Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A man has two girlfriends. Each lives at the opposite final stop of the same train. The man gets up every day at a random time. He goes to a train stop and takes the first train to come regardless of the direction and goes to the final stop to see the girlfriend which lives there. Surprisingly he sees one of the girlfriends 4 times more often than the other one. How can this be?
HintThink about train time intervals.
The trains go in 10-minute intervals. One direction, it is every hour at 0, 10, 20, 30, 40, and 50 minutes and the other direction it is every hour at 2, 12, 22, 32, 42, and 52 minutes. When the man comes to the train stop at a random time, he will take the first train if he comes within the first two minutes in any of the 10-minute intervals, i.e. with probability of 20%. The ratio of visiting girlfriends is then 20%:80% or 1:4.
Nov 21, 2001
|tricky tricky tricky|
Dec 08, 2001
| I don't think there's enough info in the question to arrive at that. What if there is only one train that makes the trip back and forth all day? |
Mar 08, 2002
|ones a model and he sees 3 posters of her on his way to see the other|
Mar 08, 2002
|He has a picture of one in his wallet|
Mar 08, 2002
|ONe of them is really kinky, and keeps the lights off 4 out of every 5 times he goes over to her house|
Mar 20, 2002
|I got it. Easy. One is great looking, fun to be with, and "saucy". The other is a dog that he just sees once in a while. He sees the first one more, because he likes her more. Go figure!|
Apr 03, 2002
|He could purposely wake up at a "random" time with an alarm clock and get to the station at a specific time.|
Apr 05, 2002
|Very good one! Well explained.|
Apr 28, 2002
|How come they change to trams in the answer? No offense, but if you had said that it might have been easier, because most trains don't come at 10 min intervals.|
Jun 12, 2002
|How about this: there is a single train which goes back and forth all day long. The train station the guy goes to is one fifth of the way from one end of the track to the other. Say it's 4 times closer to girl A than to girl B. Then four times out of five, the train will be between the man and girl B when he gets to the station, and so will be headed toward girl A when it arrives.|
Dec 04, 2002
|I had the same idea as Bender. Good question though. I liked the answer given as well.|
Mar 14, 2003
|I don't think so! Why should you count the ten minute intervals starting from 0? Why not count each 10 minute interval starting from 29, 2:19, 2:29 etc. Now according to your logic he has a 90% chance ov visiting the other one. I think if the trains run equally regularly and you arrive at a random time, by definition of randomness, he will visit both equally regularly.|
Mar 21, 2003
|jimbo, i think you're wrong. think of it this way - imagine the trains were schedued such that one train always arrived 1 minute after the other. if he goes to the station at a random time each day, he would only visit one girl friend if he happened to arrive during that one minute period. the rest of the time he'd visit the other.|
Jun 03, 2004
|I think the point is that train A arrives 2 minutes after train B and train B arrives 8 minutes after train A so there is an imbalance which causes the extra visitations. There are some alternative answers though for example more trains go in one direction than the other. Its still a good teaser because it makes you think.|
Jul 30, 2005
| my brain hurts|
besides a better question is when they find out which one will hurt him more
Jan 11, 2006
|This must surely depend on how far along the line he lives. The trams must meet at some point, and if he lives at or near that place it's 50/50 which one he gets; but if near one end he must visit the distant girl as the other tram will be gone more often than not. |
Mar 31, 2006
|i never take a train..|
Apr 07, 2006
|This teaser is correct. |
A great way to visualize this problem is to make a timeline. Draw a line, mark ten minute intervals from 0 to 50, and from 2 to 52. What you will see is two "window" sizes, 8 minute windows and 2 minute windows.
Naturally, if you arrive at a random time, you are more likely to arrive in an 8 minute window, and the 0-50 train always comes at the end of the 8 minute window.
To better visualize this, imagine two trains, one that comes 3 seconds after the other. When would you have to arrive to catch the second train? You would only have a three second window (the time after the first train leaves and the second arrives).
If the trains came at EVEN intervals, that is 0-50 and 5-55, then you would have a 50/50 chance of traveling in either direction. (this would create two equally size 5 minute windows)
Dec 15, 2008
|I came up with the same solution as Bender.|
Assume a single train and three stops on the line. The man lives at the middle stop which is one minute from the stop on end A and 4 minutes from the stop on end B. In any 10-minute period, there are two minutes during which the train is somewhere between the man's stop and stop A and eight minutes that it is between his stop and stop B.
This will produce the schedule given in the answer.
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