On Wed, Jun 10, 2015 at 11:03 AM, Thomas 'PointedEars' Lahn <pointede...@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. -- https://mail.python.org/mailman/listinfo/python-list
