By popular demand, here is a video showing you how to solve the AusCERT 2012 puzzle which featured on our conference T-shirts at this year's AusCERT conference in Queensland, Australia:
(Enjoy this video? Check out more on the SophosLabs YouTube channel.)
If you enjoy this sort of thing, keep reading Naked Security - we run puzzles fairly regularly.
You can also follow me on Twitter - @duckblog - and watch for the hashtag #sophospuzzle, where we announce new puzzles and offer hints on how to solve them.
You might also enjoy watching or trying previous #sophospuzzles!

















Or P(invalid) = nPk / (n^k) = 52!/( (52-13)! * (52^13) ) = 0.194544...
(where nPk is the Permutation function)
If you're demonstrating that longhand as 52/52 x 51/52 x ... 40/52, it's worth an aside explaining that the component probabilities are independent events, hence the multiplication for an overall result (which leads to the condensed form above).
So, did anybody actually work it out on an HP-42S? :)
You mean "P(valid) = nPk / n^k"...
I decided not to labour the explanation of the probability calculation and simply to describe them as a "chain of probabilities" - but with hindsight that could be taken to imply that they are somehow _related_, rather than independent events (on account of using a new and independently-shuffled pack for each card).
That's one of the pains of YouTube - rather than Occam's Razor (the shorter of two equivalent explanations is better), the Attention Span Razor applies instead (the shorter video is better). Sorry about that :-)
As for your question, "Did anyone work it out on an HP-42S?"
Yes. I did!
(I had a cunning plan of showing a video of the calculation done using the excellent Free42 software simulator. But Attention Span's Razor intervened again.)
Of course you're right - I wrote that first as the 1-p inverted expression and then mis-edited for more "clarity" :/
Cool puzzle, gongrats to the winners they were quick!!