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Suppose there is a coin toss game where quarters are thrown onto a checkerboard. Management keeps all of the quarters; however, if a quarter lands entirely within one square of the checkerboard the management pays a dollar. Assume that the edge of each square is twice the diameter of the quarter, and that outcomes are described by coordinates chosen at random.

I am having a hard time starting this problem. I understand the area of the quarter and the area of the square but am not sure what to do with these. Is the probability the difference in the area of the square and the circle?

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  • $\begingroup$ Start by looking at one square. Which values can the center of the quarter land on so that it's entirely within the square? Divide this by the area of the square to get the probability for one square.Use this to figure out the probability for the entire board. $\endgroup$
    – FullofDill
    Sep 16, 2017 at 3:08
  • $\begingroup$ Focus on the center of the coin. What percentage of a square could the center be on that would not overlap a border. $\endgroup$ Sep 16, 2017 at 3:08
  • $\begingroup$ So the Center of the coin would have to fall somewhere on or within the diameter of the circle centered in the square. $\endgroup$
    – Derek
    Sep 16, 2017 at 3:10

3 Answers 3

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Let's start with one square. If the radius of the coin is 1 unit, then you want your coin to land so that its center is at least 1 unit from any edge. In the figure below, this region is shown in blue (these are all of the points that are at least 1 unit from an edge).

A typical square.

Note that there are 4 blue squares, and 16 total squares, so the probability of your coin landing so that its center is in a blue square is $$ \frac{4}{16} = \frac{1}{4}. $$ Thus there is a $\frac{1}{4}$ probability of a winning, and a $\frac{3}{4}$ chance of losing. The fact that the chess board consists of 64 such squares is irrelevant---all of the squares are identical.

So, should you play this game? That is, is the game fair? To figure that you, we need to know your expected winnings. This is given by \begin{align} &(\text{payout for a blue})\cdot(\text{probability of blue}) + (\text{payout for a red})\cdot(\text{probability of red}) \\ &\qquad = (\$1)\left( \frac{1}{4} \right) + (\$0)\left( \frac{3}{4} \right) \\ &\qquad = \$0.25. \end{align} Since it costs $0.25 to play the game, you expect (on average) to break even. So, I guess the game is fair.

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Hints:

Assume the radius of the quarter is $1$ unit.

The center of the tossed coin will land at a uniformly random point in some $4{\,\times\,}4$ square.

Where do the centers need to be in order for the player to win?

Graph the region of winning centers.

The area of the region of winning centers is what fraction of the area of the $4{\,\times\,}4$ square?

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  • $\begingroup$ So if I am following correctly this would be $\frac{\pi}{16}$ for one square, correct? $\endgroup$
    – Derek
    Sep 16, 2017 at 3:14
  • $\begingroup$ First get the shape of the region (i.e., graph the region). Also, you only need to consider one square (the one the center lands in). So for example, suppose the containing square is centered at the origin. Where can the center land in order to win? The area of quarter is not relevant. What you want is the area of the region of winning centers. $\endgroup$
    – quasi
    Sep 16, 2017 at 3:20
  • $\begingroup$ Okay, I think I see it now. The region of winning centers is not a circle. It is also a square. $\endgroup$
    – Derek
    Sep 16, 2017 at 3:38
  • $\begingroup$ Yes, exactly.${}{}$ $\endgroup$
    – quasi
    Sep 16, 2017 at 3:41
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Think first of randomly dropping a line segment of length one-half inch onto a ruler of length 8 inches. (There are eight squares on a checkerboard.) You won't hit an inch mark if the leftmost part of the line segment lands from an inch mark to the next half-inch mark. So you can do it with probability 1/2. Now the quarter makes your problem twice as hard, and you have to worry about the vertical coordinate too. Assuming you toss quarters randomly on the checkerboard, you won't hit a vertical inch mark half the time. Since the events are independent, you will land in a square 1/4 of the time. The game is exactly fair.

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  • $\begingroup$ 1/4. Correct. What do you mean the game is "fair" though? Probability is whatever it is; its always fair. I would say its unfair if there is some unforeseen force altering outcome, like a digitized slot machine; or perhaps "unfair" refers to any game with less than 50% chance of winning? Is "fair" a mathematical term I dont know? $\endgroup$ Sep 16, 2017 at 3:21
  • $\begingroup$ I think fair means the expected value of your profit/loss is 0. i.e. in the long run, the house won't win. The expected value is 1/4 *\$ 0.75 - 3/4 * \$0.25 = $0 $\endgroup$
    – FullofDill
    Sep 16, 2017 at 3:23
  • $\begingroup$ Cogito, in economics, a fair bet is one that has expected value 0. So if you pay twenty-five cents for the chance to win a dollar with probability 1/4, we think of it as a fair bet. We're always looking for better than fair bets too. :) $\endgroup$ Sep 16, 2017 at 3:24

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