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

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

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.

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.

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%.

It's a piece of cake to bake a pretty cake. If the way is hazy. You gotta do the cookin' by the book. You know you can't be lazy. Never use a messy recipe. The cake will end up crazy. If you do the cookin' by the book. Then you'll have a...
Break it down! Lemme see you bake it up! Drop that down low and pick that up. Break it down! Lemme see you bake it up! Drop that down low and pick that up Now back that Hey! Now back that. Hey! Now bake that. Hey! Now bake that. Hey!
It's a piece of cake to bake a pretty cake (WHAT!) If the way is hazy (OKAY!) You gotta do the cookin' by the book, (WHAT!) You know you can't be lazy (YEAH!) Never use a messy recipe, (WHAT!) The cake will end up crazy (OKAY!) If you do the cookin' by the book, (YEEAAHHHH!) Then you'll have a...cake!
Rub that, it's yours! Grab this, it's yours! Rub that, it's yours! Grab this, it's yours! Now turn around, put that on. Grind, make it get a little bigger! Now turn around, put that on.
Grind, make it get a little bigger!
You gotta do the cookin' by the book! HEY!

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...

March 18th, 2018 at 9:44:33 PM
permalink

re RS,

no, they pick their own seat if it is not taken.

no, they pick their own seat if it is not taken.

March 19th, 2018 at 12:15:49 AM
permalink

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

It's a piece of cake to bake a pretty cake. If the way is hazy. You gotta do the cookin' by the book. You know you can't be lazy. Never use a messy recipe. The cake will end up crazy. If you do the cookin' by the book. Then you'll have a...
Break it down! Lemme see you bake it up! Drop that down low and pick that up. Break it down! Lemme see you bake it up! Drop that down low and pick that up Now back that Hey! Now back that. Hey! Now bake that. Hey! Now bake that. Hey!
It's a piece of cake to bake a pretty cake (WHAT!) If the way is hazy (OKAY!) You gotta do the cookin' by the book, (WHAT!) You know you can't be lazy (YEAH!) Never use a messy recipe, (WHAT!) The cake will end up crazy (OKAY!) If you do the cookin' by the book, (YEEAAHHHH!) Then you'll have a...cake!
Rub that, it's yours! Grab this, it's yours! Rub that, it's yours! Grab this, it's yours! Now turn around, put that on. Grind, make it get a little bigger! Now turn around, put that on.
Grind, make it get a little bigger!
You gotta do the cookin' by the book! HEY!

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.

March 19th, 2018 at 6:15:06 AM
permalink

Quote:RSDeleted — 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.

March 19th, 2018 at 7:59:11 AM
permalink

Quote:WizardHere 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.

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.

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.