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?

## 10 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.

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

Agree w/ Squïrrel also..

So for this problem, you have to figure out all the probabilities so you would start by doing the probability that the 1st guest actually got his/her own room which is 1/101 since it was randomly assigned. then I can't remember the equation you can use from stats class to get to the probability of A such that B occurred for 101 people. But each one, there is one less room available so the chance of the second person getting in to their room is 100/101 I think? So the formula for the remaining 100 guests for probability is (101-n)/101 but then you'd have to account for all the probabilities beforehand so would that be (101-n)/101 multiplied by the other probabilities? I'm not positive. this would be my first way to solve it

Any of the 100 guests that arrived after the first guest could have reserved the room taken by the first guest. However, the chance that the last guest gets her reserved room is not 99/100, because there is a 1/101 chance that the clerk randomly placed the first guest into the room that the first guest reserved. Therefore, there is a 100/101 chance that the chance for the last guest to get her reserved room is 99/100 and a 1/101 chance that the chance the last guest gets her reserved room is 100%.

If the first man gets the right room, then everybody gets the right room. If he gets wrong then there is 100% chance that the woman will not get her room. So it all depends on one probability, whether the first man got his room right. My answer is 1/101.

1/101

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