A fun probability question?

Question

100 people have been assigned seats for a lecture. There are exactly 100 seats in the room. However, the first person to enter the room has lost his seat assignment and randomly chooses a chair to sit in. The rest of the 99 people walk in one at a time and sit in their assigned seat if possible, or if someone is already in their seat they randomly choose one of the empty ones.

What is the probability that the last person to enter to room will sit in his assigned seat? And what is the answer if there are 1000 people and 1000 seats?

The probability is not 1 in 100.

Hint #2: Finding the answer does not involve any long calculations whatsoever.

Answer 1

The probability is 50%.

The first guy with the lost seat assignment, he could by dumb luck wind up in his assigned seat, right? In which case each subsequent person including me, if I'm the last person, will get the right seat.

But what's more likely is that he takes someone else's seat. It's much more likely. Now, here's the counterintuitive part. There are only two seats that count, my seat and his seat. All the other seats don't make a difference. Whoever arrives subsequent to him, those people are going to take some seat or another.

Nothing matters until either my seat gets taken or his seat gets taken. If my seat gets taken by some displaced person, then I have zero chance of getting my seat.

If his seat gets taken by some displaced person then every other person who arrives in the lecture including me, will have his assigned seat.

And how often does that happen? One time out of two. Half the time my seat is going to get taken by some displaced person, and half the time his seat is going to get taken by a displaced person.

You can see this if you start with smaller numbers of seats:
2 seats:
1/2 the time, person picks their correct seat, I get mine.
1/2 the time, person picks my seat, I don't get mine.

3 seats:
1/3 of the time, person picks their seat, I get mine
2/3 of the time, they pick the wrong seat, but there is a 50/50 chance the next person picks the first seat, so I still get mine.
1/3 x 1 + 2/3 x 0.5 = 1/2

4 seats:
1/4 of the time, person picks their seat, I get mine
1/4 of the time, person picks my seat, I don't get mine.
1/2 the time, they displace someone else... that person will either displace the other, or take my seat, or take the first seat. In the first case, it just delays the 50/50 pick. In the second case, I now have a 0% chance. In the third case, I have a 100% chance. The weighted average comes out to 50%.
1/4 x 1 + 1/4 x 0 + 1/2 x 0.5 = 0.5

This is true if there are 2 seats, or 100 seats, or 1000 seats, or a million seats. The answer is always 50%

Answer 2

I am using the concept of combination in this problem.
For the first person the probability is '100Choose1' = 100!/99! = 100 ways to sit.For the second person the probability is '99C1' = 99!/98! = 99 ways to sit, the list goes on like this and for the last person to enter in to the room the probability is '1Choose1' = 1!/(1-1)! = 1!/0! = 1/1 = 1way to sit. Hence, the probability is 1/100 (because 100 is the total no. of possible outcomes and 1 is the desired outcome) which equals 0.01%.

For 1000 people, using the same combination method, the probablity is 0.001%

Answer 3

Ok, first of all, this is not a 'fun' probability question. It's hard! My answer is this: next person who walks in has a 99% chance they get their assigned seat. But somebody is already sitting in the wrong seat, and the condition keeps worsening as more people sit in incorrect seats. There is almost a zero percent chance the last person sits in the right seat.

Answer 4

A quick guess: 0.01. There are 10,000 possible ways the seats can be assigned (or how people would sit randomly), and 100 involving last person in a particular seat., so p = 100 / 10,000.

Answer 5

While I have no interest in doing that much math, I can contribute. If you would like to see this problem in action, attend a movie showing at the ArcLight Cinemas in Hollywood. They do use assigned seating, and if you get there early, it's always entertaining to watch people attempt to sort that out. Especially when they arrive during the previews, so they have the added challenge of darness, and a desire to fix the problem quietly.

Answer 6

Odds are 1 out of 100

Answer 7

12+13+7+5= 37 shirts in entire the probability that you first grab a pink one is: 12/37 blue one is: 13/37 eco-friendly one is: 7/37 orange one is: 5/37. besides, the probability that he does no longer be conscious is 0, because no remember the colour of the shirts you grab, there's a probability more effective than 0 that he will be conscious.

Answer 8

That's a damned good question. I have a friend at the University who specializes in probability theory. Maybe I'll give her a call and see what she says

My guess would be 1 in 100! or
1 in 9.3326215443944152681699238856267e+157
but I suspect that's wrong (it 'feels' too simple )


Doug

Answer 9

It doesn't matter, as long as every body gets to sit in a chair.

Answer 10

Obviously this question is VERY easy and requires very little brainpower