

Nir Dahan Inner circle Munich, Germany 1390 Posts 
A number of people (N) are visiting a hotel. They each get a room with a key.
They go to dinner and have to put their keys in a big container. When they return from dinner (possibly drunk) they each reach in and pick a key at random. What is the probability that at least one will go back to his original room? What happens when the number of people is very large? N. 
Jonathan Townsend Eternal Order Ossining, NY 27017 Posts 
How different is this from the birthday problem?
...to all the coins I've dropped here

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Re read it.
Here each guy gets a complete different room, two people cant go back to the same room. It is completely unrelated. 
Top Hat Inner circle We peed on you! 1077 Posts 
When the number of people is very large, the number of rooms will be very large, and hence the number of keys will be very large, and hence the container will need to be very large.
What happens is that: a) it takes a very long time to put the keys into the container (maybe more than 1 week) b) the container, when full, is very heavy (and would probably need to rest on a sturdy table rather than being held in the hands) c) it takes a long time to take the keys out of the container (cf: a) above)
TH

Jonathan Townsend Eternal Order Ossining, NY 27017 Posts 
All in the container.
First one takes a key. Either theirs (1/n) or not n1/n) say not... Next one take a key. Either theirs (1/n1) or not (n2/n1) seems pretty darn likely that someone will get their key, as for this NOT to happen would require each get a wrong key and that gets less likely with each pick.
...to all the coins I've dropped here

TomasB Inner circle Sweden 1143 Posts 
Jonathan,
It's not that easy since your second probability (second person taking their key) is dependant on if the first person picked the second person's key or not. If the first person picks the second person's key the conditional probability of the second person taking his own key is 0. This is the same as the Hat Problem and the math behind the classic card trick Frequent Miracle which was discussed at http://www.themagiccafe.com/forums/viewt......forum=99 /Tomas 
Top Hat Inner circle We peed on you! 1077 Posts 
Tomas  by Hat Problem, do you mean me?
TH

MagiClyde Special user Columbus, Ohio 871 Posts 
Aren't the odds ((number of people) * (number of keys))=chance of getting your key?
Magic! The quicker pickerupper!

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Hint  it has something to do with "e" ...

Top Hat Inner circle We peed on you! 1077 Posts 
E by gum?
TH

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Stan, Tomas ?
This one's for you guys. This is a very beautiful problem with a somewhat strange outcome... nir 
stanalger Special user St. Louis, MO 996 Posts 
Tomas already answered this, didn't he?
(Or at least provided a link where the answer can be found.) 
Nir Dahan Inner circle Munich, Germany 1390 Posts 
Quote:
On 20070111 10:46, stanalger wrote: Dang, Didn't see that  that is correct. Guess I'll have to find another goodie for you guys. 
The Magic Cafe Forum Index » » Puzzle me this... » » Hotel (0 Likes) 
[ Top of Page ] 
All content & postings Copyright © 20012020 Steve Brooks. All Rights Reserved. This page was created in 0.2 seconds requiring 5 database queries. 
The views and comments expressed on The Magic Café are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic. > Privacy Statement < 