What would you guess derangement means in mathematics?
No, we are not describing mathematicians or what they do, but the same people who gave us irrational numbers, gave us deranged permutations. Imagine 5 people each wearing a different hat. They toss the hats in a box and randomly select one. What do you think it would mean for the final hat distribution to be a derangement of the original? It does make sense in an odd sort of way.
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11 Answers
Ew. I’m not mixing my hat with 4 unknown hats.
Well, the opposite of arrangement.
I’ll throw my hat in the ring. I am not afraid.
That the hats don’t end up on the original heads?
Or only a certain percentage do?
I don’t f’n know. Lol
Everyone ends up with a different hat?
It went from order to disorder.
WHOSE WIG IS THAT IN THOSE HATS??
The deranged mathematician threw it in there.
Those who said everyone ends up with a different got the correct answer, so the term must be at least somewhat descriptive. What, I am hoping at least some of you are wondering, are the chances of getting a derangement? If we had started with a larger number of people, at least 10, the probability rapidly goes to a little over ⅓, 1 / e actually, where e is Euler’s constant equal to about 2.7. The average number of people who get their own hat, for any number n, is n x 1/n = 1. Averages behave much better than probabilities.
I started to wonder that as soon as yi read your first sentence. I figred the first person’s chance of being “deranged” “is 4 /5, The second person’s chance is ¾. The thrid ⅔. The fourth ½ So my guess is 4×3x2×1/ 5×4x3×2 = .2 20%? But I don’t really know. I’ll see what your real answer gives.
The formula, which is easy to find online, is p = ½! – ⅓! + ¼! -1/5! + ... For n =5, this works out to about .37
The hat example I gave is based on a well known recreational math problem involving scrambled hat checks.
The usual proof involves the inclusion-exclusion theorem. There is also a slick derivation, based on a recurrence relation. See, for example, here
Actually, I can’t follow the above version of the recurrence proof. Here is a much easier version to follow, using a clever bit of manipulation that I read in a book.
Starting with the recurrence relationship Dn =(n-1)(Dn-1 + Dn-2),
expand the right side and do the clever step of subtracting nDn-1 from both sides,giving
Dn – nDn-1 = -Dn-1 + (n-1)Dn-2
If we let f(n) = Dn – nDn-1 then the equation says that f(n)= -f(n-1)
In other words Dn – nDn-1=f(n)= [-1]^^n k for some constant k. You can directly compute Dn for small n. D1=0, D2=1 and D3=2, so. D2 – 2D1 = 1 and D3 – 3D2 = -1, k must = 1.
Dn -n Dn-1 = [-1]^^n
To get probabilities,divide both sides by n! to get
Dn/n! – Dn-1/(n-1)! = [-1]^^n/n!
:Pn – Pn-1 = [-1]^^n/n!
See if you can take it from there. Note that P1 = 0
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