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Elevator problem

Posted: Sat Jan 27, 2018 2:27 pm UTC
by andykhang
Supposedly, there're 33 floor total in a building I live in,and I live in the 23 floor. I enter the elevator to go down into the 1st floor so that I could get out. As I go down, the elevator stop at floor 19, 14 ,9, 7 , 5 before I could go to the bottom. Statictiscally, how likely I am to encounter that situation? Remember that I only goes down from 23 to 1.

Re: Elevator problem

Posted: Sat Jan 27, 2018 3:04 pm UTC
by Soupspoon
What constraints? Is it "how likely that this specific combination of all possible five sequential stops between 23 and 1" (you're 'Merkin, so I assume that's your ground-level exit) or "…of all possible N-stop combinations…", likewise, from N 0 to 21?

The first would be based upon "Pick Any (5) From (21)", the latter with more of the "Any"s cumulatively included in the combinatorial sum.

Or maybe you were asking about how likely N=5 (every distinct N=5 counts), assuming all possibilities (all Ns) are evenly likely1. Similar calculations, differently munged together.


1 Unless I've missed anything, there's no reason to assume bias for/against various patterns, such as residential levels being even floors, business levels odd, so shifting the balances according to who/where is calling the lifts at a given time.

Re: Elevator problem

Posted: Sat Jan 27, 2018 4:25 pm UTC
by morriswalters
In the real world the controlling factor is time, or when you ride. Along with the total number of occupants and the speed at which the cars run. And of course how many wits punch all the buttons before you get on. Oh yeah, if you have a three story parking garage at the bottom of the stack and and a basement with a health club and storage units and a laundry in the basement below the garage, you have to consider that as well. The controllers calculate that on the fly. Kinda.

Re: Elevator problem

Posted: Sat Jan 27, 2018 6:51 pm UTC
by Eebster the Great
You know, the word is still "suppose," not "supposedly." The verb "suppose" means "accept this hypothetical premise for the sake of argument." The adjective "supposedly" means "allegedly," and indicates that you doubt the truth of the claim but have heard it stated as fact.

Re: Elevator problem

Posted: Sun Jan 28, 2018 4:34 am UTC
by andykhang
Soupspoon wrote:What constraints? Is it "how likely that this specific combination of all possible five sequential stops between 23 and 1" (you're 'Merkin, so I assume that's your ground-level exit) or "…of all possible N-stop combinations…", likewise, from N 0 to 21?

The first would be based upon "Pick Any (5) From (21)", the latter with more of the "Any"s cumulatively included in the combinatorial sum.

Or maybe you were asking about how likely N=5 (every distinct N=5 counts), assuming all possibilities (all Ns) are evenly likely1. Similar calculations, differently munged together.


1 Unless I've missed anything, there's no reason to assume bias for/against various patterns, such as residential levels being even floors, business levels odd, so shifting the balances according to who/where is calling the lifts at a given time.


The pick 5 from 21 thing, all likely, with the bias that you can only goes down from 23 to 1, so each level passed without stopping is eliminated.

Re: Elevator problem

Posted: Sun Jan 28, 2018 2:33 pm UTC
by measure
andykhang wrote:The pick 5 from 21 thing, all likely, with the bias that you can only goes down from 23 to 1, so each level passed without stopping is eliminated.

In how many different ways can you choose 5 elements from a set of 21 (23 floors minus the 23rd where you start and the 1st where you end = 21)?
How many of these match the list of 5?

Re: Elevator problem

Posted: Sun Jan 28, 2018 5:01 pm UTC
by morriswalters
All the math in the world won't solve this problem. Nor will knowing the combinations of the places the elevator can stop. My math is crap, but my understanding of how elevators work is not.

Re: Elevator problem

Posted: Sun Jan 28, 2018 5:04 pm UTC
by andykhang
From the descending order, meaning that, one you pick a number, the following one cannot be higher than the previous. And I was essentially asking the later question though?

Also, again, all combination from 21 to 0 is equally likely...somewhat.

Re: Elevator problem

Posted: Sun Jan 28, 2018 5:18 pm UTC
by gmalivuk
andykhang wrote:From the descending order, meaning that, one you pick a number, the following one cannot be higher than the previous.
Right, but counting combinations (rather than permutations) means we're already ignoring other possible orders. In other words, we're choosing five floors at which the elevator can stop, and then we know that it will make those stops in descending order.

21 choose 5 is 20349. That's how many different sets of five floors the elevator could stop at (in descending order) between 23 and 1. If you want to know how likely it is that your particular set of floors happens, when you already know the elevator makes exactly five intermediate stops, it's 1 in 20349.

2^21 is 2,097,152. That's how many possible sets of floors (including all and none) between 1 and 23 it's possible to stop at. If you want to know how likely it is that it stops at 5 floors (but don't care which five), it's 20,349 in 2,097,152 (about a 1% chance). If you want to know how likely it is that it stops at these particular five floors, it's 1 in 2,097,152.

Re: Elevator problem

Posted: Sun Jan 28, 2018 7:03 pm UTC
by morriswalters
If the building is empty, other than him, the elevator will never stop and that combination, as stated, will turn up, ever. Exactly what is that calculation telling him?

Re: Elevator problem

Posted: Sun Jan 28, 2018 10:49 pm UTC
by Eebster the Great
morriswalters wrote:If the building is empty, other than him, the elevator will never stop and that combination, as stated, will turn up, ever. Exactly what is that calculation telling him?

It's telling him precisely what gmalivuk said: the probability that the elevator stops at those five floors on its path from 23 to 1 under the assumption that it always stops at exactly five floors and that each combination is equally likely.

There are of course other ways to model elevator behavior, but fundamentally I don't really think Andy wants to know about elevators per se.

Re: Elevator problem

Posted: Mon Jan 29, 2018 12:51 am UTC
by morriswalters
Yeah. I get that, I guess. And I'm certain he isn't modeling elevator behavior.

Re: Elevator problem

Posted: Mon Feb 05, 2018 10:15 pm UTC
by SuicideJunkie
Also, is this one of those buildings that mislabel every floor above 12, so you've actually got 20 choose 5 chances?

Re: Elevator problem

Posted: Tue Feb 06, 2018 3:31 pm UTC
by ThirdParty
morriswalters wrote:If the building is empty, other than him, the elevator will never stop and that combination, as stated, will turn up, ever.
This is patently false. Sometimes (especially now that this thread exists) he will push those the buttons for those specific floors out of sheer perversity. Other times he will trip inside the elevator, fall against the control panel, and hit those specific buttons by accident.

These may be low-probability events, but they're definitely not zero-probability ones.

Soupspoon wrote:Unless I've missed anything, there's no reason to assume bias for/against various patterns, such as residential levels being even floors, business levels odd, so shifting the balances according to who/where is calling the lifts at a given time.
I'd just like to note that I live on Floor 6 of my apartment building and frequently take the stairs down rather than the elevator. (The elevator is faster on average, but 25% of the time ends up slower, so if I need to be somewhere--e.g. the bus stop--at a specific time, I can leave my apartment later if I take the stairs.) So I think--assuming the building has a stairwell--that the elevator is somewhat less likely to stop on Floor 5 than on Floor 19, other things equal.

Re: Elevator problem

Posted: Tue Feb 06, 2018 3:47 pm UTC
by Flumble
How often does it stop at floor 9¾?