Simple probability problem
I love probability puzzles. Here is one I liked:
There are a 100 people trying to get onto the same flight you are. The airplane has a 100 seats. You are all ready to board. You are the last one in the line of passengers at the gate. The first guy walks in to the flight and promptly realizes that he does not have his boarding pass on him and does not remember his seat number. So he picks one at random, hoping his charm will take care of the after effects. Every one else takes their assigned seat if it is available. If someone is already sitting on it they quietly look for an empty one and sit there. By the time you get in there is only 1 seat left. What is the probability that the seat which remains is indeed the one originally assigned to you?
Comments
- Anonymous
August 18, 2006
i would guess 1 in 2 - Anonymous
August 19, 2006
I would "guess" 1 in 20. Does the winner get a prize? :)
Toby - Anonymous
August 19, 2006
Sorry, I meant the other way round 8-) - Anonymous
August 19, 2006
My guess is 100% chance that the last person gets his seat.
The first passengers on a plane are usually in first class and there are only a few of those seats.
The last passenger is unlikely to be first class and thus should get his seat. - Anonymous
August 19, 2006
The comment has been removed - Anonymous
August 19, 2006
its 1/2 on the following assumption ( The first person sits in seat number 99. The 99 th person does not consider the last seat as empty .. ie seat number lesser than his number are empty ) .. So the last person either gets his seat or the first person is sitting in it. - Anonymous
August 19, 2006
1/100 % - Anonymous
August 19, 2006
1/100 - Anonymous
August 19, 2006
My guess is 100% chance that the last person gets his seat.
The first passengers on a plane are usually in first class and there are only a few of those seats.
The last passenger is unlikely to be first class and thus should get his seat. - Anonymous
August 19, 2006
Allright - lots of different answers. I will post the solution after 48 hours of no comment activity :-) - Anonymous
August 19, 2006
The comment has been removed - Anonymous
August 19, 2006
The comment has been removed - Anonymous
August 19, 2006
The comment has been removed - Anonymous
August 19, 2006
Just to clarify my partial answer, this is the chance that the final passengers seat is taken by the time the first three have taken seats:
1/100+(1/100+(1/1001/99))+((1/100+(1/1001/99))(1/98)
First Passenger: 1/100 chance of getting final passengers seat, as they don't know what seat they are in anyway.
Second Passenger: the first passenger must take their seat (1/100) AND THEN they must take the final passengers seat (1/99). AND in probability is represented by timesing the two probabilities. So the probability of second passenger getting the final passengers seat is 1/1001/99
Third Passenger: ONE OF the first two must get the third passengers seat (which is represented by addition in probability), AND THEN the third passenger must take the final passenger's seat (1/98). So the formula is ((1/100+(1/1001/99))(1/98)
Now if you just add the individual probabilities of all 99 passengers before the final passenger up, you get the probability of someone taking the final passenger's seat. Then you do 1 minus that, to get the answer. I think....... :)
And I would assume it is a very low probability that the final person gets their assigned seat. - Anonymous
August 19, 2006
Fourth passenger's probability of taking final person's seat would be (p1+p2+p3)*1/97
Fifth: (p1+p2+p3+p4)*1/96
Sixth: (p1+p2+p3+p4+p5)*1/95
Note that when they are displaced, each passenger is just as likely to take another passenger's seat as they are the final passenger's seat. - Anonymous
August 20, 2006
I think the probability is 0. There are only 100 seats and 100 pasangers besides yourself so the total number of passengers is 101. As I'm quite sure that they won't let you sit in the pilot's seat and crash the plane, or sit in anyones lap. I am pretty sure you are sol ( so outta luck)....
How did I do? - Anonymous
August 20, 2006
JohnV:
"There are a 100 people trying to get onto the same flight you are"
Not "There are a 100 OTHER people trying to get onto the same flight you are"
Surely you are a person and count in the "100 people"? - Anonymous
August 20, 2006
It's too late for math - Surely it's 50/50 - He either gets his seat or he doesn't.
(Sorry to extend time period for getting the real answer ;) - Anonymous
August 20, 2006
1/3
Satya - Anonymous
August 22, 2006
Mark Allanson: he either gets his seat or one of the remaining 99 seats, so it has to be 1/100 - Anonymous
August 23, 2006
The answer is 50%.
The key is that if any passenger after the first sits in the seat the first person was assigned to, then everyone else falls out to the seats they wanted.
So for any given displaced passenger, there are 3 possibilities: 1. sit in first person's seat, thus ending the displaced passenger problem, 2. sit in someone else's seat, 3. sit in the last person's seat.
We don't need to consider people who get to their seats fine.
Given that the general probability of sitting in any seat is equivilent (which in reality is a bad assumption), then there is an equal probability that the displaced passenger will choose the first person's seat as the last person's seat. As you permute on through, it essentially becomes a 50/50 guess.
To drive home the point, suppose there are two people left to board the plane, which means there are two seats open. The first person now has two choices: the last person's seat, or the seat that's not the last person's seat. It's clear that this is a 50/50 scenario. As argued above, this is regardless of the actual number of people. - Anonymous
August 23, 2006
My damaged brain hurts. - Anonymous
August 26, 2006
Keith's answer is the correct one. Intuitively, this is the correct proof - the last passenger always gets a 50% chance to take his place. The main idea is that you can factor out most people before the last passanger that would get their places.
The formal proof can be done through induction - I hope to post it soon. - Anonymous
September 06, 2006
I disagree that the answer is 50%. Here is why:
The next to the last person has a choice of two seats. Four possibilities exist: (1) One is his, one is someone else's; (2) one is his, one is yours; (3) one is yours, one is someone elses; (4) both are someone else's.
In the first case, the seat left belongs to someone else.
In the second case, the seat left belongs to you.
In the third case, the seat left may or may not be yours. (1:2)
In the fourth case, the seat left belongs to someone else.
So, you have a 3/8 chance of getting your seat. - Anonymous
September 07, 2006
the answer is 1/3 - Anonymous
September 07, 2006
the answer is 1/3 - Anonymous
September 14, 2006
What's the answer?