Meeting In The Park
December 24th, 2013 at 4:46:45 AM
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Dick and Jane agree to meet on a park bench sometime between 1 and 2 PM but neither wants to wait too long if the other is not there when he arrives. How long should each be willing to wait for a 90% chance that the other will arrive before the first one leaves?
This is a variation of a popular puzzle that asks the probability that they will meet if each is willing to wait for 15 minutes, and you will find solutions on the web. Work on that one first, and if you can derive an expression for the probability in terms of an arbitrary period of time, you will be on your way to solving this one.
The most common solutions to this problem are geometric, but an algebraic solution is more powerful. We'll work this on the standard problem, which asks the probability that they will meet if each is willing to wait for fifteen minutes.
The probability of anyone arriving within a particular 15-minute interval is 1/4.
If the first one to arrive arrives after 1:45 (p = 1/4) they are certain to meet. This is case I.
If the first one to arrive arrives before 1:45 (p = 3/4) they will meet only if the next one arrives during the next fifteen minutes (p= 1/4) so the net probability is (1/4)(3/4) = 3/16. This is case II
Add case I and case II and you get 7/16.
Now do it with an arbitrary fraction of an hour, f, and you get a probability of 2f -f2.
You should be able to use this expression to solve the puzzle
This is a variation of a popular puzzle that asks the probability that they will meet if each is willing to wait for 15 minutes, and you will find solutions on the web. Work on that one first, and if you can derive an expression for the probability in terms of an arbitrary period of time, you will be on your way to solving this one.
The most common solutions to this problem are geometric, but an algebraic solution is more powerful. We'll work this on the standard problem, which asks the probability that they will meet if each is willing to wait for fifteen minutes.
The probability of anyone arriving within a particular 15-minute interval is 1/4.
If the first one to arrive arrives after 1:45 (p = 1/4) they are certain to meet. This is case I.
If the first one to arrive arrives before 1:45 (p = 3/4) they will meet only if the next one arrives during the next fifteen minutes (p= 1/4) so the net probability is (1/4)(3/4) = 3/16. This is case II
Add case I and case II and you get 7/16.
Now do it with an arbitrary fraction of an hour, f, and you get a probability of 2f -f2.
You should be able to use this expression to solve the puzzle
January 12th, 2014 at 1:28:41 PM
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Do they arrive at random times? If yes, is it possible that one arrives in, say, the last few seconds?
Is the random distribution an uniform one? Are the arrival times independent random variables?
If one had decided to wait 20 minutes , should s/he arrive before 20 to 2?
Should the optimal time waited not be dependent on the time of arrival?
If so, shouldn't we have some measure of the cost of waiting?
Is the random distribution an uniform one? Are the arrival times independent random variables?
If one had decided to wait 20 minutes , should s/he arrive before 20 to 2?
Should the optimal time waited not be dependent on the time of arrival?
If so, shouldn't we have some measure of the cost of waiting?
Reperiet qui quaesiverit
January 12th, 2014 at 1:34:55 PM
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1 hour * (1-sqrt(0.1)) = 0.6838 hours = 41 minutes, 2 seconds.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 12th, 2014 at 2:18:43 PM
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The man had better get there early because women don't like to be kept waiting and will usually not wait past half a cocktail anyway.
January 12th, 2014 at 2:39:14 PM
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Quote: FleaStiffThe man had better get there early because women don't like to be kept waiting and will usually not wait past half a cocktail anyway.
If you get there early you'll be waiting a long time as women are always late.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 12th, 2014 at 3:14:46 PM
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Quote: WizardIf you get there early you'll be waiting a long time as women are always late.
Humph...beware of generalizations, Sir! 10 push-ups for that misogynistic statement. :)
If the House lost every hand, they wouldn't deal the game.
January 12th, 2014 at 3:19:13 PM
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Quote: beachbumbabsHumph...beware of generalizations, Sir! 10 push-ups for that misogynistic statement. :)
Guilty as charged. To be payable at our next meeting. Hopefully you won't be late.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 12th, 2014 at 3:25:27 PM
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Quote: beachbumbabsHumph...beware of generalizations, Sir! 10 push-ups for that misogynistic statement. :)
I'm with the wizard on this one. I find that most women are usually almost as late as I am...
January 14th, 2014 at 11:48:45 AM
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I prefer problems that are more realistic.
Barack is in love with Sarah. He hopes to ask her for a date today at lunchtime. He is free from 1 to 2 PM only. He knows she often goes to the park to eat her sandwich, but his info is scarce. Say there is a 4/5 probability that she will appear between 1 and 2, the time of her arrival being uniformly distributed, and the duration of her staying having an exponential distribution with mean 5 minutes.
He wants to maximise the probability of meeting her. When should he get to the park, and when should he leave if she has not appeared by then?
Of course, the answer is: get there at 1 and stay the whole hour. But...
It is winter in Chicago, and waiting has a detrimental effect on Barack's dick. The cost of waiting for a time t (in hour) is 2t². Barack is leaving as soon as the marginal cost of waiting an additional dt exceeds the marginal probability of Sarah appearing during that same dt.
As usual, simplify matters by considering arrivals and departures to be instantaneous.
Barack is in love with Sarah. He hopes to ask her for a date today at lunchtime. He is free from 1 to 2 PM only. He knows she often goes to the park to eat her sandwich, but his info is scarce. Say there is a 4/5 probability that she will appear between 1 and 2, the time of her arrival being uniformly distributed, and the duration of her staying having an exponential distribution with mean 5 minutes.
He wants to maximise the probability of meeting her. When should he get to the park, and when should he leave if she has not appeared by then?
Of course, the answer is: get there at 1 and stay the whole hour. But...
It is winter in Chicago, and waiting has a detrimental effect on Barack's dick. The cost of waiting for a time t (in hour) is 2t². Barack is leaving as soon as the marginal cost of waiting an additional dt exceeds the marginal probability of Sarah appearing during that same dt.
As usual, simplify matters by considering arrivals and departures to be instantaneous.
Reperiet qui quaesiverit
January 30th, 2014 at 2:20:37 PM
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Why is the Wizard the only one to give an answer? Doesn't anyone else understand the hint?
If you do, now tell me this:
A rabbi announces that he will begin services between 5 and 6 PM if ten men are present. Ten men agree to come sometime within the hour but each is willing to wait no more than thirty minutes for the others to arrive. What is the probability that services will be held?
If you do, now tell me this:
A rabbi announces that he will begin services between 5 and 6 PM if ten men are present. Ten men agree to come sometime within the hour but each is willing to wait no more than thirty minutes for the others to arrive. What is the probability that services will be held?
January 31st, 2014 at 9:10:43 AM
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Sadly, when I propose a quiz, nobody reacts.Quote: kubikulannI prefer problems that are more realistic.
When should he get to the park, and when should he leave if she has not appeared by then?
Well, here is the solution for those who still want to give it a try.
Don't wait more than 15 minutes. Arrive at a quarter to 2.
Reperiet qui quaesiverit
