The Hilton Hotel reserves all its 101 rooms for travelers and each
guest is assigned a room in advance. The first guest arrives but
has forgotten his room number. The hotel clerk, who does not have
access to the reservations book, randomly puts him in one of
the rooms. As the rest of the guests arrive they are given their
reserved room if available or if already taken, are given a random
empty room. What is the chance that the 101st guest gets
her reserved room?
6 comments:
Solution
Assumption : Every other guest knows his/her room no.!
With probability 1/101 first guest gets right room...so every other guest gets right room.
with probability 100/101 1st guest gets wrong room. 2nd guest gets either his room or a wrong room. so in system of 2 rooms only one room is allocated wrongly as either of two rooms belong to second guest.
In the same way after 1oo guest have arrived, only one could be the one which belong to 101th guest. So with probability 1/2 he gets his room.
Total probability = (1/101)*1 + (100/101)*(1/2) = 51/101
It is always 1/2, no matter how many rooms. This is essentially the same as the seats on a plane puzzle.
See http://puzzles4you.blogspot.com/2011/01/crazy-guy-on-airplane.html?showComment=1295166869087#c557275896153041844 for explanation.
Squirrel's solution looks right. I should have solved the puzzle in more structured manner.
Great puzzle!
Keep up the good work.
Check braintaming.blogspot.com for more ways to improve your mental ability.
n!= the # of possibile outcomes
(n-1)!= # of outcomes that will give you your room
# of favorable outcomes/ # possible outcomes = P(getting you room)
P(getting your room) = (n-1)!/n!
(n-1)!/n! = 1/n
P(getting your room) = 1/n... 1/101
I agree with squirrel also
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