Can You Join the World’s Biggest Zoom Call?
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8 members have voted
June 2nd, 2020 at 7:30:18 PM
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This puzzles comes to us courtesy of the Riddler Classic at fivethirtyeight.com. I think I can't improve upon their wording, so will copy and paste it:
I have what I think is the correct answer, but it involves some hand-waving logic that I don't know would get me full credit for a proper solution.
Enjoy!
Quote: fivethirtyeight.comFrom Jim Crimmins comes a puzzle about what would presumably be the largest Zoom meeting of all time:
One Friday morning, suppose everyone in the U.S. (about 330 million people) joins a single Zoom meeting between 8 a.m. and 9 a.m. — to discuss the latest Riddler column, of course. This being a virtual meeting, many people will join late and leave early.
In fact, the attendees all follow the same steps in determining when to join and leave the meeting. Each person independently picks two random times between 8 a.m. and 9 a.m. — not rounded to the nearest minute, mind you, but any time within that range. They then join the meeting at the earlier time and leave the meeting at the later time.
What is the probability that at least one attendee is on the call with everyone else (i.e., the attendee’s time on the call overlaps with every other person’s time on the call)?
I have what I think is the correct answer, but it involves some hand-waving logic that I don't know would get me full credit for a proper solution.
Enjoy!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 3rd, 2020 at 3:05:32 AM
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(I think) I didn't answer the question as asked (so it is probably wrong or only a "partial answer" at best):
Assumptions:
1. "At least one person" would have to be in the "Zoom Meeting" for the whole hour (8 to 9) to guarantee "that the attendee’s time on the call overlaps with every other person’s time on the call".
2. The "..unit of time" I used was 100 ms *** (0.1 of a second)
***: See bottom of the first answer on this link >>> link <<< where it says "...somewhere in the range of 80 ms to 125 ms..."
Using the above (probably wrong, lol) assumptions, i get: ~22.48% as the chance that "at least least one attendee is on the call with everyone else"
Proof:
Between 8 am and 9 am there are: 36,000 "1/10's of a second"
There is a 1/36,000 chance that a person will be there at 8 am and there is a 1/36,000 chance that a person will leave at 9 am.
1/36,000 x 1/36,000 = 1/1,296,000,000 = "the chance that a person will be there at 8 am and leave at 9 am".
The chance that a person will NOT " be there at 8 am and leave at 9 am" is therefore 1,295,999,999/1,296,000,000
There are 330 million people expected to be in the "zoom meeting".
"the chance that at least one person will be there at 8 am and leave at 9 am"= 1 - (1,295,999,999/1,296,000,000)^330 million
= 1 - 0.7752... = 0.2247...
Therefore the chance of at least one person being there for the meeting (whole hour###) is ~22.48%.
###: I know this was not what was asked in the OP, so this is why I think my answer is wrong, but may still be helpful.
----
At least I didn't try to use "Planck Time" or "zeptoseconds" in my attempt (The chance figure would probably have been "practically zero", if I used those).
Edit (about 325 am)
Assumptions:
1. "At least one person" would have to be in the "Zoom Meeting" for the whole hour (8 to 9) to guarantee "that the attendee’s time on the call overlaps with every other person’s time on the call".
2. The "..unit of time" I used was 100 ms *** (0.1 of a second)
***: See bottom of the first answer on this link >>> link <<< where it says "...somewhere in the range of 80 ms to 125 ms..."
Using the above (probably wrong, lol) assumptions, i get: ~22.48% as the chance that "at least least one attendee is on the call with everyone else"
Proof:
Between 8 am and 9 am there are: 36,000 "1/10's of a second"
There is a 1/36,000 chance that a person will be there at 8 am and there is a 1/36,000 chance that a person will leave at 9 am.
1/36,000 x 1/36,000 = 1/1,296,000,000 = "the chance that a person will be there at 8 am and leave at 9 am".
The chance that a person will NOT " be there at 8 am and leave at 9 am" is therefore 1,295,999,999/1,296,000,000
There are 330 million people expected to be in the "zoom meeting".
"the chance that at least one person will be there at 8 am and leave at 9 am"= 1 - (1,295,999,999/1,296,000,000)^330 million
= 1 - 0.7752... = 0.2247...
Therefore the chance of at least one person being there for the meeting (whole hour###) is ~22.48%.
###: I know this was not what was asked in the OP, so this is why I think my answer is wrong, but may still be helpful.
----
At least I didn't try to use "Planck Time" or "zeptoseconds" in my attempt (The chance figure would probably have been "practically zero", if I used those).
Edit (about 325 am)
According to this search^^^ the maximum number of participants in a zoom meeting is "...up to 500 with Large Meeting add-on...".
^^^: https://www.google.com/search?q=maximum+number+of+people+allowed+in+a+zoom+meeting&rlz=1C1CHFX_enAU710AU710&oq=maximum+number+of+people+allowed+in+a+zoom+meeting&aqs=chrome..69i57.13009j0j7&sourceid=chrome&ie=UTF-8
^^^: https://www.google.com/search?q=maximum+number+of+people+allowed+in+a+zoom+meeting&rlz=1C1CHFX_enAU710AU710&oq=maximum+number+of+people+allowed+in+a+zoom+meeting&aqs=chrome..69i57.13009j0j7&sourceid=chrome&ie=UTF-8
Last edited by: ksdjdj on Jun 3, 2020
June 3rd, 2020 at 3:16:55 AM
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Is “ hand-waving logic” similar to magic wand waving? 🤪
But seriously, there’s always going to be those people that want to be on the call for the entire duration. Towards that end, they’ll get on before 8 and get off after 9. So, 100%...?
So I‘d change the parameters of the puzzle slightly to allow for this situation, by stating that people can get on and off anytime they wish, but they can be on the call for a random duration not to exceed one hour.
So now the question is, what’s the chance one of those max duration people nails it and gets on at 8:00:00?
My response: dunno. 🤔
But seriously, there’s always going to be those people that want to be on the call for the entire duration. Towards that end, they’ll get on before 8 and get off after 9. So, 100%...?
So I‘d change the parameters of the puzzle slightly to allow for this situation, by stating that people can get on and off anytime they wish, but they can be on the call for a random duration not to exceed one hour.
So now the question is, what’s the chance one of those max duration people nails it and gets on at 8:00:00?
My response: dunno. 🤔
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
June 3rd, 2020 at 8:35:27 AM
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Quote: Wizard...I have what I think is the correct answer, but it involves some hand-waving logic that I don't know would get me full credit for a proper solution.
Enjoy!
a very small number:
1.26 x 10-6992
EDIT: I found an error. Now, I get:
1.4 x 10-13984
1.26 x 10-6992
EDIT: I found an error. Now, I get:
1.4 x 10-13984
Last edited by: ChesterDog on Jun 3, 2020
June 3rd, 2020 at 9:11:23 AM
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I haven't looked at it in great detail but if someone was able to join at 8am+a and leave at 9am-b where a and b were sufficently small then that person would cover everyone unless someone else managed to get in an out in a or b.
(e.g. Person X gets in at 8:00:05 and leaves at 8:59:54, it would need someone, a spoiler, to either join and leave before 8:00:05 or join and leave after 8:59:54 for X not cover everyone.)
The chance of person X joining at or before 8+a is about a, similar 9-b is about b (where a and b are in hours). So the chances are ab. Similarly the chances of a spoiler would be about a^2+b^2.
I guess you could then integrate the chances of various times for person X, and then look at the chances of them not having a spoiler.
(e.g. Person X gets in at 8:00:05 and leaves at 8:59:54, it would need someone, a spoiler, to either join and leave before 8:00:05 or join and leave after 8:59:54 for X not cover everyone.)
The chance of person X joining at or before 8+a is about a, similar 9-b is about b (where a and b are in hours). So the chances are ab. Similarly the chances of a spoiler would be about a^2+b^2.
I guess you could then integrate the chances of various times for person X, and then look at the chances of them not having a spoiler.
June 3rd, 2020 at 11:09:50 AM
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This is a mathematical question employing randomness, not a question involving real people.
Let's also recognize that time is almost infinitely divisible down to something like 10E-64 seconds (the plank limit on measureable time due to the Heisenberg Uncertainty principle of quantum physics).
Now let's restate the question:
Given 330 million people, we must estimate the expected values of the earliest departure time and the latest arrival time.
And then estimate what the probability is that one of the other 330 million people gets on the Zoom call before the earliest departure time and also randomly draws a departure time later than the latest arrival time?
Given 330 million people, we must estimate the expected values of the earliest departure time and the latest arrival time.
If you divide an hour into 26,000 intervals the odds of randomly getting both an arrival and departure time within a single given interval (of 1/26,000 hour) is about 1 in 337,000,000.
So, I expect that the earliest departure time will roughly be 8:00 a.m +1/26,000 of a an hour. Similarly the latest arrival time should be 9:00 p.m. - 1/26,000 of an hour.
So, what are the odds of any given person getting an arrival time in the first 1/26,000 of an hour AND a departure time in the last 1/26,000 of an hour? about 1 in 337 million!
So, as an answer, I estimate a probability of 50%. Because, even though there is one heroic person who is expected to have arrival and departure times in the same intervals as the earliest departer and the latest arriver, we must consider when in the intervals our heroic person has arrived and departed and whether is before the earliest departer and after the latest arriver. And by a quick analysis I get that 50% is the about the correct answer.
Let's also recognize that time is almost infinitely divisible down to something like 10E-64 seconds (the plank limit on measureable time due to the Heisenberg Uncertainty principle of quantum physics).
Now let's restate the question:
Given 330 million people, we must estimate the expected values of the earliest departure time and the latest arrival time.
And then estimate what the probability is that one of the other 330 million people gets on the Zoom call before the earliest departure time and also randomly draws a departure time later than the latest arrival time?
Given 330 million people, we must estimate the expected values of the earliest departure time and the latest arrival time.
If you divide an hour into 26,000 intervals the odds of randomly getting both an arrival and departure time within a single given interval (of 1/26,000 hour) is about 1 in 337,000,000.
So, I expect that the earliest departure time will roughly be 8:00 a.m +1/26,000 of a an hour. Similarly the latest arrival time should be 9:00 p.m. - 1/26,000 of an hour.
So, what are the odds of any given person getting an arrival time in the first 1/26,000 of an hour AND a departure time in the last 1/26,000 of an hour? about 1 in 337 million!
So, as an answer, I estimate a probability of 50%. Because, even though there is one heroic person who is expected to have arrival and departure times in the same intervals as the earliest departer and the latest arriver, we must consider when in the intervals our heroic person has arrived and departed and whether is before the earliest departer and after the latest arriver. And by a quick analysis I get that 50% is the about the correct answer.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
June 3rd, 2020 at 11:49:18 AM
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I expect the answer to be very close to 1.
The probability that a given person arrives between 8:00:00 and 8:00:36 and leaves between 8:59:24 and 9:00:00 is 1 / 500,000.
The probability that nobody in a group of 330,000,000 do is (499,999 / 500,000)^330,000,000; I calculate this as 1 in 10^304.
Therefore, almost certainly, at least one person will arrive before 8:00:36 and leave after 8:59:24.
The probability that anybody arrives and leaves before 8:00:36, or arrives and leaves after 8:59:24, is rather small.
If you chage it to arriving between 8:00:00 and 8:00:03.6 and leaving between 8:59:56.4 and 9:00:00, I get a 734/735 probability that at least one person out of 330 million does this.
June 3rd, 2020 at 1:05:19 PM
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Quote: ThatDonGuy
I expect the answer to be very close to 1.
The probability that a given person arrives between 8:00:00 and 8:00:36 and leaves between 8:59:24 and 9:00:00 is 1 / 500,000.
The probability that nobody in a group of 330,000,000 do is (499,999 / 500,000)^330,000,000; I calculate this as 1 in 10^304.
Therefore, almost certainly, at least one person will arrive before 8:00:36 and leave after 8:59:24.
The probability that anybody arrives and leaves before 8:00:36, or arrives and leaves after 8:59:24, is rather small.
If you chage it to arriving between 8:00:00 and 8:00:03.6 and leaving between 8:59:56.4 and 9:00:00, I get a 734/735 probability that at least one person out of 330 million does this.
I estimate that the chances are about 1 in 330,000,000 that a person will arrive and then depart within 0.2 seconds.
Bin an hour into 18,000 intervals of 0.2 seconds. The probability of arriving and departing within the first interval of 0.2 seconds is 1/ (18000^2)= 1/324,000,000 (approx). Given 330 million, we expect one person to have arrived and departed within the first 0.2 second of the Zoom video call. The same applies to the last 0.2 second interval of the Zoom call.
So what is the probability of another individual arriving within the first 0.2 second and departing in the last 0.2 second? about 1 in 326,000 million.
However, we need a person to arrive (on average) within the first 0.14 seconds and depart within the last 0.14 seconds to arrive earlier than the 1st departer and to depart later than the last arriver. So -without doing integration - I have a rough estimate that the probability is about 50%.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
June 3rd, 2020 at 5:26:34 PM
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Quote: ChesterDoga very small number:
1.26 x 10-6992
EDIT: I found an error. Now, I get:
1.4 x 10-13984
No. Not even close
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 3rd, 2020 at 5:28:29 PM
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Quote: gordonm888So -without doing integration - I have a rough estimate that the probability is about 50%.
[/spoiler]
This is my general area too.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 4th, 2020 at 1:28:34 AM
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Quote: WizardQuote: ChesterDoga very small number:
1.26 x 10-6992
EDIT: I found an error. Now, I get:
1.4 x 10-13984No. Not even close
I agree that I wasn't even close. Now, I get about 0.697 0.596354. I had to do a little hand waving, too.
I also tried the calculation with 50 people instead of 330 million and got about 0.605.
EDIT: I caught another error, and my new probability is 0.596354 for 330,000,000 people.
I also tried the calculation with 50 people instead of 330 million and got about 0.605.
EDIT: I caught another error, and my new probability is 0.596354 for 330,000,000 people.
Last edited by: ChesterDog on Jun 4, 2020
June 4th, 2020 at 4:42:40 AM
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Running a few simulations I get the following, but the variance is too high to imply there's a trend as the number of people grow.
100 people | 66.72% |
1 000 people | 66.68% |
10 000 people | 66.94% |
100 000 people | 66.87% |
1 000 000 people | 65.80% |
June 4th, 2020 at 7:34:20 AM
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Not surprisingly, the simulations show the answer is around Ln(2) or 1-1/e
1-1/e = 0.632 would be the answer for (not a) derangement
Ln(2) = 0.693 for the alternating harmonic series
1-1/e = 0.632 would be the answer for (not a) derangement
Ln(2) = 0.693 for the alternating harmonic series
Last edited by: Ace2 on Jun 4, 2020
It’s all about making that GTA
June 6th, 2020 at 4:03:38 PM
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Here is the answer from fivethirtyeight.com.
This is my answer, but I got it by doing an exact solution for two and three people and noticed the answer was the same for both. With 330 million being an arbitrary number, I figured the answer was probably the same for any number of people, much like the previous puzzle about the probability the last passenger to board a plan would sit in his correct seat.
This is my answer, but I got it by doing an exact solution for two and three people and noticed the answer was the same for both. With 330 million being an arbitrary number, I figured the answer was probably the same for any number of people, much like the previous puzzle about the probability the last passenger to board a plan would sit in his correct seat.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 20th, 2020 at 5:28:05 PM
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I went back to this problem for the four-person case. Of course, the answer was still 2/3. I will probably make an Ask the Wizard question out of this. Here is my solution. I welcome all comments.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
