andysif
Joined: Aug 8, 2011
• Posts: 430
March 18th, 2018 at 8:43:31 PM permalink
There is 100 seats and 100 persons in a room. Each person is assigned to a seat, ie person #1 to seat #1, etc. Everyone will try to sit in their own seat if they can, otherwise they will pick a random seat.

If the 1st person instead of sitting in his own seat, picks a random seat (his own included), what is the chance that the last person will sit in his own seat?

PS: given that it is a 1 star question, if forced to make a guess I will say either 0.5 or 0.99
Venthus
Joined: Dec 10, 2012
• Posts: 1039
March 18th, 2018 at 9:29:44 PM permalink
My off-the-cuff answer, bolstered by your hint about it being 1-star (presumably 'easy'...), is centered on the realization that you're actually looking at the odds that a specific chair (say, @) remains empty for the entire duration, barring the last person, which now actually makes this something like a 4th grade math problem?

Not actually sure of the correct way to represent this mathematically, but the logic runs as follows:
#1 has a 99/100 chance of leaving @ vacant.
#2 has a 98/99 chance of leaving @ vacant.
#3 has a 97/98 chance of leaving @ vacant.
etc.

The *99*/100 and the 98/*99* will cancel each other out all the way down, leaving you with 1/100, thus a 1% chance that our final individual will sit in their own seat. ...On the other hand, I also feel like I've fallen for a trap here (specifically tied to the factor that people prefer their own seats), but oh well.
RS
Joined: Feb 11, 2014
• Posts: 8226
March 18th, 2018 at 9:31:57 PM permalink

If I understand this correctly, everyone picks a random seat. If that’s the case, then the answer is 1/100 or 1%.
нет сговор. нет непроходимость. полный освобождение от ответственности.
Venthus
Joined: Dec 10, 2012
• Posts: 1039
March 18th, 2018 at 9:35:36 PM permalink
People will sit in their own seat if available, otherwise it's random.

Which is what's giving me pause; if #1 sits in s99, then everybody sits in order, until #99, whereupon he has a 50% chance of sitting in s1 or s100...
andysif
Joined: Aug 8, 2011
• Posts: 430
March 18th, 2018 at 9:44:33 PM permalink
re RS,
no, they pick their own seat if it is not taken.
RS
Joined: Feb 11, 2014
• Posts: 8226
March 19th, 2018 at 12:15:49 AM permalink
Deleted — OP is confusing. May respond later if I can figure it out.
нет сговор. нет непроходимость. полный освобождение от ответственности.
Wizard
Joined: Oct 14, 2009
• Posts: 19794
March 19th, 2018 at 6:12:13 AM permalink
This is one of my favorite problems but I ask it in the form of a 100-seat airplane.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Joined: Oct 14, 2009
• Posts: 19794
March 19th, 2018 at 6:15:06 AM permalink
Quote: RS

Deleted — OP is confusing. May respond later if I can figure it out.

Here is how I state it:

There is a plane with 100 seats. 100 passengers each have an assigned seat. The first passenger to enter the plane picks a random seat to sit in. The other 99 passengers, who enter one at a time, will try to sit in their assigned seat if it is empty. If it is taken, they pick a random seat among those still available.

What is the probability the last passenger to enter sits in his assigned seat?
It's not whether you win or lose; it's whether or not you had a good bet.
Romes
Joined: Jul 22, 2014
• Posts: 5326
March 19th, 2018 at 7:59:11 AM permalink
Quote: Wizard

Here is how I state it:

There is a plane with 100 seats. 100 passengers each have an assigned seat. The first passenger to enter the plane picks a random seat to sit in. The other 99 passengers, who enter one at a time, will try to sit in their assigned seat if it is empty. If it is taken, they pick a random seat among those still available.

What is the probability the last passenger to enter sits in his assigned seat?

I'm gonna go with 1% off the top of my head, without putting too much thought in to it.
Playing it correctly means you've already won.
Doc
Joined: Feb 27, 2010
• Posts: 6789
March 19th, 2018 at 9:41:33 AM permalink
Here are my thoughts, minus a calculated final answer:
For clarification number the passengers 1-100 based on the order they enter the plane. Number the seats 1-100 based on the passenger that each was originally assigned to.

There is a 1% chance that passenger 1 randomly picks seat 1 that he was assigned. Everything works fine, and passenger 100 gets his assigned seat.

There is a 1% chance that passenger 1 randomly picks seat 100, and passenger 100 will NOT get his assigned seat.

If passenger 1 picks seat 99 (1% chance), then all goes well until passenger 99 must randomly pick between the available seats 1 and 100, which means a 50% chance that passenger 100 gets his own seat.

It progressively gets more complicated. If passenger 1 picks seat 98 (again 1% chance), all goes well until passenger 98 makes a random selection between seats 1, 99, and 100. If he chooses 1, all is fine; if he chooses 100, there is failure; if he chooses 99, then passenger 99 makes a random selection between the remaining two seats. I believe this gives (1-(1/3)*(1+0+.5))=0.5 probability that neither of them sits in seat 100.

I think this leads to a summation of terms, each including a 1% chance of passenger 1 choosing seat number n, combined with a product of probabilities that passengers n through 99 sequentially do not randomly choose seat 100 if it is still available when they board.

I further think (1) that I don't know how to draw that summation equation on this forum, and (2) I am too lazy to calculate the total.

I'll take a wild guess that 98 of the 100 terms wind up with a (.01*0.5) probability with the other two being (.01*1) and (.01*0) for a grand total of 50% chance that passenger 100 gets his own seat.