{"record":{"id":"c0ea10903ada95ce","repo":"TheAlgorithms/Python","slug":"at-least-one-simulation-is-necessary-to-estimate-p","errorCode":null,"errorMessage":"At least one simulation is necessary to estimate PI.","messagePattern":"At least one simulation is necessary to estimate PI\\.","errorType":"exception","errorClass":"ValueError","httpStatus":null,"severity":"error","filePath":"maths/pi_monte_carlo_estimation.py","lineNumber":47,"sourceCode":"    The estimate is generated by Monte Carlo simulations. Let U be uniformly drawn from\n    the unit square [0, 1) x [0, 1). The probability that U lies in the unit circle is:\n\n        P[U in unit circle] = 1/4 PI\n\n    and therefore\n\n        PI = 4 * P[U in unit circle]\n\n    We can get an estimate of the probability P[U in unit circle].\n    See https://en.wikipedia.org/wiki/Empirical_probability by:\n\n        1. Draw a point uniformly from the unit square.\n        2. Repeat the first step n times and count the number of points in the unit\n            circle, which is called m.\n        3. An estimate of P[U in unit circle] is m/n\n    \"\"\"\n    if number_of_simulations < 1:\n        raise ValueError(\"At least one simulation is necessary to estimate PI.\")\n\n    number_in_unit_circle = 0\n    for _ in range(number_of_simulations):\n        random_point = Point.random_unit_square()\n\n        if random_point.is_in_unit_circle():\n            number_in_unit_circle += 1\n\n    return 4 * number_in_unit_circle / number_of_simulations\n\n\nif __name__ == \"__main__\":\n    # import doctest\n\n    # doctest.testmod()\n    from math import pi\n\n    prompt = \"Please enter the desired number of Monte Carlo simulations: \"","sourceCodeStart":29,"sourceCodeEnd":65,"githubUrl":"https://github.com/TheAlgorithms/Python/blob/f5988cc09713315817df6a7e327e258013a94440/maths/pi_monte_carlo_estimation.py#L29-L65","documentation":"estimate_pi() in maths/pi_monte_carlo_estimation.py estimates pi as 4 * (points in unit circle / total points) over number_of_simulations draws. If number_of_simulations < 1 it raises ValueError('At least one simulation is necessary to estimate PI.') because the estimator's ratio m/n is undefined for n = 0 and meaningless for negative n. This is a statistical pre-condition, not a performance knob: even 1 is statistically worthless but technically allowed.","triggerScenarios":"Calling estimate_pi(0), estimate_pi(-100), or passing a computed count (e.g. int(request.args['n']) defaulting to 0, or a variable that underflowed to 0) as number_of_simulations.","commonSituations":"Config/default values left at 0; a loop or formula producing 0 simulations for tiny inputs; CLI flag parsing that yields 0 when the flag is omitted.","solutions":["Pass at least 1; realistically pass a large count (e.g. 100_000+) since accuracy grows with sqrt(n).","If the count is user/config supplied, clamp or validate it (max(1, n) or explicit error) before calling.","Guard callers that compute n dynamically so n = 0 fails loudly upstream with a clearer message."],"exampleFix":"# before\nestimate_pi(num_points)  # ValueError when num_points == 0\n\n# after\nif num_points < 1:\n    raise ValueError(f\"need >= 1 simulation, got {num_points}\")\nestimate_pi(num_points)","handlingStrategy":"validation","validationCode":"if number_of_simulations < 1:\n    raise ValueError('simulation count must be >= 1')\nestimate_pi(number_of_simulations)","typeGuard":"def is_valid_simulation_count(v) -> bool:\n    return isinstance(v, int) and v >= 1","tryCatchPattern":null,"preventionTips":["Treat simulation/iteration counts as validated config, not free-form.","Default counts to a large sane number (e.g. 100_000), never 0.","For Monte Carlo, remember accuracy scales with sqrt(n) — prefer big n."],"tags":["python","value-error","monte-carlo","maths"],"backgroundTag":null,"analyzedSha":"f5988cc09713315817df6a7e327e258013a94440","analyzedAt":"2026-08-14T17:30:07.041Z","schemaVersion":2},"datasetVersion":"2026-08-15T22:17:37.221Z"}