Bertrand paradox calculator
With both ends picked at random on the circle, a chord is longer than the side of the inscribed equilateral triangle with probability 1/3, while two other fair-looking methods give 1/2 and 1/4. Choose the method and radius to see the answer for each. Free in your browser.
- Free
- Runs in your browser
- No sign-up
- Your data stays on your device
Run — free
Worked example
- Radius
- 1
- Probability
- 0.3333333333
- Triangle side
- 1.7320508076
- Critical distance
- 0.5
Bertrand's paradox is the classic demonstration that the phrase random chord does not name a single probability until a selection method is chosen. Given a circle of positive radius, an equilateral triangle inscribed in that circle has a definite side length. The geometric measure is evaluated in closed form from the radius and the chosen method, with no sampling and no language model. A non-positive radius is rejected as invalid input because there is then no circle and no inscribed triangle to compare against.
What Bertrand's paradox asks and why three answers appear
Joseph Bertrand posed a deceptively short question in eighteen eighty-nine: a chord is drawn at random in a circle; what is the probability that the chord is longer than the side of an equilateral triangle inscribed in the same circle? The side length itself is not ambiguous.
In a circle of radius r the inscribed equilateral triangle has side r times the square root of three, because each central angle is one hundred twenty degrees and the chord formula two r sine of half the angle reduces exactly to r times root three. A chord whose midpoint sits at distance d from the center has length two times the square root of r squared minus d squared, so it exceeds that side exactly when d is smaller than r over two.
The difficulty is not the comparison; it is the meaning of at random. Bertrand exhibited three procedures that look equally fair and obtained one third, one half, and one quarter. The paradox is resolved by naming the measure, not by declaring one of the three numbers wrong.
This calculator takes that resolution seriously: you choose a method, it evaluates the corresponding measure, and it returns the probability that the random chord exceeds the inscribed triangle side.
What you can do with it
Classroom demonstration of geometric probability
Teachers contrast the three Bertrand methods on the same circle so students see that random chord is incomplete until a measure is named, and they check the classic one-third, one-half, and one-quarter answers without drawing chords by hand.
Textbook and homework verification
Students confirm that the inscribed equilateral side is r times root three, that the long-chord event is the disk of radius r over two, and that each named method produces its published probability for any positive radius.
FAQ
What happens if the circle radius is not positive?
Zero, negative, infinite, and missing radius values are rejected and not charged, because they do not define a circle.
Why do the three methods give three different answers?
Each method induces a different measure on chords. Radial is uniform along a radius and yields one half. Midpoint is uniform in the disk and yields one quarter.
Is the Bertrand paradox calculator free?
The browser tool runs free on your device.
Does the probability depend on the circle radius?
No. Once the radius is positive, each of the three measures is scale-invariant, so the probability is one third, one half, or one quarter.
How long is the side of the inscribed equilateral triangle?
It is the radius times the square root of three, so 1.732050807569 for a radius of 1. A chord beats that side exactly when its midpoint lies closer to the center than half the radius.
Related tools
For developers — API access
Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.
API endpoint
Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.
Call it from your stack
curl -X POST https://api.kit.forhosting.com/stat/betrand-paradox \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"radius":1,"method":"endpoints"}'const res = await fetch("https://api.kit.forhosting.com/stat/betrand-paradox", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"radius": 1,
"method": "endpoints"
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/stat/betrand-paradox",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"radius": 1,
"method": "endpoints"
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/stat/betrand-paradox", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"radius":1,"method":"endpoints"}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"radius":1,"method":"endpoints"}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/stat/betrand-paradox", body)
req.Header.Set("Authorization", "Bearer "+os.Getenv("KIT_KEY"))
req.Header.Set("Content-Type", "application/json")
res, _ := http.DefaultClient.Do(req)Example request
{
"radius": 1,
"method": "endpoints"
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "stat.betrand_paradox",
"status": "queued",
"_links": {
"result": "/tasks/tsk_…/result"
}
}The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.
Pricing
Published price — no tokens, no invented credits. A failed task is never charged.
Errors
| HTTP | Code | Meaning |
|---|---|---|
401 | unauthorized | Missing or invalid API key. |
402 | insufficient_balance | Your balance doesn't cover the task price. |
404 | unknown_type | That task type doesn't exist. |
429 | rate_limited | Too many requests. Use the webhook instead of polling. |
Technical notes
This calculator reports the probability that a random chord is strictly longer than that side under one of the three historical Bertrand methods: random endpoints on the circumference, a random point on a random radius, or a random midpoint drawn uniformly from the disk.
What Bertrand's paradox asks and why three answers appear
Different natural procedures for picking a chord induce different measures on the space of midpoints or of endpoints, and therefore different probabilities for the same geometric event.
How this calculator validates input and evaluates the three measures
The engine first computes the inscribed triangle side s equal to r times the square root of three and the critical midpoint distance d-star equal to the square root of r squared minus s over two squared, which simplifies to r over two. Under radial a radius is chosen and a point is taken uniformly along it as the midpoint of the perpendicular chord: the favorable segment has length d-star, so the probability is one half. Under midpoint the midpoint is uniform in the disk: the favorable set is the concentric disk of radius d-star, so the probability is the area ratio one quarter. Call the capability with two required fields. radius is the circle radius: a finite number that must be strictly positive. Zero, negative values, NaN, and infinities fail with invalid_input and are not charged, because a non-positive radius does not define a circle and therefore does not define an inscribed triangle. Aliases r, R, and circle_radius are accepted. method is one of three closed values: endpoints, radial, or midpoint. Aliases selection and bertrand_method are accepted. Under endpoints the second point on the circumference is treated as uniform: the chord exceeds s when the central angle lies in the open interval from two pi over three to four pi over three, so the favorable arc divided by the full circumference is one third. The response reports probability as the primary value, together with radius, method, triangle_side, critical_distance, and formula.
Practical use, scale invariance, pricing, and delivery channels
Use the browser tool when you are teaching geometric probability, checking a textbook claim about Bertrand's three methods, or showing why the instruction pick a random chord is incomplete until a measure is named. Results are fully deterministic: the same radius and method always yield the same probability, the same side, and the same critical distance. There is no language model, no external geometry service, and no random sampling of chords, so the capability is safe to pin in continuous-integration checks and reproducible lesson materials. Each successful request is one billable item at $0.002; failed validation does not consume balance. Radius may be any positive finite number: the three probabilities do not depend on the particular radius, which is a theorem of the scale invariance of each measure, but the field is still required so that triangle_side and critical_distance are dimensionally honest and so that a non-positive radius is rejected rather than silently ignored. Wire the endpoint into probability tutorials, interview oracles, or notebooks that contrast rotation-invariant, radial, and area-uniform constructions of a random chord without ever drawing a Monte Carlo histogram that never quite settles.
Integration use cases
Deterministic oracles for quizzes and tests
Curriculum platforms and CI suites call a stable oracle that rejects a non-positive radius, accepts only endpoints, radial, or midpoint, and returns identical probability fields for identical inputs.
Technical questions
How do I compute the Bertrand paradox probability?
Send JSON with a positive radius and method set to endpoints, radial, or midpoint. The response includes probability that the random chord exceeds the inscribed equilateral side, plus triangle_side and critical_distance.