Testing random

[random832 at fastmail.us]

On Fri, Jun 12, 2015, at 17:32, Thomas 'PointedEars' Lahn wrote:
> Ian Kelly wrote:
> 
> > The probability of 123456789 and 111111111 are equal. The probability
> > of a sequence containing all nine numbers and a sequence containing
> > only 1s are *not* equal.
> 
> There is a contradiction in that statement.  Can you find it?

There is not.

The probability of 111111111 is ~0.000000002581, given each digit is
chosen from 1 to 9.

The probability of 123456789 is ~0.000000002581.
The probability of 123456798 is ~0.000000002581.
The probability of 123456879 is ~0.000000002581.
The probability of 123456897 is ~0.000000002581.
The probability of 123456978 is ~0.000000002581.
The probability of 123456987 is ~0.000000002581.
The probability of 123456897 is ~0.000000002581.
The probability of 123457689 is ~0.000000002581.
The probability of 123457698 is ~0.000000002581.
And so on, for a total of 362,880 combinations.

The probability of getting one of these combinations is ~0.000936656708

~0.000936656708 is not equal to ~0.000000002581. QED.

Your problem is that you _continually insist_ on interpreting
"probability of a member of this set" as meaning "probability of one
particular member of this set" rather than "probability of any member of
this set".

It's like claiming the probability of getting a digit from 1 to 3 when
you roll a six-sided die is ~0.1667 instead of 0.5, and playing stupid
when people point out why you're wrong. And it's not even defensible as
an honest misunderstanding because the linguistic ambiguity doesn't
exist in the original claim about the limit probability of full
coverage.