Testing random

[Ian Kelly]

On Wed, Jun 10, 2015 at 11:03 AM, Thomas 'PointedEars' Lahn
<PointedEars at web.de> wrote:
> Jussi Piitulainen wrote:
>
>> Thomas 'PointedEars' Lahn writes:
>>> Jussi Piitulainen wrote:
>>>> Thomas 'PointedEars' Lahn writes:
>>>>>   8 3 6 3 1 2 6 8 2 1 6.
>>>>
>>>> There are more than four hundred thousand ways to get those numbers
>>>> in some order.
>>>>
>>>> (11! / 2! / 2! / 2! / 3! / 2! = 415800)
>>>
>>> Fallacy.  Order is irrelevant here.
>>
>> You need to consider every sequence that leads to the observed counts.
>
> No, you need _not_, because – I repeat – the probability of getting a
> sequence of length n from a set of 9 numbers whereas the probability of
> picking a number is evenly distributed, is (1∕9)ⁿ [(1/9)^n, or 1/9 to the
> nth, for those who do to see it because of lack of Unicode support at their
> system].  *Always.*  *No matter* which numbers are in it.  *No matter* in
> which order they are.  AISB, order is *irrelevant* here.  *Completely.*

Order is relevant because, for instance, there are n differently
ordered sequences that contain n-1 1s and one 2, while there is only
one sequence that contains n 1s. While each of those individual
sequences are indeed equiprobable, the overall probability of getting
a sequence that contains n-1 1s and one 2 is n times the probability
of getting a sequence that contains n 1s.

The context of this whole thread is about the probability of getting a
sequence where every number occurs at least once. The order that they
occur in doesn't matter, but the number of possible permutations does,
because every one of those permutations is a distinct sequence
contributing an equal amount to the total overall probability.

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.