I just encountered this interesting combinatorics problem that seemed easy but actually stumped me for a while. My combinatorics and statistics has definitely rusted a bit over the years .. ;-)
If you pick N random numbers, what is the probability that they are sorted in ascending order ? (no dups possible)
1/N! = 1/(1*2*3*4*..*N)
- Q2) How would you check that the list is sorted ?
How many checks will you do in a) best case b) worst case c) avaerage
a) 1 b) N-1 c) aehhhhh… hmmm.
On average, how many checks do you have to do to establish a list being sorted or not ?
This took me a while to work through, i found it deceptive – but here’s a simple proof i finally came up with.
The expectation value of X, E(X) is the sum of all the possible Xes times their respective probabilities, right:
E(X) = SUM[ X * pX ] for X=1,2,3,...N
In this case X is the number of tests needed to be performed, and take values 1,2,3,…,N where N is the size of the list. pX, the probability that any given list will require exactly X tests. Now, consider a given list of numbers, [a,b,c,d,e,f,..] For us to end up needing exactly, say, 3 tests the following 2 things must be true: a,b and c must be in order while d will NOT be in order. the first two tests pass while the third fails.
1) the probability that a,b and c are in order is 1/3! (one over 3 factorial) 2) the probability that a,b,c and d are in order is 1/4!
thus the probability that 1 is true but 2 is not true is p3 = 1/3! – 1/4! or more generally the probability that we will require X tests is
pX = 1/X! - 1/(X+1)!
So, plugging that into the expectation value formula we get
E(X) = SUM[ X * (1/X! - 1/(X+1)! )] = SUM[ X/X! ] - SUM [ X/(X+1)! ]
now lets write that out so we can collapse some terms:
E(X) = 1/1! + 2/2! + 3/3! + 4/4! + ... - 1/2! - 2/3! - 3/4! - ...
Summing the terms that are above each other we get:
E(X) = 1/1! + 1/2! + 1/3! + 1/4! + ... = e - 1 = 1.712..
THe reason this is actually = e – 1 is that e itself is defined as a series that looks just like that with an extra 1 at the beginning
e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ....
see http://en.wikipedia.org/wiki/E_(mathematical_constant) P.S. A much simpler way to show this occurred to me later: (it circumvents the need to calculate the probabilty that a list required exactly X tests, only that it requires at least X tests.
Given the set of possible lists:
- Each will require the first test
- 1/2! of them will have the first two numbers in order, and thus require a second test
- 1/3! of them will have the first 3 numbers in order, and thus require a third test
Thus the average will be
1 + 1/2! + 1/3! + 1/4! + … = e – 1