Coin Dropped on a Chessboard
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A table has a chessboard integrated into the center of the table. A round coin fits exactly inside of a square on a chessboard. The coin is dropped on the table and lands face-down, with at least part of the coin on the chessboard.
What is the probability that the coin covers the corner of four chessboard squares?
Answer
49Pi/(320+Pi) ~ .47638
The diameter of the coin is equal to the length of the side of a square on the chessboard. For part of the coin to land on the chessboard, the center of the coin must be no more than the radius of the coin from the chessboard. If the length of a side of a square on the chessboard is one, then this describes a 9 x 9 square with rounded corners. The area missing from the rounded corners is equal to the size of a chessboard square minus the size of an inscribed circle (the coin). So the size of the area that the center of the coin must land in is:
9 x 9 - (1 - Pi/4) = 80 + Pi/4
For part of the coin to cover a corner, the center of the coin must be within a circle centered over the corner that has a radius equal to the radius of the coin. The area of this circle is
(1/2)^2 x Pi = Pi/4
There are 7 x 7 = 49 corners that are shared by four squares, so the total area that the center of the coin can land in to cover the corner of four squares is:
49 * Pi/4
The probability is then:
(49Pi/4) / (80 + Pi/4)
= 49Pi / (4 x (80 + Pi/4))
= 49Pi / (320 - Pi) ~ .47638
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Comments
bradon182001   
May 02, 2009
| Wish I could say I figured it out. Had fun working on it, but I didn't get the answer.Thanks for posting.  |
gnarldroot
Feb 04, 2010
| too easy |
zembobo 
Jul 07, 2010
| "this describes a 9 x 9 square with rounded corners". A square with rounded corners? What kind of square is that? Are you saying in the answer that the table is round? Maybe you should have put that in the question. Exact dimensions would have been nice. I pictured a 9X9 table with a 9x9 chessboard on it. The only way the coin will NOT cover at least 4 corners is if it happens to land exactly in the center of one of the squares (infinitesimal chance of that), or off the table by less than it's radius (any more and it would fall off). So we have two areas we're looking at. 81 for the large table, and 9.25 for the small amount which the coin is allowed to be off the table (9.5^2 - 9^2). The chances in this scenario is 100-[(9.25/81)*100] which is 88.58% chance that the coin will cover the corners of 4 squares. |
zembobo
Jul 07, 2010
| Wow. Never mind all. My calculations are way off. I wish you could delete posts from here. I believe the answer as posted is correct. I get what he says now about round corners. |
krishnan 
Aug 22, 2010
| Fun, tricky puzzle. I missed out on the rounded corner part and so my answer was slightly off. |
abharath27
May 11, 2013
| Problem statement is not very clear. |
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