(Followup directed to sci.stat.edu) In article <[EMAIL PROTECTED]>, DeLa <[EMAIL PROTECTED]> wrote: >Maybe it is a bit off topic, but I couldn't find a ng that better matches my >issue. Which is... > >We are a group of nine people. We want each of us to buy a present for one >other member. We have a hat with names and each one picks a name. What is >the chance of at least one picking his or her own name? > >I have come close two times: > >- Calculate the chance for anybody to pick another name and adding those >chances, which resluts in a calculation like this: (8/9) * (7/8) * etc. and >then reversing the chnace. This results in a chance of 1/9. Not really >correct I'm afraid. Also using the idea of 1/n! does not get me closer. > >- We can write down all possible outcomes and then count the ones in which >at least one has his own name. If it was really important I'd do that but >I'm looking for the rationale. > >The point seems to be that we have to take into account the possibility that >is the first person draws the name of the second person, he has a chance of >8/8 to draw another name than his own.
As another poster suggested, the key is to find the probability that no-one gets his/her own name. Take a step back for a minute and think about permutations of 9 objects. Assuming nothing is rigged :-) drawing names from a hat corresponds to picking a permutation at random with all permutations equally likely, so the problem can be solved by counting permutations. Let A be the set of permutations in which the first object gets mapped to itself, B be the set in which the second object gets mapped to itself, and so on. Write out a simple expression for the set of permutations in which no object gets mapped to itself. Then write out an expression for the number of elements in the set. (I'm assuming you are taking a course with some elementary probability content, and the you've seen things like #(A union B) = #(A) + #(B) - #(AB).) . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
