Here is an extremely easy problem which some of you may find interesting. If you hate easy problems, please skip to the next article.
- Kanad Chakraborty
Easy Problem Alert:
A man has two mistresses. The man lives with neither. They live at points A and B, and the man goes to work every day at C, located somewhere between A and B. His job is most unusual -- there are no fixed check-in and check-out times. In fact, he chooses his arrival and departure times with uniform probability over the 24-hour period of a day. Also, there are A-bound and B-bound trains, running 24 hours a day, arriving at fixed intervals at C, with the same frequency (that is, both A-bound and B-bound trains run round the clock and arrive every t minutes at C).
The man has just had a fight with his mistresses, each of whom complains that he spends more time with the other. He makes a deal with each of them -- namely, he will board the first train that comes to C as soon as he gets off work every day, and go whichever direction that train takes him. Feeling happy that he has been able to satisfy both mistresses, he follows this practice for one month, but finds to his astonishment (?) that he has ended up going 28 days to one of his mistresses, and only 2 days to the other. This did not happen by fluke -- in other words, if the man were to repeat this practice for the next month, he would probably obtain similar results. How did this happen ? Note that the man's arrival time at C after work was chosen with uniform probability over a 24-hour period, and he did not make any deliberate effort to avoid going to one of his mistresses.