Two friends agree to meet up between 6 PM and 7 PM, and they agree that whoever arrives first will not wait for the other for more than 20 minutes. What is the probability that they will meet up?
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Probability Question for Fun (Again)
- Thread starter Debater Slayer
- Start date
Here's a question for you, Sir. It was given to me in a university course some 50 years ago, and the prof said if anyone solved it, he'd give them an A, no questions asked.
Problem: In any random closed curve, does there exist 4 points that make a square? If so, provide a proof. (Simple question, right?)
Problem: In any random closed curve, does there exist 4 points that make a square? If so, provide a proof. (Simple question, right?)
ouch!
after trying to think it thru for a few minutes I decided it would be easier to write a little code: run thru a couple thousand iterations using two random numbers between 1 and 60, count the "+- 20 hits" and compare the hits to the total iterations to get the percentage.
i anxiously await the non-code explanation, this one seems hard!
after trying to think it thru for a few minutes I decided it would be easier to write a little code: run thru a couple thousand iterations using two random numbers between 1 and 60, count the "+- 20 hits" and compare the hits to the total iterations to get the percentage.
i anxiously await the non-code explanation, this one seems hard!
I initially said that they were of 4/9 , about 44%. But I was wrong, and here is why.
Let's call the two people A and B.
Assuming that they are equally likely to arrive at any point in the one-hour interval (including right there at the tail end, exactly at 7 o'clock) and that they will leave at 7 o'clock or after 20 minutes (whichever happens first).
For ease of analysis, let's just consider that they will both arrive at some random minute in that range, from 0 to 60. That means that they will meet if the difference between A's minute and B's minute is in the range from -20 to +20 (inclusive). 20 minutes is one third of the one-hour interval.
If A's minute of arrival is in the range from 0 to 19, the chance of overlap with B's presence rises gradually from 1/3 to (about) 2/3. I am unfortunately too weak in math to know how to deal with what appears to be a calculus problem, so I will instead just use the average chance of overlap. So for that range we have 1/3 * 1/2 (the average between 1/3 and 2/3), resulting in 1/6.
There is also a 1/3 chance that A's minute of arrival will be in the range 20 to 40. If it is in that range, the odd of overlap will be (nearly) constantly at 2/3. This third part has therefore a probability of 1/3 * 2/3 = 2/9.
Finally, if A arrives in the last third part of the interval, a similar rationale shows the odds of overlap running down from 2/3 at 40 minutes to approximately zero (or 1/60) at the 60 minutes mark. For this last third part of the range we therefore have 1/3 * the average of 2/3 and zero, meaning 1/9.
Adding it up, the end result is 1/6 + 2/9 + 1/9, or (3+4+2)/18, or 9/18.
So the odds are of 50%.
Let's call the two people A and B.
Assuming that they are equally likely to arrive at any point in the one-hour interval (including right there at the tail end, exactly at 7 o'clock) and that they will leave at 7 o'clock or after 20 minutes (whichever happens first).
For ease of analysis, let's just consider that they will both arrive at some random minute in that range, from 0 to 60. That means that they will meet if the difference between A's minute and B's minute is in the range from -20 to +20 (inclusive). 20 minutes is one third of the one-hour interval.
If A's minute of arrival is in the range from 0 to 19, the chance of overlap with B's presence rises gradually from 1/3 to (about) 2/3. I am unfortunately too weak in math to know how to deal with what appears to be a calculus problem, so I will instead just use the average chance of overlap. So for that range we have 1/3 * 1/2 (the average between 1/3 and 2/3), resulting in 1/6.
There is also a 1/3 chance that A's minute of arrival will be in the range 20 to 40. If it is in that range, the odd of overlap will be (nearly) constantly at 2/3. This third part has therefore a probability of 1/3 * 2/3 = 2/9.
Finally, if A arrives in the last third part of the interval, a similar rationale shows the odds of overlap running down from 2/3 at 40 minutes to approximately zero (or 1/60) at the 60 minutes mark. For this last third part of the range we therefore have 1/3 * the average of 2/3 and zero, meaning 1/9.
Adding it up, the end result is 1/6 + 2/9 + 1/9, or (3+4+2)/18, or 9/18.
So the odds are of 50%.
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Here's a classic why I don't bow down before ChatGPT's "omniscience":
Let's assume that the arrival time of the first friend is uniformly distributed between 6 PM and 7 PM. That is, any time between 6 PM and 7 PM is equally likely to be the arrival time of the first friend.
The second friend will arrive at some time between (6:20 PM and 7 PM) if the first friend arrives at or before 6:40 PM (since the first friend will not wait for more than 20 minutes). The probability of the first friend arriving at or before 6:40 PM is:
Probability = (Time for the first friend to arrive before 6:40 PM) / (Total time between 6:00 PM and 7:00 PM) = (40 minutes) / (60 minutes) = 2/3
So the probability that the second friend will arrive at some point between 6:20 PM and 7 PM is also 2/3.
Therefore, the probability that the two friends will meet up is the probability that the second friend arrives between 6:20 PM and 7 PM given that the first friend has arrived before 6:40 PM. This conditional probability is:
Probability = (Probability of the second friend arriving between 6:20 PM and 7 PM) / (Probability of the first friend arriving before 6:40 PM) = (2/3) / (2/3) = 1
Therefore, the probability that the two friends will meet up is 1 or 100%
The second friend will arrive at some time between (6:20 PM and 7 PM) if the first friend arrives at or before 6:40 PM (since the first friend will not wait for more than 20 minutes). The probability of the first friend arriving at or before 6:40 PM is:
Probability = (Time for the first friend to arrive before 6:40 PM) / (Total time between 6:00 PM and 7:00 PM) = (40 minutes) / (60 minutes) = 2/3
So the probability that the second friend will arrive at some point between 6:20 PM and 7 PM is also 2/3.
Therefore, the probability that the two friends will meet up is the probability that the second friend arrives between 6:20 PM and 7 PM given that the first friend has arrived before 6:40 PM. This conditional probability is:
Probability = (Probability of the second friend arriving between 6:20 PM and 7 PM) / (Probability of the first friend arriving before 6:40 PM) = (2/3) / (2/3) = 1
Therefore, the probability that the two friends will meet up is 1 or 100%
Let's assume that the arrival times of the two friends are uniformly distributed between 6 PM and 7 PM. This means that the probability of any arrival time within that one-hour window is the same.
If friend A arrives first, there are two possibilities:
If friend B arrives first, the same probabilities apply, so the probability of them meeting up is:
P(meeting up) = P(A arrives first and B arrives within 20 minutes) + P(B arrives first and A arrives within 20 minutes) P(meeting up) = 2 x (1/2) x (1/3) = 1/3
The factor of 2 comes from the fact that there are two ways for one friend to arrive first (either A arrives first or B arrives first). Therefore, the probability that they will meet up is 1/3 or approximately 0.333.
If friend A arrives first, there are two possibilities:
- Friend B arrives within 20 minutes, so they meet up.
- Friend B arrives after 20 minutes, so they don't meet up.
If friend B arrives first, the same probabilities apply, so the probability of them meeting up is:
P(meeting up) = P(A arrives first and B arrives within 20 minutes) + P(B arrives first and A arrives within 20 minutes) P(meeting up) = 2 x (1/2) x (1/3) = 1/3
The factor of 2 comes from the fact that there are two ways for one friend to arrive first (either A arrives first or B arrives first). Therefore, the probability that they will meet up is 1/3 or approximately 0.333.
I won't spoil the answer yet, but what was your methodology?
ChatGPT is wrong here. Just clarifying so that no other members think it has solved the question!
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This is an open problem in mathematics, so it has no conclusive answer yet. I came across it years ago and didn't bother trying to solve it.
Inscribed square problem - Wikipedia
Yes, I researched it recently, and apparently it was solved in 2021 during Covid, but nowhere could I find a simple solution, saying yes or No.This is an open problem in mathematics, so it has no conclusive answer yet. I came across it years ago and didn't bother trying to solve it.
Inscribed square problem - Wikipedia
Being a hurried silly man, apparently. Please verify my revised answer, which I posted in the original post.
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50%
I would guess no, on the grounds that
an equilateral triangle with slightly rounded ends would count as a closed curve, and in my mind's eye it could not possibly include a square with all four corners being part of that triangle.
Edited to add: then again, of course it could if one of the sides of the square is part of one of the triangle's sides.
But I will maintain that the answer is still "no", on the grounds that the sides could conceivably be wavy so that they do not touch the square's vertices. Of course, by now I am grasping at guesswork straws...
Edited to add: then again, of course it could if one of the sides of the square is part of one of the triangle's sides.
But I will maintain that the answer is still "no", on the grounds that the sides could conceivably be wavy so that they do not touch the square's vertices. Of course, by now I am grasping at guesswork straws...
Your thought process of looking at it as a calculus problem is excellent. I won't give the correct answer yet, but the idea of gradual change in the probability depending on time of arrival is very solid.
Ok, here's some horrible code I hacked together:
// diners
import java.lang.Math.*;
public class diners {
public static void main(String[] args) {
float hits = 0.0f;
int time1 = 0;
int time2 = 0;
float percentage = 0.0f;
int iterations = 10000; // an arbitrary big number
for(int x = 0; x < iterations; x++) {
time1 = (int)(Math.random() * 60);
time2 = (int)(Math.random() * 60);
// System.out.println(time1 + " " + time2);
if(Math.abs(time1 - time2) <= 20)
hits++;
}
percentage = hits/iterations;
System.out.println("hits = " + hits);
System.out.println("percentage is: " + percentage);
}
}
I ran the program several times and got very similar answers (within a percentage).
// diners
import java.lang.Math.*;
public class diners {
public static void main(String[] args) {
float hits = 0.0f;
int time1 = 0;
int time2 = 0;
float percentage = 0.0f;
int iterations = 10000; // an arbitrary big number
for(int x = 0; x < iterations; x++) {
time1 = (int)(Math.random() * 60);
time2 = (int)(Math.random() * 60);
// System.out.println(time1 + " " + time2);
if(Math.abs(time1 - time2) <= 20)
hits++;
}
percentage = hits/iterations;
System.out.println("hits = " + hits);
System.out.println("percentage is: " + percentage);
}
}
I ran the program several times and got very similar answers (within a percentage).
56.x %
Heyo
Veteran Member
No spoiler because this isn't a mathematical answer:
The probability is (close to) 1. Since both friends want to meet and they know the conditions, they are most likely to arrive at 6:20 as that maximises the chance to meet.
The most crucial question IMO is when exactly those who arrive past 6h40 leave. Will they stay for 20 minutes, or will they leave when it is seven o'clock?
I took a literal (admitedly implausive) reading of the problem.
I took a literal (admitedly implausive) reading of the problem.
I would say anytime between 620 and 640, no?