Chance from Areas

Nothing to count

A dart is thrown at a rectangular board and lands somewhere on it, and every spot on the board is equally likely. The board has infinitely many points, so the chance cannot be found by counting outcomes.

Areas can still be compared. If every spot is equally likely, then two parts of the board with the same area are equally likely to be hit. Rule the board into 4 rows of 5 equal squares. That makes 20 squares of the same area, so the dart is equally likely to land in each of them, and each square has probability $\frac{1}{20}$1/20.

The board ruled into 4 rows of 5 equal squares. Each of the 20 squares has the same chance, $\frac{1}{20}$1/20.

The target

The target is a patch 2 squares wide and 2 squares tall, so it covers 2 × 2 = 4 of the 20 squares. The chance of hitting it is the chance of landing in one of those 4 squares: $4\times \frac{1}{20}=\frac{4}{20}$4 × 1/20 = 4/20.

The 2-by-2 target covers 4 of the 20 equal squares.

Area over area

So the probability of a hit is the area of the target divided by the area of the board: $P\left(\text{hit}\right)=4÷20=\frac{1}{5}$P(hit) = 4 ÷ 20 = 1/5. Both areas must be in the same unit, squares here, and the unit cancels in the division.

The position of the target does not matter, only its area. Move the 2-by-2 patch to the far corner or the middle of the board and it still covers 4 of the 20 squares, so the chance is still $\frac{1}{5}$1/5.

The chance of missing is the rest of the board: 16 of the 20 squares, so $\frac{16}{20}=\frac{4}{5}$16/20 = 4/5, and $\frac{1}{5}+\frac{4}{5}=1$1/5 + 4/5 = 1.

Any shape of target

The target need not be made of whole squares. A board 10 cm by 10 cm has a circular target of radius 3 cm drawn on it. The target's area is $\pi \times 3^2=9\pi$π × 3² = 9π, about 28.3 cm², and the board's area is 10 × 10 = 100 cm². So the chance of hitting the circle is $\frac{9\pi }{100}$9π/100, which is about 0.283.

A 10 cm by 10 cm board with a circle of radius 3 cm: the circle covers $9\pi \approx 28.3$9π ≈ 28.3 of the 100 square centimeters.

Where a point must land

Sometimes the target has to be worked out first. A coin 2 cm across is rolled onto a square 5 cm by 5 cm, and it wins if it lands wholly inside the square. Follow the coin's center: the coin touches no edge when its center is at least 1 cm, the radius, from all four sides.

That leaves a smaller square in the middle, 5 − 1 − 1 = 3 cm wide. Its area is 3 × 3 = 9 cm² out of 5 × 5 = 25 cm², so the chance of a win is $\frac{9}{25}$9/25.

The 5 cm square in 1 cm cells. The coin's center must land in the middle 3-by-3 block, 1 cm from every side: 9 of the 25 cells.

The usual mistakes

Comparing one side with one side. The target is 2 squares wide on a board 5 squares wide, but $\frac{2}{5}$2/5 is not the chance: area needs both the width and the height, $\frac{4}{20}$4/20.

Putting the board over the target. $\frac{20}{4}=5$20/4 = 5 is more than 1, and no probability is more than 1. The target goes on top.

Comparing the target with the rest of the board. 4 squares against the other 16 gives $\frac{4}{16}$4/16, but the whole board, 20 squares, is what the dart lands on.

Mixing units, such as a target in square centimeters on a board measured in square meters. Convert both areas to the same unit first.

A coin at a fair

In the application below, a coin rolled onto a board of squares wins when its center lands in the smaller square inside each one, and the probability is that square's area over the whole square's. Then the size of square is found that gives a chance of $\frac{4}{9}$4/9.