Probability of seven people sharing a room in a 25 floor hotel

1. Jun 10, 2014

Dostre

1. The problem statement, all variables and given/known data

Seven persons are staying at a hotel. The hotel has twenty-five floors, each with the same number of rooms. Each of the seven persons has been randomly assigned to one of the rooms in the hotel.

a. What is the probability that at least two people have rooms on the same floor? Show and explain how you calculate the answer.

b. Suppose you are one of the seven people. What is the probability that at least one of the other six persons has a room on the same floor as you? Assume there are at least 6 rooms on each floor of the hotel. Show and explain how you calculate the answer

c. Explain qualitatively why the answer to part b. is smaller than the answer to part a. In other words, give an intuitive argument or reason for the difference in probabilities.

2. Relevant equations
$P(A)=1-P(A^c)$

3. The attempt at a solution

Part a. Probability that at least two people have a room on the same floor is the complement of probability that no one has rooms on the same floor.
Event A=at least two people have rooms on the same floor
Event $A^{c}$=everybody lives on distinct floors.

$P(A^c)=(25*24*23*22*21*20*19)/25^7=0.397$
$P(A)=1-P(A^c)=0.603$

I think it is correct.

Part b. Using the same logic and taking into account that there are at least 6 rooms on each floor I get that there 24*6=184 rooms to choose, since I occupy the 25th floor.
Event A=at least one of the other six people has a room on the same floor with me
Event $A^{c}$=nobody lives on the same floor with me
Hence,
$P(A^c)=(184*183*182*181*180*179)/184^6=0.92$
$P(A)=1-0.92=0.08$

Is this correct?

Part c. I think it is because in part b order matters, but I am not sure. Help appreciated. Thanks.

2. Jun 10, 2014

Staff: Mentor

That indeed seems correct, although I think that your answer should be more detailed to fulfill the "explain how you calculate the answer" criterion.

What if there actually 7 rooms on each floor? 8? Your answer shouldn't change: the problem mentions at least 6 rooms, not exactly 6. That said, I first assumed that this meant that there are enough rooms on each floor to accomodate everyone, which is not exactly true if there are 7 guests if you include yourself. There is no single solution in that case, so I would still assume that this is a mistake, and that you are to consider that all guests can fit in a single floor.

Is it really order, or is it because one guest is different from the others?

3. Jun 10, 2014

Dostre

@DrClaude sounds good. What do you suggest should be my approach to part b then? An equation would be really helpful.

4. Jun 10, 2014

Staff: Mentor

Your basic approach is good. So, considering that you are on a given floor, what is the probability that the 6 other guests are all on other floors?

5. Jun 10, 2014

Dostre

(24/25)^6

oh ok makes sense. In this way I do not need to account for rooms at all then. For part c I am still struggling.

6. Jun 10, 2014

Staff: Mentor

That answers my question, but not the orginal question

Which probability is highest, the one in a or the one in b? From that, figure out which cases satisfy the higher probability that do not satisfy the lower probability.

7. Jun 10, 2014

D H

Staff Emeritus
There are two problems here:
1. 24*6 = 144, not 184.
2. The denominator 184^6 is wrong, even after you correct for the 24*6 error. There are five rooms on your floor that you aren't taking into account.

Suppose the hotel only has four floors and there are only three rooms per floor and the hotel only has three floors. That makes 12 rooms. You've got one of them, so there are 11 left. There are 11^6 ways to assign those 6 people to those 11 rooms. That's the denominator in the four floors, three rooms per floor problem.

I'll leave correcting the denominator in the 25 floor, 6 rooms per floor case up to you.

8. Jun 10, 2014

Dostre

@D H I subtract those five rooms cause I try to calculate the probability of none of those six people being on the same floor with me.

9. Jun 10, 2014

D H

Staff Emeritus
Those five rooms still need to be in the denominator. The denominator represents *all* of the ways those six people can be assigned rooms. You've taken care of them being on some other floor than yours by the numerator.

10. Jun 10, 2014

D H

Staff Emeritus
There are two parts to the answer for part c. One is the answer DrClaude is moving you towards.

The other is finite number of rooms per floor. Your calculation for part a implicitly assumed an infinite number of rooms per floor.

Let's take the extreme case of a finite number of rooms in the hotel: What would your answer to part b be if there was only one room per floor?

11. Jun 10, 2014

Dostre

then the probability of at least one other person being on the same floor with me would be zero.

12. Jun 10, 2014

D H

Staff Emeritus
Exactly. All other things being equal, that a floor has a finite number of rooms reduces the probability you will be have a co-occupant on your floor. All other things not being equal, what if your floor has 20,000 rooms and the other 24 floors only have 6 rooms each? (Note that this extreme example shows why you have to adjust the denominator as indicated earlier.)

In the limit of every floor having an infinite number of rooms, you'll still get a lower probability that at least one of the other 6 is on your floor than the probability that at least two people have rooms on the same floor.

13. Jun 10, 2014

Dostre

I get what you are saying. In the problem statement they say that each floor has the same number of rooms. I would think this assumption holds for part b where they say at least 6 rooms per floor. Hence, I think it is reasonable to say the probability that one of the other 6 people are on the same floor with me is 1-(24/25)^6. To address the changing number of rooms I may use the equation below

p=1-C(24*n,6)/C(25*n-1,6)

I think it sounds like a reasonable answer. What is your opinion?

Last edited: Jun 10, 2014
14. Jun 10, 2014

haruspex

While I agree with what you say, my impression is that the problem setter did not intend the exact number of rooms per floor to be considered. Indeed, the statement
is nonsense. The formula is not going to change magically at that threshold.
I suspect the problem ought to have said, for all parts,
assume there is a very large number of rooms on each floor​

15. Jun 10, 2014

Dostre

uhh guys. I am still cracking my head to come up with an answer to part c. I know this problem is very similar to the Birthday Problem. So one person is fixed. Now, there is a pool of six people that can be on my floor. I am stuck there.

16. Jun 10, 2014

D H

Staff Emeritus
That's the answer for a very large number of rooms on every floor.

That's rather different from what you had before. But yes, that looks correct.

You'll have to justify this expression, of course.

Indeed. Using Dostre's 1-C(24*n,6)/C(25*n-1,6), there's nothing magical at n=6:
http://www.wolframalpha.com/input/?i=Plot[1-((24*n)+choose+6)+/+((25*n-1)+choose+6),{n,1,20}]

17. Jun 10, 2014

Ray Vickson

For part (a) the answer depends on the number of rooms per floor. Say there are $R$ rooms on each floor. The first person to register gets some room on some floor, $F_1$. That leaves $24R$ rooms on floors different from the first one, and $25R-1$ rooms altogether for the next person to occupy. So (given a random choice of rooms) the probability that the second person to register gets a room on a different floor ($F_2$) from the first is $P(F_2 \neq F_1) = 24R/(25R-1)$. The probability that the third person's floor $F_3$ is different from the first two is $23R/(25R-2)$, etc. So, in terms of the total number of rooms $N = 25R$ we have
$$P\{ \text{all different} \} = \frac{24 \cdot 23 \cdot 22 \cdots \cdot 19 R^6}{(25R-1)(25R-2) \cdots (25R-6)} = \frac{19381824}{48828125}\frac{1}{(1-\nu)(1-2\nu) \cdots (1-6\nu)}, \: \nu \equiv \frac{1}{N}$$
The first factor in the above is $19381824/48828125 \doteq .3969397555$ so gives your 0.397 answer; the second factor is a correction that is always > 1 but which will be pretty close to 1 for a large hotel. In any case, if you know $R$ you can compute it exactly. Of course, if $R \to \infty$ then the correction factor = 1 exactly, and your answer obtains. Your answer would also be correct if floors, rather than rooms, are selected at random, but that is not what the question implied.

You can do part (b) in terms of a hypergeometric distribution: from your point of view there are two types of rooms left for the other 6 to occupy: rooms of type I = other rooms on your floor, and rooms of type II = rooms on other floors. There are $N_1 = R-1$ rooms of type I and $N_2 = 24R$ rooms of type II. The complementary probability in (b) is the probability that in selecting 6 rooms without replacement (from the total $25R-1$ rooms) we obtain $0$ type I rooms. This is just the hypergeometric probability $P(X = 0)$, where $X$ is a 'hypergeometric' random variable with parameters $(N_1,N_2,n) = (R-1,24R,6)$.

18. Jun 10, 2014

haruspex

If you accept my suggestion that the question was posed badly, and should have said to assume a large number of rooms on each floor, the best hints you've been given for part (c) are those from DrClaude. Yes, it's to do with the reason that the answer to the Birthday problem surprises people.
pace Ray & D H, but trying to include the actual number of rooms per floor is probably not relevant to the task in hand.

19. Jun 10, 2014

Dostre

Figured it all out guys.

Ray Vickson thanks for an elaborate answer. I did similar calculation using combinations which gave me the same results.

For part c, assuming there is a large number of rooms on each floor, question in part b is a subset of question in part in short. All answers may be considered answered and the thread closed.