Teaching objectives
A floor of boards with parallel lines the same distance d apart, and a needle that falls spinning, bounces a little and stays where it lands; the crossings light up in red. The student drops needles (and later bent pieces and a hoop) and counts how many lines they cross; at the end, an ant does the same in a room in the dark. In every room the student bets before each experiment and then checks whether they were right. Each student starts with 10 Ᵽ; a won bet adds 5 Ᵽ and a lost bet takes away 3 Ᵽ. In all six rooms the bet is settled with a long series (100 to 3000 repetitions) that plays itself, so deciding well pays off.
There are almost no explanation screens: the theory comes as challenges, with a short theory box in the rooms that introduce a new idea (geometric probability, additive expectation, the hoop, relative frequency, estimating by proportions). Every room follows the same script: a bet, a few drops by hand on the parquet floor, a series in which the mean settles onto a line, and comprehension challenges (choose between options, complete a fraction, write a number) that lead from the experiment to the formula. Everything is done without integrals. The rooms are ordered so that π does not appear until it is justified: the first three work with the 0.637 of the first room and it is the hoop, in the fourth, that explains where it comes from.
What you’ll learn
The lab takes the student from “how many needles cross a line?” to seeing why π appears and why the mean number of crossings depends only on the length.
- Seeing that the chance of crossing depends on how the needle is turned, using geometric probability (favourable measure over total measure) and keeping the question of where the 63.7 % comes from (room 1).
- Moving from “does it cross or not?” to counting how many crossings there are, and seeing that the mean of a sum is the sum of the means, even when the pieces are joined; so a whole needle gives the same as two halves and one of 2d gives double, with more than 1 crossing on average (room 2).
- Checking that a needle bent into a V or a wire of any shape gives the same mean crossings as a straight one of the same length (room 3).
- Proving the formula with a hoop, without integrals, as Barbier did: π appears there for the first time, and the result is applied to a straight needle (the 2/π of room 1) and to a square frame (room 4).
- Using the frequency of crossings to estimate π, reading the estimate, and being suspicious of results that are “too good” (room 5).
- Using the same formula backwards to measure an area by counting crossings, like the scout ants that choose a nest, and recognising estimation by proportions (lentils, marked fish) (room 6).
Key mathematical ideas
- The turn matters: a needle lying across the lines spans d vertically and always crosses; a parallel one spans nothing and almost never crosses. At 30° it spans d/2 vertically: it crosses with probability (d/2)/d = 1/2. This is geometric probability: “at random” means uniform and the probability is favourable measure over total measure (like waiting less than 5 minutes for a bus that comes every 20: 5/20 = 1/4). With many needles the percentage that cross settles at 63.7 %, and where it comes from is seen in room 4.
- Expectation is additive: a piece can cross 0, 1, 2… lines, and the mean crossings add up. A die gives 3.5 on average; two dice, 3.5 + 3.5 = 7; and on a single die the top face and the bottom face, 3.5 + 3.5 = 7 even though they are joined (opposite faces always add up to 7). A half needle (d/2) gives half as much as a whole one; two halves, the same as a whole one, and glued into a V they still give the same: the mean of a sum is the sum of the means, whether the pieces are loose or joined. This is the linearity of expectation. A needle of 2d gives double, 4/π ≈ 1.27: it goes above 1, so it is no longer a probability.
- Only the length matters: for any shape of length ℓ, the mean number of crossings is 2ℓ/(πd) (Buffon’s “noodle”, proved by Barbier in 1860). The V crosses no line at all more often than the straight needle (≈ 52 % against ≈ 36 %) and sometimes crosses two (≈ 16 %), and the two effects cancel out. A wire of length 3d, whatever its shape, gives three times as much as a needle of d; a square frame of side d, 4 · 2/π = 8/π ≈ 2.55.
- The hoop proves it: a hoop of radius d is 2d tall and the lines are d apart, so there are always two lines within that height and each crosses it twice: always 4 crossings, even when it is tangent to a line (2 + 1 + 1). It is a bent needle of length 2πd. If the mean number of crossings is proportional to the length, each unit of length gives 4/(2πd) = 2/(πd) crossings. A straight needle of length d then gives d · 2/(πd) = 2/π ≈ 0.637, Buffon’s result (1777), without integrals, and since it crosses at most one line, that is its probability of crossing. A hoop of radius d/2 always crosses 2 times. This is where π appears for the first time, and it comes from the length of the circle.
- π as a frequency: if C of N needles cross, C/N ≈ 2/π, hence π ≈ 2 · N/C. The relative frequency approaches the probability but does not equal it, and that is why the estimate of π wobbles: with 3000 needles, in the hundredths. It is a Monte Carlo method: chance is used to compute a number, but slowly, because the error shrinks like 1/√N; each extra decimal needs about a hundred times more needles (hundreds of thousands to get 3.14 right).
- Lazzarini (1901) claimed to have dropped 3408 needles of length 5/6 of d and obtained 355/113 = 3.1415929, six correct decimals. With that needle, π ≈ (5/3) · N/C, and N = 3408 = 213 · 16, C = 1808 = 113 · 16 gives exactly 355/113: it was enough to keep notes and stop at the right moment.
- Ants that measure their home (Mallon and Franks, 2000): a Leptothorax albipennis scout wanders aimlessly in the dark around a candidate room, twice; on the first visit it leaves a scent trail of length L₁ and on the second, of length L₂, it counts the number n of times it crosses it. On the parquet there is 1/d of line per unit of area and a needle of length ℓ crosses on average (2/π) · ℓ · (1/d); in the room the “line” is the trail, with L₁/A per unit of area, so the mean number of crossings is 2L₁L₂/(πA). It measures area, not width: with twice the area it crosses half as often (in the lab, about 12.8 crossings per visit in 60 × 40 mm and 6.4 in 80 × 60 mm, with 22 cm per visit). It is an estimate by proportions, like the lentils in a sack or the fish in a lake (50 are marked, 40 are caught and 8 carry a mark: about 250 fish). If half the floor is swapped for clean floor between visits, the ant crosses half as often and thinks the room is twice as big: with that trick, colonies chose a half-size room 8 times out of 20, which without it they almost never chose (1 out of 16).
Level: ESO and Bachillerato (from the 4th year of ESO, age 15). It is a short lab in the probability series, alongside “Probability”, “Expectation” and “Monty Hall”; room 2 reuses the idea that expectation adds up. In class it works well to project room 1 and ask the class to bet before any needle has fallen and to keep the question of what that 63.7 % is; room 4, with its hoop that always crosses 4 times, is the one that surprises most and the one that answers it; room 5 serves to compare the estimates of π from different groups and discuss Lazzarini’s case, and room 6 shows the formula outside mathematics.
Room-by-room contents
Room 1 · The needle
The Count of Buffon proposed this game in 1777: a floor of boards with parallel lines a distance d apart and a needle of length d that falls spinning onto it. Bet: what fraction of the needles will cross a line (less than half, between half and three quarters, or more than three quarters)? Eight needles are dropped by hand, with a «Crossing: c of n» tally, and the bet is settled with a series of 300 needles. Then a theory box on geometric probability: when the outcomes cannot be listed, “at random” means uniform and the probability is favourable measure over total measure (example of a bus that comes every 20 minutes: waiting less than 5 is 5/20 = 1/4). A needle lying across the lines spans d vertically and always crosses; a parallel one spans nothing and almost never crosses; at 30° it spans d/2 vertically and crosses with probability 1/2. π does not appear here: the room ends with the series settled at almost two needles out of three (63.7 % with far more needles) and the question of what percentage that is and where it comes from, which the next rooms answer.
Student tasks
- Bet what fraction of the needles will cross a line.
- Choose which needles cross a line more often (those that land across the lines).
- Write the probability that a needle at 30° crosses (1/2).
Room 2 · Two half needles
Now we do not ask whether the needle crosses, but how many crossings there are in total: a piece can cross 0, 1, 2 lines or more. A needle of length d is compared with two half needles of length d/2, each falling on its own. Bet: which gives more crossings on average, the whole needle, the two halves, or the same? Four rounds are played by hand (in each one the whole needle falls and then the two halves, on the same floor), with a tally of the crossings of each; the bet is settled with a series of 3000 drops of each, and if the means differ by less than 0.05 they count as equal. Then a theory box: the expectation is what comes out on average over many repetitions (a die gives 3.5) and it is additive, the expectation of a sum is the sum of the expectations (two dice, 3.5 + 3.5 = 7), even when the things are joined (on a single die the top face and the bottom face give 3.5 + 3.5 = 7 on average, and indeed opposite faces always add up to 7). A whole needle is two halves stuck together, so it gives the same as two loose halves. π does not come out yet: the 0.637 of the first room is used. The room closes with the mean crossings being proportional to the length.
Student tasks
- Bet whether the whole needle, the two halves or the same give more crossings on average.
- Choose why a whole needle gives on average the same crossings as two loose half needles (it is two halves stuck together, and each crosses as often on average stuck as loose).
- Choose how many crossings a needle of length 2d gives on average (about 1.27: double; it goes above 1, so it is no longer a probability).
Room 3 · The bent needle
A straight needle of length d against another of the same length bent into a V, with two pieces of d/2. The straight one crosses at most one line; the V can cross two. Bet: which gives more crossings on average, the straight one, the V, or the same? Four rounds are played by hand, with a tally of the crossings of each and of the times the V crosses two lines, and the bet is settled with a series of 3000 drops of each (if the means differ by less than 0.05 they count as equal). This room has no theory box, because it is the same idea as the previous one. Then a table with the share of drops with 0, 1 or 2 crossings: the V crosses nothing more often (≈ 52 % against ≈ 36 %) and sometimes crosses two (≈ 16 %), and the two effects cancel out. The V is two half needles stuck together, and each crosses as often on average as when loose. A wire of length 3d, twisted into any shape, gives three times as much as a straight needle of d; it is Buffon’s “noodle”: the mean crossings depend only on the length.
Student tasks
- Bet whether the straight needle, the V or both give more crossings on average.
- Explain how the straight needle and the V can give the same (the V crosses nothing more often and that makes up for its doubles).
- Choose how many crossings a wire of length 3d gives on average, whatever its shape (three times a straight needle of d).
Room 4 · The hoop proves it
Instead of a needle, a hoop: a circle of radius d, equal to the distance between lines, which is a needle bent into a circle. Bet: when it is dropped, how many times does it cross the lines (it depends on how it lands, 2 to 4; always 4; or always 2)? Five hoops are dropped by hand, with a tally of the crossings of each, and they settle the bet, with no series, because it does not depend on chance: it is always 4. The hoop is 2d tall and the lines are d apart, so within that height there are always two lines and each crosses it twice; even when it is tangent to a line, it is 2 + 1 + 1 = 4. Then a theory box: the hoop is a bent needle of length 2πd that gives 4 crossings, so each unit of length gives 4/(2πd) = 2/(πd) crossings on average, and that is where π comes in, from the length of the hoop; if a needle crosses at most one line, its mean crossings are its probability of crossing. This is Barbier’s proof (1860), without integrals; Buffon had worked it out in 1777 with integrals.
Student tasks
- Bet how many times a hoop of radius d crosses.
- Choose how many times a hoop of radius d/2 crosses (always 2: it always has one line inside).
- Calculate the mean crossings of a straight needle of length d (d · 2/(πd) = 2/π ≈ 0.637, the 63.7 % of the first room).
- Calculate the mean crossings of a square frame of side d (4 · 2/π = 8/π ≈ 2.55).
Room 5 · π falling from the sky
After the hoop, 2/π of the needles of length d cross, and this can be used the other way round to work out π by counting: with N needles and C crossings, C/N ≈ 2/π. Bet: with 3000 needles, where will the estimate of π land (below 3.0, between 3.0 and 3.3, or above 3.3)? Six needles are dropped by hand with a meter of «needles», «cross» and «π ≈», and the bet is settled with a series of 3000 needles whose estimate is drawn on a graph, with π as a dashed line. Then a theory box on relative frequency: it approaches the probability but does not equal it, and it lets us estimate a quantity that appears in it (example of darts thrown at a square of side 2 with a circle of radius 1: π/4, and 785 out of 1000 gives π ≈ 3.14). Solving, π ≈ 2 · N/C. In 1901 Lazzarini reported 3408 needles (of length 5/6 of d) and 355/113 = 3.1415929, six correct decimals: with that needle π ≈ (5/3) · N/C, and 3408 and 1808 are 213 and 113 times 16, so he stopped just when the count gave him 355/113. With 3000 needles the estimate wobbles in the hundredths: each extra decimal needs about a hundred times more needles. It is the first Monte Carlo method.
Student tasks
- Bet where the estimate of π will land with 3000 needles.
- Solve for π in C/N ≈ 2/π (π ≈ 2 · N/C).
- Decide what is most likely in Lazzarini’s experiment (he stopped dropping just when the count gave 355/113).
- Choose how many needles are needed to be almost sure of two decimals (hundreds of thousands).
Room 6 · Ants that measure their home
Leptothorax albipennis ants live in flat cracks in rocks. When the colony moves, a scout enters a candidate room in the dark, seen from above, and wanders around it aimlessly, from wall to wall: on the first visit it leaves a trail with its scent and on the second it counts the times it crosses that trail (Mallon and Franks, 2000). It is measuring the area of the floor, not the width. Two rooms, one of 60 × 40 mm and another of twice the area (80 × 60 mm), and on each visit the ant walks 22 cm. Bet: in the big one, does she cross twice as often, the same, or half as often? Two explorations are watched by hand in the two rooms, with the crossings lighting up, and the bet is settled with 100 visits to each (about 12.8 crossings per visit in the small one and 6.4 in the big one). Then a theory box: to estimate a total that cannot be counted we use a proportion (example of the lentils in a 3 kg sack: 100 weigh 6 g, so there are about 50,000). It is Buffon’s needle: on the parquet there is 1/d of line per unit of area and a needle of length ℓ crosses on average (2/π) · ℓ · (1/d); in the room the “line” is the first trail, of length L₁ in an area A, and the mean crossings are 2L₁L₂/(πA). It is seen with another case of proportions, capture and recapture of fish: 50 are marked, 40 are caught and 8 carry a mark, so there are about 250. Finally, the trap of the experiment: if half of the floor is swapped for clean floor, the scout crosses half as often and thinks the room is twice as big; with a trap like that, colonies chose a room of half the size 8 times out of 20, when without it they almost never did (1 out of 16).
Student tasks
- Bet how many times the ant crosses her trail in a room of twice the area.
- Choose why she crosses half as often (the same trail, over twice the area, is half as dense).
- Choose the formula for the mean crossings, 2L₁L₂/(πA), from the one for the parquet.
- Estimate how many fish there are in a lake if 50 are marked and a catch of 40 has 8 with a mark (about 250).
- Predict what the scout concludes if half of the floor is swapped for clean floor between visits (that the room is twice as big).
Rooms to project
The most striking ones to show and discuss in class.