Testing random
sohcahtoa82 at gmail.com
sohcahtoa82 at gmail.com
Wed Jun 10 13:52:01 EDT 2015
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.
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