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Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
You have 13 cards numbered ace through king (or numbered cards 1-13)
They are shuffled randomly. What is the probability that not one of the cards (none) lies at a position corresponding to its value? Another way of stating this is, I am looking for the probability that a random permutation of thirteen cards results in no fixed point (a derangement). My calculation is 2.8%. Thanks, Larry |
Dale A. Hildebrandt Special user 637 Posts |
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Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
Thanks Dale for the link. Very helpful. I made an error in reading off a list of derangements from https://oeis.org/search?q=derangement&language=english&go=Search.
I was off by one because they begin counting at 0 with zero having 1 derangement and 1 having 0 derangements. For a set of 13, there are 2290792932 derangements out of 6227020800 possible permutations. 2290792932/6227020800 = .3678 or about 3.7%. For any large set greater than 10 the number of derangements approaches 1/e = 0.367879441... Larry |
Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
Correction of my typo:
The percent of derangements is about 37%, not 3.7% Interestingly, the chance of having 1 fixed point (one right) is about the same as having no fixed points (a derangement). It's amazing how e, the natural log base is so ubiquitous in mathematics. |
Dale A. Hildebrandt Special user 637 Posts |
If you think of the fixed point "arrangement" as another derangement variation (one that is deranged to fixed points) it helps, at least for me, to understand that the chances are the same as other available (non-fixed) derangements.
Sincerely, Dale |
Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
Good point Dale. It's an argument based on symmetry which can be used to simplify complex problems.
Larry |
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