Comments on: math-and-logic-filter: only two more floors - I'm almost there! or am I?

Question: math-and-logic-filter: only two more floors - I'm almost there! or am I?

krautland — Fri, 20 Jul 2007 10:27:19 -0800

math-and-logic-filter: only two more floors - I'm almost there! or am I?

I currently reside on the 23rd floor of a 50-floor apartment building. my elevators are slow and I am notorious for my tardiness. so every other morning I find myself staring at the little digits counting down the floors as I decent into the lobby. "8..7..6..almost there." I think to myself, "surely nobody will stop the elevator NOW."

but do my chances of making it to the lobby without being stopped actually increase with every floor I pass (because there are less people who could stop the elevator with every floor)? or do they stay exactly the same (1:2) because just like the dice, the game starts over again with every new floor?

I'm almost ashamed to post this question since it will undoubtedly reveal that a gerbil on vicodin has a better grasp of basic math or logic problems than I do. my stalker would almost certainly like to assure you I have plenty of redeeming qualities.

By: Horace Rumpole

Horace Rumpole — Fri, 20 Jul 2007 10:35:06 -0800

I won't attempt the larger question, but wouldn't the likelihood of a person choosing to walk rather than take the elevator increase as you get closer to the ground floor? If so, then the odds of the elevator being stopped would decrease for that reason alone.

By: bruce

bruce — Fri, 20 Jul 2007 10:35:50 -0800

your chance of reaching the lobby unstopped increases as you approach it. your comparison with dice is inapposite, because if every floor is treated as a separate dice roll, it's still easier to roll just one seven from the 2nd floor than it is to roll 22 consecutive sevens from the 23rd floor.

By: krautland

krautland — Fri, 20 Jul 2007 10:37:33 -0800

wouldn't the likelihood of a person choosing to walk rather than take the elevator increase as you get closer to the ground floor?

ah, I should have mentioned that: there is no re-entry to the lobby from the fire stairs. they all have to use the elevator.

By: cmiller

cmiller — Fri, 20 Jul 2007 10:38:04 -0800

It's obviously a function of the likelihood of someone waiting between you and the ground. And that is pretty well couple to the number of people between you and the ground; some fraction of them will be waiting for the elevator.

To see that the likelihood is a function of height, think backward. What is the probability that you will be stopped on the way down, overall? How about for someone ten stories above you?

Also, consider the extremes. Ten below you? Second floor? Floor one-bazillion?

By: fallenposters

fallenposters — Fri, 20 Jul 2007 10:44:02 -0800

Have you tried this to help you out?

By: krautland

krautland — Fri, 20 Jul 2007 10:44:59 -0800

yes, that works with the security card in my building but not withouth. sadly.

By: sanko

sanko — Fri, 20 Jul 2007 10:45:08 -0800

It's been a while since I've done stats\probability, but I think this would follow a Poisson Distribution.

By: otherwordlyglow

otherwordlyglow — Fri, 20 Jul 2007 10:50:31 -0800

they have to use the elevator to go down? people on the 2nd floor are required to take the elevator one floor down? or is it just that where the fire stairs exit the building is a gross alley that no one wants to walk through? i've never heard of a place that required you to use the elevator to exit.

By: vacapinta

vacapinta — Fri, 20 Jul 2007 10:57:58 -0800

i've never heard of a place that required you to use the elevator to exit.

I work in such a building. Its a security policy. They want a common entrance/exit for everybody. Obviously there are exceptions in the case of a fire emergency.

By: DevilsAdvocate

DevilsAdvocate — Fri, 20 Jul 2007 11:00:18 -0800

or do they stay exactly the same (1:2) because just like the dice, the game starts over again with every new floor?

bruce and cmiller are right. Your chances increase with each floor you pass. If you want to think of it as "the game starts over again," you can do that, but only if you don't think of the entire journey as a single game, but instead think of each individual floor as a game. The chance of you "losing" at least one of the two remaining "games" when you're at the third floor and almost down is much smaller than the chance of you losing at least one out of the twenty-two games you have to play when you're starting out.

By: DevilsAdvocate

DevilsAdvocate — Fri, 20 Jul 2007 11:04:03 -0800

Or another way to look at it: the person starting on the 24th floor necessarily has a greater chance of getting stopped than you do, because his chances of being stopped somewhere on floors 2-22 are the same as yours, plus he can also get stopped at your floor.

By: sjuhawk31

sjuhawk31 — Fri, 20 Jul 2007 11:32:40 -0800

Your chances of it stopping on any particular floor is 50/50...either it will or it won't. But as you gauge the likelihood over an entire trip, the logic follows what many people here said.

By: grumblebee

grumblebee — Fri, 20 Jul 2007 11:34:21 -0800

Wait, are we assuming that AS you're descending, new people may be added to the queue of people-waiting-for-the-elevator? Or are we assuming that once you've started down, the system is closed? (And does it make a difference? I'm bad a statistics, so maybe I should quit while I'm ahead.)

I realize that either scenario COULD be true. There's nothing to stop new people from joining the queue at any time, but I'm guess that if the elevator is relatively fast, that doesn't happen all that often (in an apartment building, as opposed to in a busy office building).

I'm asking, because if you're on the 20th floor, and you've calculated that you have a 15% chance of getting interrupted before you get to the bottom, given a closed system, wouldn't that percent hold all the way down?

Let's say that every day, I put ten mugs on a table, upside down. I tell you that 50% of days, I will place a dollar in one of the ten cups. Every day, you look under all of the mugs. You have a 50% chance of getting a dollar each day, right?

On a specific day, you've looked under nine of the ten mugs. What's are the odds that a dollar will be under the tenth cup? 50%, right?

Math people, what am I getting wrong?

By: suedehead

suedehead — Fri, 20 Jul 2007 11:35:42 -0800

Seconding DevilsAdvocate. And sjuhawk31, just because there are two outcomes doesn't mean that the chances are 50/50.

By: Cool Papa Bell

Cool Papa Bell — Fri, 20 Jul 2007 11:40:23 -0800

Also, remember to NOT assume an even distribution of people across all the floors. You may be "almost there" in terms of number of floors, but you may be passing through the most populated area of the building, increasing the odds of people there.

By: vacapinta

vacapinta — Fri, 20 Jul 2007 11:41:19 -0800

On a specific day, you've looked under nine of the ten mugs. What's are the odds that a dollar will be under the tenth cup? 50%, right?

No. With each empty mug, the probability goes up that there is no dollar there at all.

By: adamrice

adamrice — Fri, 20 Jul 2007 11:43:40 -0800

Probability is not my strong point, but I think we're missing a few givens that we'd need to get at an answer.

Let's say that during any given period of time equal to an elevator passing a floor, there's a consistent N% chance that a person on any floor will press the elevator-call button. Now, this isn't the only way to set this problem up, but let's go with it. Just for the sake of argument, let's say N=5.

This means that you have a 95% chance of passing each floor unimpeded. Since you are on Floor 23 and are trying to get to Floor 1, there are 21 intervening floors you're playing Russian Roulette with. Your chances of getting past all of them unimpeded are .95 x .95... (21 times) = 35.8%.

But let's say you've already made it down to the 8th floor. That means your chances of getting past the remainder are .95 x .95... (8 times) = 73.5%.

Actually, you know what? I think my math is wrong. Those 35.8% and 73.5% probabilities represent the odds that someone will press the call button just during the time you are passing one floor, not each of the remaining floors. To work out the real offs, you'd work out the probability for the floor your on, then multiply that by the probably at the next floor down, and so on.

But the principle still holds: if you've made it most of the way down without being stopped, you're less likely to get stopped on one of the remaining floors.

By: DevilsAdvocate

DevilsAdvocate — Fri, 20 Jul 2007 11:47:07 -0800

Let's say that every day, I put ten mugs on a table, upside down. I tell you that 50% of days, I will place a dollar in one of the ten cups. Every day, you look under all of the mugs. You have a 50% chance of getting a dollar each day, right?

Yes, before I begin looking under the cups I have a 50% chance of getting a dollar each day.

However, if on a given day I've already turned up nine of the ten cups, and haven't seen a dollar yet, there's only a 5% chance that there's a dollar under the remaining cup.

By: vacapinta

vacapinta — Fri, 20 Jul 2007 11:50:27 -0800

Those 35.8% and 73.5% probabilities represent the odds that someone will press the call button just during the time you are passing one floor, not each of the remaining floors

I think people are getting this too confused. If I start on the 23rd floor and am on my way down, I will have to stop if someone on the 5th floor presses the button while I'm still going past the 20th floor.

In fact, I will have to stop if they pressed their button while the elevator was on the 30th floor on my way down to me still. So its all irrelevant...

By: DevilsAdvocate

DevilsAdvocate — Fri, 20 Jul 2007 11:53:37 -0800

However, if on a given day I've already turned up nine of the ten cups, and haven't seen a dollar yet, there's only a 5% chance that there's a dollar under the remaining cup.

Scratch that, it's actually about 9% that there's a dollar under the remaining cup. Durn Bayesian probabilities.

By: dondiego87

dondiego87 — Fri, 20 Jul 2007 12:00:06 -0800

(IANA statistician, but I play one on AskMeFi.)

As you continue down the elevator shaft, your likelihood of being stopped any given number of times (or, similarly, for any given amount of time) decreases. This is because as the elevator passes floors, the elevator cannot return to those floors without turning around, by which time you will have gotten off.

For example, the probability of you getting stopped 5 times decreases as you descend from the 23rd floor, reaching zero once you pass the 6th floor.

However, the likelihood of stopping at any one floor remains the same until you pass that floor, at which point it becomes zero.

By: grumblebee

grumblebee — Fri, 20 Jul 2007 12:13:34 -0800

Okay, I see my problem. I was confusing two things:

Probability A: the chance that the elevator will stop between floor 20 and the lobby.

Probability B: the chance that the elevator will stop between whatever floor I'm currently on and the lobby.

Even though this doesn't quite map onto the real world, presumably when I'm passing the fifth floor, the odds that the elevator will stop between floor 20 and the lobby remain unchanged.

By: sjuhawk31

sjuhawk31 — Fri, 20 Jul 2007 12:14:19 -0800

suedehead - I'm channeling Chuck Klosterman, perhaps.

By: damn dirty ape

damn dirty ape — Fri, 20 Jul 2007 12:39:08 -0800

I see problem here.

Elevator systems use logic. For all you know the banks may be balanced. Like "only so many stops going down." Or allow only so many stops going down during rush hour. Or go into express mode (do not make a stop) if elevator hasn't been to ground floor in the last 9 minutes. Or "bank 1 has been to the the top thus allow bank 2 and 3 to go express down." Or "bank 2 is going to 6 so it might as well go to 5 instead of breaking bank 3's express."

So, you can have a lot of fun with probability, but in real life the logic will dictate whats going on.

By: metahawk

metahawk — Fri, 20 Jul 2007 13:46:56 -0800

grumblebee - the odds don't remain unchanged because you now know more information than you did when you started. When you got in the elevator you were looking at the odds of not getting stopped at all of 22 floors. Now you know you didn't get stopped on any of the first 12 floors so you need to update the odds. In other words, the odds a person on the 23th floor not getting stopped anywhere are worse the odds of a person starts on the 8th floor, right? [Baring special elevator programming] Once he has made down to floor 8 without being stopped then you update his odds with this information and now his odds are same as if he just got on the 8th floor.

It's like people who count in blackjack - they keep recalculating the odds based on the information that they learn as the hands are played. Here he is recalulating after he has successful passed each floor.

By: grumblebee

grumblebee — Fri, 20 Jul 2007 13:54:48 -0800

Thanks, metahawk.

By: Durin's Bane

Durin's Bane — Fri, 20 Jul 2007 14:27:02 -0800

I'll assume the logic system works like most elevators. You hit the down button and the elevator starts going to your floor. Anyone below you from that point on can hit their down button and your elevator will then stop at their floor. No more going up. I'll also assume that people use the elevator in a stochastic manner equally distributed among floors.

The probability of you stopping at a given floor is proportional to how long it has been since you pressed the button and the elevator started going to your floor.
P(floor)=ct where c is an unknown constant
You pass the higher floors in less total time so the probability of stopping at that floor, P(floor), is lower than the lower floors. In other words, you are most likely to stop at the 2nd floor since you give that floor the most time between you starting and you passing that floor.

However, you're concerned with whether your total probability of stopping is getting higher or lower as you go down. The total probability of stopping on any remaining floors -- let's call it P(below) -- is the sum of the P(floor)'s below you.
P(below) = P(2nd floor)+P(3rd)+P(4th)+...
And obviously the time, t, past a floor is higher for lower floors so as an example,
P(below) = c(20) + c(19) + c(18)...
Then with each floor you pass going down, your chance of making it to the lobby improves, however it improves less and less with each floor since there's a higher probability of stopping at the lower floors.

By: vacapinta

vacapinta — Fri, 20 Jul 2007 14:37:56 -0800

Great, Durin's Bane! Although that only applies to places with one elevator. Otherwise, the people on lower floors also had more time to be picked up by another elevator (which were below you after you started taking yours)

By: Durin's Bane

Durin's Bane — Fri, 20 Jul 2007 14:43:32 -0800

Oy, having multiple elevators does complicate things. I'll leave that to someone smarter than me.

By: krautland

krautland — Fri, 20 Jul 2007 16:22:40 -0800

whouff... I am glad I am not the only one who had to think for a moment about this.

I hadn't even considered the possibility that people might have more time to summon the elevator the farther away they are from my floor. yikes.

as for the fire stairwell: the lobby is actually on the third floor (don't get me started), whereas the stairs lead you to the loading dock on the first floor. show up there without a "I am moving in, please reserve me an elevator - slip" will get you angry comments from security guards. somehow they don't like people passing through on their way to work.

By: Durin's Bane

Durin's Bane — Fri, 20 Jul 2007 18:01:31 -0800

Well, I've been overthinking this plate of beans and confusing myself to the point where I think I may not be exactly correct in my P(below) expression. For the sake of correctness, I thought I'd point out this technicality.

I don't think we can calculate the total probability of stopping, P(below), as a sum. Instead let's first work with the total probability of not stopping,
1-P(below),
and the probabilities of not stopping on an individual floor
1-P(floor).
Now we should be multiplying the individual floor probabilities, 1-P(floor), instead of summing the P(floor)'s.
1-P(below) = [1-P(2nd floor)][1-P(3rd)]...[1-P(n)]
If you multiply the right side of the equation out, you get something like this:
1-P(below)= 1 - P(2nd) - P(3rd)...- P(n) + P(2nd)P(3rd)... many, many higher order terms.

If the P(floor)'s are small, we can approximate the higher order terms as going to zero and ignore them. This gives back the P(floor) expression I had in my original answer, so I was kind of right.

But if we want to be even more precise and not make assumptions, let me first change my nomenclature in an obvious way (n is the floor you're on), we can regroup terms and get
1-P_below(n) = [1-P_below(n-1)][1-P_floor(n)].
From this, you can see that [1-P_floor(n)] is less than 1 so
1-P_below(n) < 1-p_below(n-1),br> then
P_below(n) > P_below(n-1), which agrees with everything above that each floor you go down, you have a better chance of reaching the lobby without stopping.

Whew. I sure hope I'm right this time.