, how
can you make a sensible assertion about what it might possibly mean in
the infinite case?

In the sequence of rolls of an n-sided die I gave above:

> > (1,2,3,...,n,1,2,3,..,n,1,2,3,...n,...)

each number 1 to n appears with asymptotic frequency 1/n. Would you
think such a die was a fair, random die?

Can /you/ define what you mean, mathematically, in at least the finite
case - what do you mean when you say an n-sided die is a fair, random
die? Or even simpler - what does it mean to say a coin flip is fair and
random?

Would you then say a sequence of fair coin tosses is equivalent to a
sequence of selections from a set of two elements, chosen at random?

Cheers - Chas

PS: Define your terms, declare your premises, proceed logically to the
conclusions.

.

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