Derangement - Complete FAIL!

Source: Interview at one of the quant firms

Problem:
As posted in problems this and this, this is an extension to the derangement problem.
There are n men, n hats, one hat belonging to each person. A random permutation of hats is picked by the men. What is the probability that no person gets the correct hat?

Update (March 6, 2011)
Solution: Posted by Vipul Verma (Engineer at Portware, IIT KGP and University of South California Alumnus) in comments!

Comments

  1. Vipul6:31 AM, March 06, 2011

    The number of arrangements so that no one gets his own hat = n!*(1 - 1/1! + 1/2! - 1/3! + 1/4! ..... + (-1)^n/n!)

    Total number of arrangements = n!

    The prob that this happens = (1 - 1/1! + 1/2! - 1/3! + 1/4! ..... + (-1)^n/n!)


    Easy one....

    ReplyDelete
  2. Pratik Poddar4:23 PM, March 06, 2011

    Yes, easy one. Direct application of Inclusion Exclusion Principle. :)

    But your answer is wrong.

    Prob that no one gets his own hat = 1 - (Prob that at least one correct hat) = 1 - (nC1*1/n - nC2*1/(n*n-1) +..) = 1 - (1 - 1/2! + 1/3! .. ) = 1 - 1/e
    Prob: 1-1/e

    Your solution gives the probability that at least one person gets the correct hat.

    ReplyDelete
  3. Vipul3:52 AM, March 08, 2011

    Pratik, how is my answer wrong? How my soln is giving the prob. that at least one person gets the correct hat.

    If you look closely the expression you have got is same as what I got. The only difference is that expression has infinite terms which I don't understand why.

    Your series should not go to infinity. It should stop at n.

    ReplyDelete
  4. anujjain10:18 AM, March 08, 2011

    Pratik, if you see yours and vipul's answer are actually same.

    ReplyDelete
  5. Pratik Poddar11:31 AM, March 08, 2011

    I am sorry. My apologies.
    Yes your solution is correct. Our answers are same. Also, I should not have taken infinite terms. Thanks.

    ReplyDelete
  6. Unknown1:32 AM, March 11, 2011

    1 - 1/2! + 1/3! .. = 1 - 1/e, and so the prob for no one get it right should be 1/e, right...?

    ReplyDelete
  7. Pratik Poddar1:15 PM, March 13, 2011

    @Yun.
    Yes you are correct.

    For large n

    Prob that no one gets his own hat = 1 - (Prob that at least one correct hat) = 1 - (nC1*1/n - nC2*1/(n*n-1) +..) = 1 - (1 - 1/2! + 1/3! .. ) = 1 - (1-1/e)
    Prob: 1/e

    ReplyDelete

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