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?

i would guess 1 in 2

I would "guess" 1 in 20. Does the winner get a prize? ðŸ™‚

Toby

Sorry, I meant the other way round ðŸ˜Ž

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.

I think this is the probably that the third person in the line took the last person’s seat:

1/100+(1/100+(1/100*1/99))+((1/100+(1/100*1/99))*(1/98)

Now if only I hadn’t skipped that class on recurssion.

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.

1/100 %

1/100

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.

Allright – lots of different answers. I will post the solution after 48 hours of no comment activity ðŸ™‚

I remember this from when I was studying Hash Tables in the early ’80s. I don’t remember the exact formular, but I think the probability approached 0 as the % full increases, at 99% full I thing the chance is less then 2/100 but greater then 1/100.

Of course I could be totally off base ðŸ™‚ (I have been known to be in the wrong ballbark )

Tom Vande Stouwe

MCT, MCSD

In this case, the first person in the line makes the mistake, I believe the chance is 99% that my seat is free.

Why:

– 1% is covered by: because the guy made 1 choice from 100 seats available that could be my seat from the first time.

– 99% is covered by: If he would chose other seat than my seat then all other people that walked in front of me would make him change the seat to (probably) his seat. ( supposition S1= I suppose that once somebody tells him about the mistake he goes to his seat).

For the case that anyone would make the mistake to sit in the wrong place or if my supposition S1 is wrong (the guy makes again the mistake) then I have 50% chance to find my seat free because the guy nr 99 could make the mistake (he has to chose between 2 sits).

That was not the question, but in that situation the odds are 50/50. Yes he would be pushed into one of 2 seats at the end, and it is 50/50 to be yours.

But the question said the passengers displaced found NEW seats, and therefore displaced others. Like a Hash table, the person moves to the first available, and makes the chances of your seat being unoccupied unlikely. So the odds of your seat being not taken is based upon the first selection. The lost passenger either picked his own seat (1%) or not (99%). If the 1% happens, then you should get your seat, (no one was displaced). There is still a chance that the seat will be available at the end if he does not pick his seat, In the subsequent 98 seatings, if one of the ‘relocating’ passengers occupies his seat, further displacement will not be needed, but your seat may still be occupied already. So that raises the odds just above 1% (I’m sure there is a formular for that 1 in 98 etc).

Well it will be interesting to see what the ‘real ‘ answer is ðŸ™‚

Tom Vande Stouwe

MCT, MCSD

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+(1/100*1/99))+((1/100+(1/100*1/99))*(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 1/100*1/99

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+(1/100*1/99))*(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……. ðŸ™‚

And I would assume it is a very low probability that the final person gets their assigned seat.

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.

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?

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"?

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 ðŸ˜‰

1/3

Satya

Mark Allanson: he either gets his seat or one of the remaining 99 seats, so it has to be 1/100

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.

My damaged brain hurts.

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.

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.

the answer is 1/3

the answer is 1/3

What’s the answer?