Wild About Math!

MMM #17: The bishop and the king

Doug Hull was randomly chosen as the winner for MMM #16 over at Blinkdagger. Congratulations, Doug!

It’s time for a new contest problem. As you may have noticed, I like counting and probability problems. I found an elegant one in an old book. I won’t reveal what the book is until the end of the contest. Here’s the problem:

If you randomly place a bishop and a king on a normal 8×8 chessboard, what is the probability that the king will be in check?

For those of you who aren’t chess players, here’s an equivalent problem description: If you randomly select two squares on an 8×8 chessboard, or checkerboard, or simply in an 8×8 grid, what is the probability that the two squares will share a diagonal?

For special recognition, but no increased chance of being selected as winner, generalize your solution for an MxM chessboard, where M is even. For extra special recognition, generalize your solution for an NxN chessboard, where N is odd. And, for super duper extra special recognition, come up with a general solution for an AxB chessboard, where A and B are arbitrary integers, both greater than 0. I’ve not explored the AxB generalization and it’s not mentioned in the book that I got the problem from; it might be tough but I believe the problem and the three generalizations will make for great mathematical exploration.

I have a Rubik’s Revolution, courtesy of Techno Source (or $10 Amazon.com gift certificate) to give to the winner. I’ll give more than one prize if I get lots of correct submissions.

Here are the rules for the contest:

1. Email your answers with solutions to mondaymathmadness at gmail dot com.
2. Only one entry per person.
3. Each person may only win one prize per 12 month period. But, do submit your solutions even if you are not eligible.
4. Your answer must be explained. You must show your work! Wild About Math! and Blinkdagger will be the final judges on whether an answer was properly explained or not.
5. The deadline to submit answers is Tuesday, October 21, 12:01AM, Pacific Time. (That’s Tuesday morning, not Tuesday night.) Do a Google search for “time California” to know what the current Pacific Time is.)
6. The winner will be chosen randomly from all timely well-explained and correct submissions, using a random number generator.
7. The winner will be announced Friday, October 24, 2008.
8. The winner (or winners) will receive a Rubik’s Revolution or a $10 gift certificate to Amazon.com. For those of you who don’t want a prize I’ll donate $10 to your favorite charity.
9. Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
10. I may post names and website/blog links for people submitting timely correct well-explained solutions. I’m more likely to post your name if your solution is unique.

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