In a common carnival game, a player tosses a penny from a distance of about 5 feet onto the surface of a table ruled in 1-inch squares. If the penny (3/4 inch in diameter) falls entirely inside a square, the player receives 5 cents but does not get his penny back; otherwise, he loses his penny. If the penny lands on the table, what is his chance to win?
Solution:
When we toss the coin onto the table, some positions for the center of the coin are more likely than others, but over a very small square we can regard the probability distribution as uniform. This means that the probability that the center falls into any
region of a square is proportional to the area of the region, indeed, is the area of the region divided by the area of the square. Since the coin is 3/8 inch in radius, its center must not land within 3/8 inch of any edge if the player is to win. This restriction generates a square of side 1/4 inch within which the center of the coin must lie for the coin to be in the square. Since the probabilities are proportional to areas, the probability of winning is \displaystyle \left( \frac{1}{4} \right)^2 = \frac{1}{16}. Of course, since there is a chance that the coin falls off the table altogether, the total probability of winning is smaller still. Also, the squares can be made smaller by merely thickening the lines. If the lines are 1/16 inch wide, the winning central area reduces the probability to \displaystyle \left( \frac{3}{16} \right)^{2} = \frac{9}{256} or less than \displaystyle \frac{1}{28}.