How many posts? The number of posts changes the necessary number of bumps. Unless you're looking for an equation. In which case I'll digress from the conversation.

n-1 is my guess. you dont have to bump the most recent post, it will automatically be last after every other post has been bumped. do i get a +?

Number of posts(factoral) ez math

n= number of posts.
n!
------
(n-1)!

it's O(n), (you remove constants in big Oh notation-- worst case is n-1). no factorial. worst case for 7 posts is 7654321

step 1- bump 6 = 6754321
2- bump 5 = 5674321
3- bump 4 = 4567321
4- bump 3= 3456721
5- bump 2= 2345671
6- bump 1 = 1234567

QED bitches.

when skrum used the word "necessary" i interpreted that as "worst case". my bad.

P = [(n-1)^AIM]/[(n-1)^2 /MSN]

right?

it depends entirely on how things are ordered. if the 2nd post ever posted is still ordered after the first post and the third post is not still after the 2nd, then it would require N-2 bumps.
I will use < and > to denote order in with the posts are listed. If pos A> Post B then that means that post A has been bumped more recently.
the total number of bumps needed to put things back in the order they where posted would then be.
B=N-X where B is the number of bumps required, N is the total number of posts, and X is the number of posts, starting at the first post, that are ordered correctly.

so in the case of this 20 post example ordering where post are numbered 1-20

Oldest 5, 12, 1, 7, 9, 2, 3, 13, 11, 10, 4, 6, 8, 15, 18, 20, 19, 14, 17, 16 newest
X=4 because 1<2<3<4>5

One more example

Oldest 2, 1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 newest
X=1 because 1>2

B=N-X

I could be wrong, but I doubt it.