## Welcome to BrainDen.com - Brain Teasers Forum

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-) |

# The probability of getting my seat

### #1

Posted 14 July 2013 - 03:18 AM

Balcony seating at the opera is by ticket only, and all the tickets are sold and all the ticketholders are in line to enter. But it turns out that the first person through the door is rather inebriated by the time the hall is open for seating and, instead of taking the properly assigned seat, chooses a chair at random (presumably the one with the lowest-seeming relative velocity to his or her person). The next people come in one at a time—and if their seat is available, they take it. But if someone is already sitting there, rather than disrupt things, they just pick some other random seat. Which raises the question, "what is the probability that the last opera lover ends up in his or her assigned seat?"

### #2

Posted 14 July 2013 - 04:14 AM

**Edited by bonanova, 14 July 2013 - 08:42 AM.**

Analysis added

*Vidi vici veni.*

### #3

Posted 15 July 2013 - 03:21 PM

### #4

Posted 16 July 2013 - 05:53 AM Best Answer

*Vidi vici veni.*

### #5

Posted 16 July 2013 - 10:25 AM

Spoiler for Detailed and different version

Bonanova is right. In the above solution, the number of cases where the last person does not get his seta is much higher than n-1. It is actually 2^(n-2) considering all possibilities where the "loop is closed" with n-2 persons instead of n-1 so that the last person's seat is always available.

The probability will then be as suggested by Bonanova.

### #6

Posted 16 July 2013 - 11:57 AM

Kudos to DeGe.

I started a recursive / summation approach, and ran quickly out of steam

(read: lacked ability to think through to conclusion.)

If you're familiar with the Hole in the Sphere problem (a classic that I shared a while back),

when I was first asked about it, I ran away from the calculus and made the assumption that

gave the answer immediately. And I impressed the female colleague who posed it to me.

I looked for the easy way out here as well.

*Vidi vici veni.*

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users