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. -- https://mail.python.org/mailman/listinfo/python-list
