The bus waiting time paradox
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| | 2 votes (33.33%) | ||
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6 members have voted
September 27th, 2024 at 10:47:43 AM
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Imagine there are two busses, as follows:
Assuming no passengers own a watch and simply choose a bus and then wait for the next one to arrive. Your questions are as follows:
- Bus A arrives at the bus stop exactly once an hour.
- Bus B arrives at a random time, uniformly distributed, every hour, with the range starting and ending on the hour (for example 3:00 PM to 4:00 PM).
Assuming no passengers own a watch and simply choose a bus and then wait for the next one to arrive. Your questions are as follows:
- What is the average time between arrivals for bus A?
- What is the average time between arrivals for bus B?
- What is the waiting time for bus A?
- What is the waiting time for bus B?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
September 27th, 2024 at 10:57:34 AM
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You ask that like you've never taken a bus, or you just take questions from exams from school.
September 27th, 2024 at 11:14:30 AM
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Quote: WizardImagine there are two busses, as follows:
- Bus A arrives at the bus stop exactly once an hour.
- Bus B arrives at a random time, uniformly distributed, every hour, with the range starting and ending on the hour (for example 3:00 PM to 4:00 PM).
Assuming no passengers own a watch and simply choose a bus and then wait for the next one to arrive. Your questions are as follows:
- What is the average time between arrivals for bus A?
- What is the average time between arrivals for bus B?
- What is the waiting time for bus A?
- What is the waiting time for bus B?
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The answer to every question is, I own a car I never take the bus. The last time I rode a bus was in 1965 in high school and the next year I had a car and drove to school. Batman was my favorite TV show in 1965. Any more questions?
"It's not called gambling if the math is on your side."
September 27th, 2024 at 12:44:59 PM
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I'm a bigot.
All animals are equal, but some are more equal than others
September 27th, 2024 at 1:38:05 PM
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Quote: Wizard
- What is the average time between arrivals for bus A?
- What is the average time between arrivals for bus B?
- What is the waiting time for bus A?
- What is the waiting time for bus B?
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Did somebody say buses?
1. 60 minutes
2. 60 minutes
3. 30 minutes
4. f(a,b), where a is the fraction of an hour past the start of an hour when you arrived, and b is the fraction of an hour past the start of an hour when that hour's bus arrives
If a <= b, then it is b - a; otherwise, it is (1 - a) + 1/2, or 3/2 - a
The sum of the first one is INTEGRAL(0,1) {INTEGRAL(x,1) (t-x) dt} dx = INTEGRAL(0,1) (x^2/2 - x + 1/2) dx = 1/6
The sum of the second one is INTEGRAL(0,1) (3/2 - x) dx = 1/2
The sum of the waiting times = 1/6 + 1/2 = 2/3
Since the domain is 1 hour squared, the average is 2/3 hour, or 40 minutes
September 27th, 2024 at 3:38:00 PM
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Quote: ThatDonGuy
1. 60 minutes
2. 60 minutes
3. 30 minutes
4. f(a,b), where a is the fraction of an hour past the start of an hour when you arrived, and b is the fraction of an hour past the start of an hour when that hour's bus arrives
If a <= b, then it is b - a; otherwise, it is (1 - a) + 1/2, or 3/2 - a
The sum of the first one is INTEGRAL(0,1) {INTEGRAL(x,1) (t-x) dt} dx = INTEGRAL(0,1) (x^2/2 - x + 1/2) dx = 1/6
The sum of the second one is INTEGRAL(0,1) (3/2 - x) dx = 1/2
The sum of the waiting times = 1/6 + 1/2 = 2/3
Since the domain is 1 hour squared, the average is 2/3 hour, or 40 minutes
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The error is probably on my end, as I have the greatest of respect for your math skills, but I'll show my answer in spoiler tags and further comments. Hoping to get a third opinion.
I get 7/12 of an hour, or 35 minutes. I did this as a double integral between two buses in consecutive hours. It seems to me the answer is the square of the sum of these two waiting times*, thus they shouldn't be calculated separately and added.
* Divided by 2 (footnote added later)
Again, not saying you're wrong, as it is probably me in error.
I look forward to another math genius on the forum chiming in.
* Divided by 2 (footnote added later)
Again, not saying you're wrong, as it is probably me in error.
I look forward to another math genius on the forum chiming in.
Last edited by: Wizard on Sep 27, 2024
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
September 27th, 2024 at 4:36:28 PM
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Quote: WizardQuote: ThatDonGuy
1. 60 minutes
2. 60 minutes
3. 30 minutes
4. f(a,b), where a is the fraction of an hour past the start of an hour when you arrived, and b is the fraction of an hour past the start of an hour when that hour's bus arrives
If a <= b, then it is b - a; otherwise, it is (1 - a) + 1/2, or 3/2 - a
The sum of the first one is INTEGRAL(0,1) {INTEGRAL(x,1) (t-x) dt} dx = INTEGRAL(0,1) (x^2/2 - x + 1/2) dx = 1/6
The sum of the second one is INTEGRAL(0,1) (3/2 - x) dx = 1/2
The sum of the waiting times = 1/6 + 1/2 = 2/3
Since the domain is 1 hour squared, the average is 2/3 hour, or 40 minutes
link to original post
The error is probably on my end, as I have the greatest of respect for your math skills, but I'll show my answer in spoiler tags and further comments. Hoping to get a third opinion.I get 7/12 of an hour, or 35 minutes. I did this as a double integral between two buses in consecutive hours. It seems to me the answer is the square of the sum of these two waiting times, thus they shouldn't be calculated separately and added.
Again, not saying you're wrong, as it is probably me in error.
I look forward to another math genius on the forum chiming in.
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My calculation on #4 is under the assumption that the mean time to wait if you missed the current hour's bus is 30 minutes plus the time remaining in the current hour. Perhaps if I add a third variable of just when the next bus arrives, I get a different answer.
Update: I did a simulation, and
The answer does appear to be 35 - perhaps my assumption is invalid? Although in that case, it may be a triple integral
September 27th, 2024 at 5:42:17 PM
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Quote: WizardQuote: ThatDonGuy
1. 60 minutes
2. 60 minutes
3. 30 minutes
4. f(a,b), where a is the fraction of an hour past the start of an hour when you arrived, and b is the fraction of an hour past the start of an hour when that hour's bus arrives
If a <= b, then it is b - a; otherwise, it is (1 - a) + 1/2, or 3/2 - a
The sum of the first one is INTEGRAL(0,1) {INTEGRAL(x,1) (t-x) dt} dx = INTEGRAL(0,1) (x^2/2 - x + 1/2) dx = 1/6
The sum of the second one is INTEGRAL(0,1) (3/2 - x) dx = 1/2
The sum of the waiting times = 1/6 + 1/2 = 2/3
Since the domain is 1 hour squared, the average is 2/3 hour, or 40 minutes
link to original post
The error is probably on my end, as I have the greatest of respect for your math skills, but I'll show my answer in spoiler tags and further comments. Hoping to get a third opinion.I get 7/12 of an hour, or 35 minutes. I did this as a double integral between two buses in consecutive hours. It seems to me the answer is the square of the sum of these two waiting times, thus they shouldn't be calculated separately and added.
Again, not saying you're wrong, as it is probably me in error.
I look forward to another math genius on the forum chiming in.
link to original post
It turned out that my assumption was correct, but the error was somewhere else.
f(a,b), where a is the fraction of an hour past the start of an hour when you arrived, and b is the fraction of an hour past the start of an hour when that hour's bus arrives
If a <= b, then it is b - a; otherwise, it is (1 - a) + 1/2, or 3/2 - a
The sum of the first one is INTEGRAL(0,1) {INTEGRAL(x,1) (t-x) dt} dx = INTEGRAL(0,1) (x^2/2 - x + 1/2) dx = 1/6
The sum of the second one is INTEGRAL(0,1) {INTEGRAL(0,y) (3/2 - y) dt} dy = INTEGRAL(0,1) (3/2 y - y^2) dy = 3/4 - 1/3 = 5/12
The sum of the waiting times = 1/6 + 5/12 = 7/12, or 35 minutes
September 27th, 2024 at 8:13:11 PM
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Quote: ThatDonGuy
f(a,b), where a is the fraction of an hour past the start of an hour when you arrived, and b is the fraction of an hour past the start of an hour when that hour's bus arrives
If a <= b, then it is b - a; otherwise, it is (1 - a) + 1/2, or 3/2 - a
The sum of the first one is INTEGRAL(0,1) {INTEGRAL(x,1) (t-x) dt} dx = INTEGRAL(0,1) (x^2/2 - x + 1/2) dx = 1/6
The sum of the second one is INTEGRAL(0,1) {INTEGRAL(0,y) (3/2 - y) dt} dy = INTEGRAL(0,1) (3/2 y - y^2) dy = 3/4 - 1/3 = 5/12
The sum of the waiting times = 1/6 + 5/12 = 7/12, or 35 minutes
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Great minds think alike. We may have approached it a little differently, but all's well that ends well.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
