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        riddles >> easy >> TWO GIRLFRIEND PARADOX
        
(Message started by: S. Owen on Sep 28^th, 2002, 5:32pm)

Title: TWO GIRLFRIEND PARADOX
Post by S. Owen on Sep 28^th, 2002, 5:32pm

One train arrives always arrives 1 minute after the other. That is, if the East train arrives at 12:00, 12:10, 12:20, ..., and the West train arrives at 12:01, 12:11, 12:21, ..., then he'll catch the West train if he arrives between 12:00 and 12:01, 12:10 and 12:11, etc., and the East train any other time.

Title: Re: TWO GIRLFRIEND PARADOX
Post by Brett Danaher on Oct 1^st, 2002, 8:08am

That was my response too, S. Owen. Interestingly, you could have it that the train to the West leaves at 12:00:30 (30 seconds after 12) and every 10 minutes thereafter, while the train to the East leaves at 12:01 and every 10 minutes thereafter. In that case, It's even less likely he'll ever go to the East. Or try 12:00:59 and 12:01. As the time between the two gets smaller and smaller, the chances of ever hitting the train to the East get smaller and smaller.

So the answer to the second part of the question, "how many times more," is really a limit problem, such that the ratio of going West to going East approaches 1:0.

Title: Re: TWO GIRLFRIEND PARADOX
Post by James Fingas on Oct 1^st, 2002, 8:18am

It's even a little more convoluted than that. The question says he gets on whichever train arrives first. Again, the arrival times of the trains may vary independently from their departure times.

Train East:
arrives 12:00:01
departs 12:05:00

Train West:
arrives 12:00:00
departs 12:05:01

He'll get on the West train nearly every time, even though it departs a second later than the East train.

Title: Re: TWO GIRLFRIEND PARADOX
Post by Brett Danaher on Oct 2^nd, 2002, 7:50am

on 10/01/02 at 08:18:39, James Fingas wrote:

I don't really see how that added to the answer - I mean, just take everywhere I mistakenly said "departs" and change it to "arrives." I guess what you're saying is that he could always be on a train which is actually leaving later than the other. Ok, well.......yeah. The question is how much more likely might he be to end up with one girlfriend. The answer is.....likelihood can approach 100%.

Title: Re: TWO GIRLFRIEND PARADOX
Post by yogesh mahajan on Oct 6^th, 2002, 3:36am

If the west bound train arrives at times a:00:00,a:10:00,a:20:00 and so on
and the east bound train arrives at
a:0x:yz,a:1x:yz,a:2x:yz and so on
then the probabiltity of Bill's catching west bound train=
(600-(60*x+yz))/600 and that of catching the east bound trian =(60*x+yz)/600. Now depending upon the value of x and yz Bill may or may not catch one train more often.
and if he does then:
Assuming the total no of times he travels is N, then he'll got o Hillary N*(600-2(60*x+yz))/600 time more often than Monica. ;D
tinymahajan@rediffmail.com

Title: Re: TWO GIRLFRIEND PARADOX
Post by Brett Danaher on Oct 7^th, 2002, 10:42am

Woah. Now THAT added something.

Title: Re: TWO GIRLFRIEND PARADOX
Post by unicorn on Oct 27^th, 2002, 1:02am

I think everyone has missed an important point here. The question says that "Once every morning at a RANDOM TIME, Bill arrives at the train station at the center of the city". Does this not create another parameter that we should consider. I think with this parameter it is really difficult to judge which train he is going to get in. This ultimately leads to the fact that he could go to any of his girlfriend. Correct me if my assumption is wrong. ???

Title: Re: TWO GIRLFRIEND PARADOX
Post by S. Owen on Oct 27^th, 2002, 5:26am

True, we could never predict which train he would get on in any particular day, because he arrives at a random time.

What we're really considering is the probability that he gets on one train versus the other, and the expected number of times he gets on both trains over a long period of time.

So the random variable's value itself isn't important... I guess it should be noted that we're assuming that the time of arrival is distributed uniformly over some period of time that starts and ends on an hour, like 8:00 to 11:00.

Title: Re: TWO GIRLFRIEND PARADOX
Post by rajatiet on Dec 21^st, 2008, 12:08am

We are told that Bill arrives at a train station at a random time each day. One train leaves for the east every 10 mins and one train leaves for the west every 10 mins—his strategy is to jump on whichever train arrives first. It turns out on average that he sees Monica nine times more often than Hillary. Why is this so? This seems a little hard to believe given that he arrives at a time random time each day.
Imagine a scenario where the eastbound train leaves every 10 mins on the hour, and the westbound train leaves every 10 mins one minute later. If Bill arrives after, say, 10:11 am he will have a nine minute window that captures the eastbound train, but if he arrives after 10:10 am there is a one minute window in which the westbound train will arrive first. Thus if he arrives randomly, he is more
likely to end up in the nine minute window, and thus sees Monica nine times more
often.
If If west bound train arrives first then he will see Hillary 9 times more than Monica.
Ref : Developments in Parrondo’s Paradox