Question

Probability (counting): Each one of n persons, indexed by 1,2,…,n, has a clean hat and throws it into a box. The persons then pick hats from the box, at random. Every assignment of the hats to the persons is equally likely. In an equivalent model, each person picks a hat, one at a time, in the order of their index, with each one of the remaining hats being equally likely to be picked. Find the probability of the following:

Each one of persons 1,…,m gets back a hat belonging to one of the last m persons (persons n−m+1,…,n), where 1≤m≤n.

Answer 1

This is a question from the edX probability class I just got wrong. If you are in that class, please don't try to get the answers here.

Answer 2

I can't :(

Answer 3

ok, can people please stop telling me why they can't help so I can just wait for someone who knows the answer?

Answer 4

Cuz Im in 8th :D

Answer 5

clearly the denominator is \(\large\binom nm\), but what the heck is the numerator?

Answer 6

Wouldn't the numerator be the number of people... I'm brainstorming >_<

Answer 7

unfortunately I don't know the answer, so I need some pretty clear logic to back up whatever answer y'all throw at me. I can only tell you two answers that are wrong (the one's I used in my attempts :P)

Answer 8

Hmm, what was your attempt?

Answer 9

\[\large\frac m{\binom nm}~~and~~\frac{n-m}{\binom nm}\]

Answer 10

so not those

Answer 11

Hm, why did you use M? It says the number of people is n, my attempt would have been something similar to the second, where did the 1 go?

Answer 12

I used some logic like "the probability of the remaining n-m people getting the hats of the previous m people is \[(\frac m{n-m}\cdot\frac{m-1}{n-m-1}\cdots )\] as I'm typing it I realize this logic is flawed. I think the wording of the question throws me.

Answer 13

I erased my previous work, so I don't remember what I did exactly, but it was multiple choice, and none of my work led to a possible answer, so I looked for the ones that looked close :P

Answer 14

in other words, I'm lost as to how to approach this

Answer 15

The wording of the last part also threw me off D; "in the other of their index" ... I'm not sure how to set this up ;-;

Answer 16

I can ask the tutors at my school, though I'm worried they start using complicated equations and I'm just looking at random numbers >_<

Answer 17

I was trying to think of special cases, like when m=n, but I couldn't get that to look like anything I wanted

Answer 18

sounds good

Answer 19

well I'm not crying about it; it was 1 point of 30 and the only I missed. I'll patiently await a clear approach.
for now, I am off for pizza. feel free to debate the q without me.
later!

Answer 20

kk bye!!

Answer 21

Have fun ;p

Answer 22

i was thinking P(n-m,m) / P(n,m) but i'm pretty sure it's wrong XD

Answer 23

I would think the total probability space would be \(nPn=n!\)

Answer 24

There are \(mPm=m!\) ways to ensure that the first \(m\) people get the last \(m\) hats. Then there are \((n-m)P(n-m)=(n-m)!\) ways to assign the remaining hats.

Answer 25

yes, but we need only consider the subset of the sample space, which is the number of ways we get m subsets of a particular value, 3.
then all we have to do is consider the number of ways that can occur for the numerator

Answer 26

Do you happen to know the answer?

Answer 27

so you say the answer is \[\large {m!(n-m)!\over\binom nm}\]?
and not, I know it is one of two possibilities I did not choose

Answer 28

no*

Answer 29

I'm getting \[
\frac{m!(n-m)!}{n!}
\]And I could be wrong, but I'm not sure where my analysis is flawed.

Answer 30

ok, or \[\frac 1{\binom nm}\] which was another option and my thrid choice, though I kinda forgot how I derived that atm...
I thinkyou are right

Answer 31

no thinking of the numerator only...

Answer 32

Well, there might be a way to derive \[{n \choose m}^{-1}\] through analysis.

Answer 33

I was thinking we could say... dangit I can't recall how I cane to the smae answer right now
.. I'm kinda trying to eat dinner lol, sorry

Answer 34

yeah I came to that conclusion at one point I recall, and dismissed it. Sorry right now is not the best time for me to try to recall how exactly I got that answer

Answer 35

you are essentially ignoring the order of assignment, in which case there is only one ordering which would work.

Answer 36

no, I used ordering
ok I think my idea was, that the remaining m people have m-1 hats to choose from, hence the product (m-1/m)(m-2/m-1)(...)=1/m, and ... that there were m ways this could happen... I'm sorry I don't recall how I came to that last conclusion

Answer 37

yeah I gotta eat, sorry
I think you are right @wio , if you could elaborate that would be appreciated. I have to go for now, thanks!