The unfair subway This is problem 24, "The Unfair Subway", in Mosteller's Fifty Challenging Problems in Probability with Solutions 

Marvin gets off work at random times between 3 and 5 P.M. His mother lives uptown, his girl friend downtown. He takes the first subway that comes in either direction and eats dinner with the one he is first delivered to. His mother complains that he never comes to see her, but he says she has a 50-50 chance. He has had dinner with her twice in the last 20 working days. Explain.

The accompanying solution says that it's because the uptown train always arrives one minute after the downtown train, which in turn arrives nine minutes after the uptown train, in this time span. So there's a nine-to-one chance that Marvin will get on the downtown train and not the uptown one.
Huh? Then what happened to the "50-50 chance" part of the problem?
The problem seemed to be posed as a probabilistic inference problem, i.e. one where the goal is to calculate:
$$\binom{20}{2} (0.5)^2 (1-0.5)^{18} \approx 0.00018$$
but it turns out it was a statistical inference problem (one based on maximum likelihood estimates at that) that contradicts information in the problem itself.
So my question is: is this a valid problem in probability? Am I missing something that would make this a valid problem?
 A: The 50-50 chance is a red herring.  As the solution points out, from the fact that there are the same number of trains in each direction, you cannot infer that there is a 50% chance that the next train will be going in each direction.  If you randomly selected a train from the list and then met it, you would have a 50% chance of going either direction.  I would say it is a valid problem, but not one that requires any calculation at all.  The problem is just to figure out how the next train can not be 50-50 if there are the same number of trains in each direction.
A: There's a better solution to this. Suppose there are ten equidistant stops, and one train that shuttles back and forth. Then if Marvin's station is the second, there is a 90% chance that the train is uptown of him when he arrives, so he will go downtown.  
In fact, this solution is so much better than Mosteller's (IMHO) that I'm tempted to think he heard it from a friend a long time ago and garbled it.
A: here an illustration of the problem:
Assume there is one train per hour (time goes clock-wise in the picture) for each direction. In the pictured case, the GF is roughly three times more likely to be visited (in spite of equally many trains per hour): 
