Introduction
Consider a plane ruled into parallel strips of uniform width (t). A needle of length (l) is dropped at random; the event of interest is whether the needle intersects one of the parallel bounding lines (illustratively, one needle may straddle a line while another may lie entirely within a strip). Posed by Georges‑Louis Leclerc, Comte de Buffon in the 18th century, this is the earliest solved problem in geometric probability and is naturally treated within integral geometry.
When the needle is no longer than a strip ((l\le t)) the crossing probability admits the simple closed form
[
p=\frac{2}{\pi}\,\frac{l}{t}.
]
This expression follows from the natural probabilistic model in which the needle’s orientation (\theta) is uniformly distributed on ([0,\pi]) (rotational symmetry) and the signed distance (x) from the needle’s center to the nearest parallel line is uniformly distributed on ([0,t/2]). The needle crosses a line precisely when (x\le \tfrac{l}{2}\sin\theta); integrating this condition over the angular variable yields the displayed formula. The constant (\pi) appears because the angular integral extends over a half‑turn and reflects the underlying rotational symmetry.
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A practical consequence is that, for (l\le t), empirical measurement of the crossing frequency furnishes a Monte Carlo estimate of (\pi) via (\pi=(2l)/(t p)), although Buffon did not introduce the problem for this purpose. The simple form above is restricted to (l\le t); when (l>t) the probability is more intricate but can be obtained by analogous integral‑geometric reasoning.
A needle of length l is dropped at random on a plane ruled by parallel lines a distance t apart. Denote by x the transverse distance from the needle’s center to the nearest line; by symmetry x is confined to the interval 0 ≤ x ≤ t/2. Let θ be the acute angle between the needle and the parallel lines, so 0 ≤ θ ≤ π/2. Under the usual uniform-random assumptions the center is uniformly distributed in [0,t/2] with density f_X(x)=2/t, and the orientation is uniform in [0,π/2] with density f_Θ(θ)=2/π; independence of X and Θ yields the joint density
f_{X,Θ}(x,θ) = (4)/(tπ), for 0 ≤ x ≤ t/2, 0 ≤ θ ≤ π/2.
Geometrically the needle intersects a line exactly when the transverse half-projection (l/2) sin θ exceeds the center’s distance to the nearest line, i.e. when
x ≤ (l/2) sin θ.
Consequently the crossing probability is the integral of the joint density over the region {(x,θ): 0 ≤ x ≤ min(t/2,(l/2) sin θ), 0 ≤ θ ≤ π/2}, equivalently
P(cross) = ∫{0}^{π/2} ∫{0}^{min(t/2,(l/2) sin θ)} (4)/(tπ) dx dθ.
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The presence of the minimum leads to two geometric regimes. If l ≤ t then (l/2) sin θ ≤ t/2 for all θ and the inner limit is (l/2) sin θ for every θ; if l > t there exist orientations for which (l/2) sin θ exceeds t/2 and the inner limit becomes t/2 on that subinterval of θ. This bifurcation—arising from how (l/2) sin θ compares with the maximum transverse distance t/2—governs the separate analytic treatments of the short-needle (l ≤ t) and long-needle (l > t) cases.
Case 1 (l ≤ t): Consider a needle of length l dropped at random on a plane ruled by parallel lines a distance t apart. Let θ ∈ [0, π/2] denote the acute angle between the needle and the lines, and let x ∈ [0, t/2] denote the perpendicular distance from the needle’s midpoint to the nearest line. Because position and orientation are uniform, the pair (θ, x) is uniformly distributed on the rectangle [0, π/2] × [0, t/2]; the area of this domain is (t/2)(π/2) = tπ/4, so the joint density is constant and equals 4/(tπ).
For a fixed orientation θ the needle intersects a line exactly when the midpoint distance satisfies x ≤ (l/2) sin θ. Under the short-needle assumption l ≤ t this upper bound never exceeds t/2 for θ ∈ [0, π/2], so the event that the needle crosses a line corresponds to the region {(θ, x) : 0 ≤ θ ≤ π/2, 0 ≤ x ≤ (l/2) sin θ}. The crossing probability is the integral of the constant density over this region:
P = (4/(tπ)) ∫{0}^{π/2} ∫{0}^{(l/2) sin θ} dx dθ
= (4/(tπ)) ∫{0}^{π/2} (l/2) sin θ dθ
= (2l/(tπ)) ∫{0}^{π/2} sin θ dθ
= 2l/(tπ).
Thus, when l ≤ t the intersection probability is P = 2l/(tπ); it increases linearly with needle length and decreases inversely with line spacing, attaining its maximum value 2/π ≈ 0.6366 at l = t.
Assuming the needle length l exceeds the line spacing t (l > t) and restricting orientations to θ ∈ [0, π/2], the probability that a randomly dropped needle crosses at least one line can be written by integrating the uniform joint density of orientation and lateral midpoint position. With joint density 4/(tπ) on the relevant domain, one convenient double integral is
P = ∫{0}^{π/2} ∫{0}^{m(θ)} (4/(tπ)) dx dθ,
where the integration cutoff m(θ) = min((l/2) sin θ, t/2) encodes the geometry: for a given θ the midpoint distance x need only be integrated up to the smaller of the half-projection (l/2) sin θ and half the interline spacing t/2.
This cutoff induces a natural angular partition at θ0 = arcsin(t/l). For 0 ≤ θ ≤ θ0 the half-projection (l/2) sin θ is ≤ t/2, so the inner integral runs to (l/2) sin θ and the crossing probability for that θ agrees with the short-needle formula (i.e., depends on x). For θ0 < θ ≤ π/2 the half-projection exceeds t/2, so every allowed midpoint produces a crossing and the integrand integrated over x yields the constant 2/π; equivalently the second angular contribution is (2/π) ∫_{θ0}^{π/2} dθ = (2/π) arccos(t/l).
Evaluating the two-piece integral explicitly reproduces the closed-form probability
P = 1 + (2l)/(tπ) − (2/(tπ))[ √(l^2 − t^2) + t·arcsin(t/l) ],
which may be algebraically rearranged to the equivalent expression
P = (2/π) arccos(t/l) + (2/π)(l/t) [1 − √(1 − (t/l)^2)].
In the latter form the first term corresponds to orientations that guarantee a crossing regardless of lateral position, while the second term measures the additional probability coming from orientations where crossing depends on the midpoint location. All identities above assume l > t so that t/l ∈ (0,1) and the inverse trigonometric functions arcsin(t/l) and arccos(t/l) are well defined for the stated ranges.
Using elementary calculus one treats the short-needle case (needle length l no greater than strip spacing t, so a needle can cross at most one line) by decomposing the crossing probability into two factors: the probability that the needle’s center falls close enough to a bordering line, and the conditional probability of a crossing given the center is so positioned. Writing P = P1·P2, the first factor is obtained geometrically: the center must lie within horizontal distance l/2 of a line on either side, so the total favorable band width is l and hence
P1 = l/t.
For the conditional probability P2 one may proceed by fixing the center-to-line horizontal distance x and the needle angle θ measured from the horizontal. For a given x only orientations within an angular interval of length 2θ(x) produce a crossing (orientations are taken over π radians), so the conditional proportion at fixed x is 2θ(x)/π. Choosing units so that l = 2 (half-length 1) makes x range over [0,1]; the boundary angle satisfies θ(x) = arccos(x) and therefore
P2 = ∫0^1 (2θ(x)/π) dx = (2/π) ∫0^1 arccos(x) dx.
The integral ∫0^1 arccos(x) dx evaluates exactly to 1, yielding P2 = 2/π. Thus, in these scaled coordinates the conditional probability is independent of l, and combining factors gives P = (l/t)·(2/π).
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An alternative, more compact derivation averages directly over orientations. By symmetry it suffices to consider θ ∈ [0,π/2]; for a given θ the far end of the needle can lie at horizontal distance up to l cos θ from the nearest line, so averaging the available crossing distance over θ produces
P = (∫0^{π/2} l cos θ dθ) / (∫0^{π/2} t dθ) = (l/t) · (∫0^{π/2} cos θ dθ) / (π/2).
Since ∫0^{π/2} cos θ dθ = 1, this yields the same result:
P = 2l/(tπ),
the exact crossing probability for the short-needle case.
The short-needle version of Buffon’s problem asks for the probability p that a segment of length L, thrown at random onto a plane ruled by parallel lines a distance t apart, will meet one of the lines. An elementary geometric solution avoids integration by relating the crossing behavior of any curve to that of a particular circle: a circle whose diameter equals the strip spacing t always meets the family of parallel lines in exactly two points, regardless of its placement or orientation.
Because that circle has circumference πt and yields two intersections, the average number of intersections produced per unit length of any randomly oriented curve is 2/(πt). For a segment of length L the expected number of intersections is therefore (2/(πt))·L. In the short-needle regime (L ≤ t) a segment cannot intersect more than one line, so its expected number of intersections coincides with the probability of at least one crossing. Hence
p = 2L/(πt).
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This integration-free reasoning was published by Joseph-Émile Barbier in 1860 and is commonly termed “Buffon’s noodle”: the expected number of crossings depends only on the curve’s length, not on its specific shape.
Estimating π from Buffon’s needle
Buffon’s needle relates a simple geometric probability to π. For a needle of length l dropped at random onto a plane ruled with parallel lines distance t apart (with l ≤ t), the probability that the needle crosses a line is P = 2l/(tπ). Solving for π gives π = (2l)/(tP), so an empirical estimate of π follows directly from an observed crossing frequency.
In practice one observes h crossings in n independent tosses and substitutes P ≈ h/n to obtain the estimator π̂ ≈ (2 l n)/(t h). This procedure is illustrated both by small manual experiments (for example, dropping matches of length l = 9 on rows spaced t = 9 with n = 17 and h = 11 yielding π̂ = (2·9·17)/(9·11) = 34/11 ≈ 3.0909) and by computer simulations. A Python/Matplotlib simulation with t = 5.0 and l = 2.6 shows the running value of π̂ converging toward ≈3.14 as n increases, reflecting the law of large numbers.
A historically notable report by Mario Lazzarini (1901) claimed n = 3,408 throws and an estimate equal to the rational 355/113. In his setup l/t = 5/6, which changes the theoretical crossing probability to P = 5/(3π) and hence yields the specialized estimator π̂ ≈ (5 n)/(3 x), where x is the observed number of crossings. Algebraically, reproducing 355/113 from that estimator requires x = (113 n)/213; thus choosing n as a multiple of 213 makes x an integer (for n = 213 one finds x = 113, and Lazzarini’s n = 3,408 = 213×16 would require x = 113×16 = 1,808).
Statistical scrutiny and historical context cast doubt on the experiment’s credibility. The exactness and sustained agreement of intermediate partial results with 355/113 are extremely unlikely under unbiased random sampling, suggesting selective reporting, confirmation bias, aggregation or fabrication rather than a straightforward physical experiment. The fraction 355/113 (Milü) is a well‑known exceptionally accurate rational approximation to π; reproducing it exactly from a modest number of trials without deliberate selection is therefore improbable. Contemporary commentators, including Dutch science journalist Hans van Maanen, have argued that Lazzarini’s publication—appearing in a pedagogical magazine—was not intended as a rigorous experimental claim and that readers familiar with the apparatus would have noted implausibilities in the description.
Laplace’s short-needle extension of Buffon’s problem considers a rectangular grid formed by two orthogonal families of parallel lines with cell side lengths a and b. A needle of length l is assumed shorter than both cell sides (l < a, l < b); its midpoint (x,y) lies uniformly in one representative cell 0 ≤ x ≤ a, 0 ≤ y ≤ b and its (undirected) orientation φ is taken uniformly on −π/2 ≤ φ ≤ π/2. With x, y, φ independent and uniformly distributed, the total sample-space “volume’’ is V = πab.
For a fixed orientation φ the set of admissible midpoint positions that avoid all grid lines is a rectangle of side lengths a − l cosφ (in x, since cosφ ≥ 0 on the chosen φ-interval) and b − l|sinφ| (in y), so the allowed area is
F(φ) = (a − l cosφ)(b − l|sinφ|) = ab − b l cosφ − a l |sinφ| + (l^2/2)|sin2φ|.
The measure V of all non-crossing configurations is V = ∫_{−π/2}^{π/2} F(φ) dφ, and evaluating the elementary integrals ∫ cosφ dφ = 2, ∫ |sinφ| dφ = 2, ∫ |sin2φ| dφ = 2 yields
V* = πab − 2bl − 2al + l^2.
Hence the probability that the needle does not meet any grid line is P(no intersection) = V*/V = 1 − (2l(a + b) − l^2)/(πab), and the desired probability of at least one intersection is
P(intersection) = (2l(a + b) − l^2) / (πab),
valid under the short-needle conditions l < a and l < b and the stated uniform, independent sampling of midpoint and orientation.
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In a normalized comparison between Buffon’s original (single family of parallel lines) experiment and Laplace’s grid extension (two perpendicular families), set the spacing parameters to $a=b=t=2\ell$ so that the marginal probabilities of a needle intersecting either a horizontal or a vertical line satisfy $P(A)=P(B)=1/\pi$. For the Laplace (grid) experiment take $N=100$ independent drops; note that a direct total-drop comparison to Buffon’s single-family setup would use $2N=200$ drops for the parallel-lines case.
Denote by $x,y\in{0,1}$ the indicators that a given drop intersects a horizontal ($A$) or vertical ($B$) line, respectively. From the general expression for the probability of missing both families, $P(A’!B’)=1-[2\ell(a+b)-\ell^2]/(\pi ab)$, the specialization $a=b=2\ell$ yields the exact value $P(A’!B’)=1-7/(4\pi)$. Combining this with the marginals $P(A)=P(B)=1/\pi$ and the identities $P(A)=P(AB)+P(AB’)$, $P(B)=P(AB)+P(A’B)$ determines the joint probability structure exactly: $P(AB)=1/(4\pi)$ and $P(AB’)=P(A’B)=3/(4\pi)$.
For the grid estimator define
$\hat p=(1/100)\sum_{n=1}^{100}(x_n+y_n)/2=(1/200)\sum_{n=1}^{100}(x_n+y_n)$; the efficiency question reduces to its variance. For a single drop, $\mathrm{Var}(x+y)=\mathrm{Var}(x)+\mathrm{Var}(y)+2\mathrm{Cov}(x,y)$. With $E(x)=E(y)=1/\pi$ the marginal variances are $\mathrm{Var}(x)=\mathrm{Var}(y)=(1/\pi)(1-1/\pi)$. Using $E(xy)=P(AB)=1/(4\pi)$ gives the covariance
$\mathrm{Cov}(x,y)=1/(4\pi)-(1/\pi)^2=(\pi-4)/(4\pi^2)$, which is strictly negative (since $\pi<4$). Substituting these terms yields the closed form
$\mathrm{Var}(x+y)=(5\pi-8)/(2\pi^2)$, and hence the grid estimator variance
$\mathrm{Var}(\hat p)=(1/200^2)\cdot 100\cdot\mathrm{Var}(x+y)=(5\pi-8)/(800\pi^2)\approx 0.000976$.
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For Buffon’s single-family estimator with $M$ independent drops, $\hat q=(1/M)\sum_{m=1}^M x_m$, the variance is
$\mathrm{Var}(\hat q)=(1/M)(1/\pi)(1-1/\pi)\approx 0.217/M$. Equating this to the grid variance $0.000976$ yields $M\approx 222$, which exceeds $2N=200$. Thus, to match the precision attained by $N=100$ grid drops, one needs roughly $222$ drops against a single family of lines, so the Laplace grid estimator is more efficient in this comparison.
The efficiency gain arises from the negative covariance between the two intersection indicators for a single drop: the joint probability structure induces negative correlation, which reduces $\mathrm{Var}(x+y)$ below the sum of marginal variances. This mechanism is an instance of antithetic variates—using negatively correlated components to achieve variance reduction.