Thank you for the heads-up.

The story is a little bit more complicated. A few months ago, Ubuntu One announced that because of new security systems all users would have to de-register their machines and then re-register them. Because of a bug somewhere, I was unable to de-register my laptop. So I simply copied everything out of the Ubuntu One directory and waited. Ubuntu One promised that they would tell me what to do to get round the problem, but they never did. So when they announced the shut-down I was already prepared; they can delete all my files now.

I decided in the end to buy a subscription to Dropbox. The reason for hestitating was that you could buy multiples of 20Gb from Ubuntu One whereas the minimum order from Dropbox was 100Gb. But I now have enough stuff to fill up a fair proportion of 100Gb. Of course, I keep my options open by regular backups onto a portable hard disk.

]]>I hope you would be able to (or already have) downloaded your stuff from there!

]]>then where

is the indicator function for the event that

lies in the image of the function. The

are Bernoulli variables with mean approximately

but are only approximately independent. I would have to

swot up on my probability theory to see if there’s a version

of the central limit theorem liberal enough to apply here,

but it does seem plausible that asymptotic normality holds. ]]>

The probability that the image has size k is

where is a Stirling number of of the second kind. The

mean size of an image is exactly

(n times the probability the image contains ).

Similar arguments give exact formulas for the variance etc.

The probability that the image has size k is $(n!/(n-k)!) S(n,k)/n^n$

where $S(n,k)$ is a Stirling number of of the second kind. The

mean size of an image is exactly $n(1-(n-1)^n/n^n)$

(n times the probability the image contains $1$).

Similar arguments give exact formulas for the variance etc.