Dropping needles to find π
Tonight's idea
A floor has parallel lines a distance \(d\) apart. Drop a needle of length \(L \le d\) at random. The chance that it lands across a line is
\(P = \dfrac{2L}{\pi d}\).
No circle is in sight, so why \(\pi\)?
Drop a bent wire, a "noodle", of length \(L\). Let \(E\) be its expected number of crossings.
Step 1. Cut the noodle into tiny straight pieces. Its crossings are the sum of theirs, and expectation adds even though the pieces are joined. So \(E\) depends only on length, and twice the length gives twice the crossings: \(E = cL\) for some constant \(c\).
Step 2. A circle of diameter \(d\) always crosses exactly twice. Its length is \(\pi d\), so \(c \times \pi d = 2\) and \(c = \dfrac{2}{\pi d}\).
Step 3. A needle with \(L \le d\) crosses once or not at all, so its expected number of crossings is its chance of crossing: \(P = cL\).
Watch one: a needle as long as the gap
Take \(L = d\). Then \(P = \dfrac{2}{\pi} \approx 0.637\), so about 64 needles in 100 cross a line.
Turn it round. In \(N\) drops you expect about \(\dfrac{2N}{\pi}\) crossings, so
\[\pi \approx \frac{2 \times \text{drops}}{\text{crossings}}.\]
Suppose 500 drops give 320 crossings. The estimate is \(1000/320 = 3.125\). That is close, but not very. Needles are a slow way to find \(\pi\): each extra digit takes about 100 times as many drops.
Drop 10,000 needles in a second: mathdetours.com/demos/buffon.html.
Your turn
- The lines are 5 cm apart and the needle is 2 cm long. Find the chance that it crosses a line. About how many crossings do you expect in 400 drops?
- With \(L = 3\) cm and \(d = 4\) cm, 2,000 drops give 950 crossings. Rearrange the formula and estimate \(\pi\).
- (a) A wire of length \(3d\) is bent into a square. Find its expected number of crossings. (b) Do the same for a straight needle of length \(2d\). Why is the answer to (b) not the chance of crossing a line?
- The rounded triangle in the figure is made from three arcs. Each arc joins two corners of an equilateral triangle of side \(d\) and is centred on the third corner. Like the circle, it is \(d\) wide in every direction, so it always crosses exactly twice. What does the noodle argument say its perimeter must be? Check by working out the three arcs.
Stretch: square tiles
Now the floor has lines both ways, making squares of side \(d\). Drop a needle of length \(d\). Find the expected number of crossings. The chance of crossing both a horizontal and a vertical line is \(\tfrac1\pi\). Find the chance of crossing at least one line.
Answers
- \(P = \dfrac{2 \times 2}{\pi \times 5} = \dfrac{4}{5\pi} \approx 0.255\). In 400 drops, about \(400 \times 0.255 \approx 102\) crossings.
- \(\dfrac{\text{crossings}}{\text{drops}} \approx \dfrac{2L}{\pi d}\), so \(\pi \approx \dfrac{2L \times \text{drops}}{d \times \text{crossings}} = \dfrac{2 \times 3 \times 2000}{4 \times 950} = \dfrac{12000}{3800} \approx 3.16\).
- (a) Only the length matters: \(E = \dfrac{2 \times 3d}{\pi d} = \dfrac{6}{\pi} \approx 1.91\). (b) \(E = \dfrac{4}{\pi} \approx 1.27\). This is bigger than 1, so it cannot be a chance. A needle longer than \(d\) can cross two lines at once, so the expected number of crossings is more than the chance of at least one.
- It always crosses twice, so \(c \times \text{perimeter} = 2\), the same equation as for the circle. The perimeter must be \(\pi d\). Check: each arc has radius \(d\) and turns through \(60^\circ\), a sixth of a full circle, so its length is \(\tfrac16 \times 2\pi d = \tfrac{\pi d}{3}\). Three arcs make \(\pi d\). Every curve of constant width \(d\) has this perimeter.
Stretch. Expectation adds: \(\tfrac2\pi\) for the horizontal lines plus \(\tfrac2\pi\) for the vertical ones, so \(E = \tfrac4\pi \approx 1.27\). For the chance of at least one, add the two chances and take away the overlap, which was counted twice: \(\tfrac2\pi + \tfrac2\pi - \tfrac1\pi = \tfrac3\pi \approx 0.955\).