🚀 go-pugleaf

In article <3356c9d1.4969186@158.152.254.10>, Hugh Steers
<hugh@flounder.demon.co.uk> writes
>
>Ok,
>
>Ten hat wearers go to a party, they all put their hats on the
>(initially empty) hatstand. After the party they all leave drunk,
>taking a hat at random from the hatstand.
>
>What is the probability that nobody gets his own hat?
>
>-- Hugh.
>
This is the classic problem of putting n letters into n envelopes such
that every letter goes into a wrong envelope.

If P(A) is the probability that letter A goes into envelope A, and if
 n = 10 then   P(A) = 1/10

            P(A and B) = 1/10 x 1/9

  P(A and B and C and ... and K) = 1/10 x 1/9 x 1/8 x ... x 1/2 x 1/1

To ensure that no letter is correctly placed we must however calculate
the probability that some at least are correct, and then subtract this
result from 1.

So what we require is   1 - P(A or B or C or .... or K)

P(A or B or C or .. or K) = P(A) + P(B) + P(C) + ... + P(K)
                            - 10_C_2 x P(A and B)
                            + 10_C_3 x P(A and B and C) 
                            - 10_C_4 x P(A and B and C and D)
                            ----------------------------------
                            - 10_C_10 x P(A and B and ... and K)  

               = 10(1/10) - 10*9/2!(1/10 1/9) + 10*9*8/3!(1/(10*9*8)) 
               = 1 - 1/2! + 1/3! - 1/4! + .... - 1/10!

So required probability is  1 - [1 - 1/2! + 1/3! - ...   - 1/10!]

                     1/2! - 1/3! + 1/4! .... + 1/10!

Note that as n gets larger this tends to the value e^(-1) = 0.36788

-- 
Anthony H.Back