If the possible fairness of a coin is a continuum, the probability of the coin being exactly fair (50H/50T) is 0. To begin exploring, let’s say the coin can only be rigged heads (100H/0T), fair (50H/50T), or rigged tails (0H/100T).
Here’s an arrangement of the numbers 1 to 4, and what it looks like plotted into a 4x4 grid such that each row corresponds to a number in the arrangement. Notice that the 3 dots on the bottom right form a straight line.
Here’s a way to put 6 dots into a 3x3 grid such that no 3 dots form a straight line.
There’s a combination lock with a 3 digit code with no repeated digits, and for each of our attempts to guess the code it tells us the number of correct digits and ones in the correct places.
If we draw a line through the center of a square, we can fold the square in half along this line. Depending on the angle this line makes with the horizontal, as shown in the diagram, the folded square will have different looks.
Here’s one way to put a pair of 1s, a pair of 2s, and a pair of 3s in a row such that the pair of 1s have one number between them, the pair of 2s have two numbers between them, and the pair of 3s have three numbers between them:
For any point inside a square, the sum of the distances from this point to the four sides is half of the perimeter.
Let’s draw a smaller square inside a unit square by dividing out a fixed fraction of each side and connecting the dividing point to an opposite angle.
Starting with 3, let’s double it and add one (or 2n+1), then modulate the answer by 14 (divide by 14 and take the remainder). If we repeat these two steps, we get the first cycle.
Starting with a natural number at the top, we can split it at each level such that the two lower neighbours always add up to the higher neighbour. Let’s repeat this process until a row can’t be split using whole numbers in the following row anymore.