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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?

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    The problem doesn't say there's a 50-50 chance; it says that Marvin says there is :-)2011-05-18
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    Another aspect to consider is that around 5pm there will probably be more trains going uptown than going downtown.2011-05-18
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    @joriki Yes, I was aware that the 50-50 was an indirect quote. My basic issue is that this problem goes against my notion of what a "math problem" is. If it were a puzzle or riddle, I wouldn't mind weaselly bits like that; but the fact that I have to doubt the truth of a statement (even in an indirect quote) in a math problem is what gets my goat.2011-05-18
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    This is also problem 6 in Chapter Three, Nine Problems, in Martin Gardner's first column collection, The Scientific American book of Mathematical Puzzles and Diversions, which dates it to somewhere between 1956 and 1958. Dunno whether Mosteller got it from Gardner, or vice versa, or both got it from somewhere else.2011-05-19

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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.

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    I guess this then turns into something of an ontological question but: is it valid to have a red herring in a math problem? Math problems shouldn't aspire to Agatha Christie. On the other hand, do we have to assume it's a red herring? A likelihood of 0.00018 is not outside the realm of possibility. It really might be 50-50 and Marvin just happened to fall under a tail case.2011-05-18
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    I think it would be better to state the problem as, "Marvin argued that" there was a fifty-fifty probability, and give the baseline argument - there are the same number of trains heading in each direction at the same frequency. Then the problem becomes, "Why is Marvin's reasoning incorrect?"2011-05-18
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    @Thomas That's a definite improvement over how the problem is currently stated.2011-05-18
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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.

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    Mosteller's solution, verbatim: "Downtown trains run past Marvin's stop at, say, 3:00, 3:10, 3:20, ..., etc., and uptown trains at 3:01, 3:11, 3:21, ..."2011-05-18
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    @Jason: OK, I've edited out the baseless libel.2011-05-18
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    This solution is similar to the one described by OP, in that the length of time between uptown and downtown train arrivals is asymmetric.2014-01-08
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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): enter image description here