Testing random

[sohcahtoa82 at gmail.com]

On Wednesday, June 10, 2015 at 10:06:49 AM UTC-7, Thomas 'PointedEars' Lahn 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.*
> 
> This is _not_ a lottery box; you put the ball with the number on it *back 
> into the box* after you have drawn it and before you draw a new one.
> 
> > One of those sequences occurred. You don't know which.
> 
> You do not have to.
> 
> > When tossing herrings […]
> 
> Herrings are the key word here, indeed, and they are deep dark red.
> 
> > Code follows. Incidentally, I'm not feeling smart here. 
> 
> Good.  Because you should not feel smart in any way after ignoring all my 
> explanations.
> 
> > [nonsense]
> 
> -- 
> PointedEars
> 
> Twitter: @PointedEars2
> Please do not cc me. / Bitte keine Kopien per E-Mail.

To put it another way, let's simplify the problem.  You're rolling a pair of dice.  What are the chances that you'll see a pair of 3s?

Look at the list of possible roll combinations:

1 1     1 2     1 3     1 4     1 5     1 6
2 1     2 2     2 3     2 4     2 5     2 6
3 1     3 2     3 3     3 4     3 5     3 6
4 1     4 2     4 3     4 4     4 5     4 6
5 1     5 2     5 3     5 4     5 5     5 6
6 1     6 2     6 3     6 4     6 5     6 6

36 possible combinations.  Only one of them has a pair of 3s.  The answer is 1/36.

What about the chances of seeing 2 1?

Here's where I think you two are having such a huge disagreement.  Does order matter?  It depends what you're pulling random numbers out for.

The odds of seeing 2 1 are also only 1/36.  But if order doesn't matter in your application, then 1 2 is equivalent.  The odds of getting 2 1 OR 1 2 is 2/36, or 1/18.

But whether order matters or not, the chances of getting a pair of threes in two rolls is ALWAYS 1/36.

If this gets expanded to grabbing 10 random numbers between 1 and 9, then the chances of getting a sequence of 10 ones is still only (1/9)^10, *regardless of whether or not order matters*.  There are 9^10 possible sequences, but only *one* of these is all ones.

If order matters, then 7385941745 also has a (1/9)^10 chance of occurring.  Just because it isn't a memorable sequence doesn't give it a higher chance of happening.

If order DOESN'T matter, then 1344557789 would be equivalent, and the odds are higher.