good chemistry is complicated,
and a little bit messy -LW
Re^3: [OT] Stats problemby syphilis (Bishop)
|on Feb 26, 2015 at 14:04 UTC||Need Help??|
What are the odds of a card ending up in its standard position?
That's easy enough - it's 1 in 52.
A card can be in any one of 52 positions, only one of which is its "standard position" - and there's no bias that favours one position over another.
UPDATE What follows doesn't scale correctly ... it aint right IOW - probably the linkage that salva was getting at.
The chance that none of the cards are in their standard position is simply (51/52) ** 52, which is around 0.3643 or approximately 1 in 3.
Therefore the chance that 1 or more cards are in their standard position is 1 - ((51/52) ** 52) which is around 0.6357 or approximately 2 in 3.
(You could turn this into a rather boring game of patience/solitaire.)
As for what you're actually trying to calculate, I'm still trying ot get my head around it.
(Please stick to cards in future ;-)