Buffon's needle
Press a button to drop needles on a lined floor. Each needle lands in a random place at a random angle. A needle that crosses a line turns orange. Now work out 2 × drops ÷ crossings. Drop 100 and look at the answer. Then guess what 10,000 will give before you press it.
The estimate after each drop. The dashed line is π. The drop count is on a scale where each step is ten times more.
What's going on
Start with the needle as long as the gap between the lines. Then the chance that a needle crosses a line is exactly 2/π, about 0.637. So crossings ÷ drops is close to 2/π, and turning that around, 2 × drops ÷ crossings is close to π.
Why 2/π? Tilt a needle of length L at an angle θ to the lines. Measured straight across the lines, it reaches a distance L sin θ. The lines are d apart, so the chance it covers one is L sin θ ÷ d. Every angle from 0° to 180° is equally likely, and the average of sin θ over that range is 2/π. So the chance of a crossing is 2L/(πd). The π comes from averaging over a half turn.
Move the slider to make the needle shorter. The chance drops in proportion, so the estimate becomes 2 × (needle ÷ gap) × drops ÷ crossings. The slider stops at 1 because a longer needle can cross two lines at once and the rule changes.
The estimate is slow to settle. The error shrinks roughly like 1 over the square root of the number of drops, so one more correct digit costs about 100 times more needles. Sweep the floor and drop 10,000 a few times. You will get a different answer each time, and with the full-length needle most of them will be within 0.05 of π.
Sheet: Dropping needles to find π. A cousin of this demo: measure π with random darts.