How is the at-least-once probability calculated?

The chance of failing every run is (1 − p) raised to the number of runs N, so the chance of at least one drop is 1 − (1 − p)^N. This is the correct cumulative formula for independent attempts.

Why isn't 100 runs at a 1 percent rate a guaranteed drop?

Each run is independent, so luck never carries over. At 1 percent over 100 runs the at-least-once odds are 1 − 0.99^100 ≈ 63 percent, not 100 percent — a common and counter-intuitive result.

What is the expected number of drops?

Expected drops equal the drop rate times the number of runs. At 5 percent over 40 runs you expect 2 drops on average, though any single farm session can return zero or several.

How many runs do I need to be 90 percent sure?

Solve 1 − (1 − p)^N ≥ 0.90 for N, which gives N ≥ ln(0.10) / ln(1 − p). The tool computes this for 50, 90, and 99 percent confidence so you can plan a realistic farming target.

Does this work for bonus or guaranteed drops?

It models pure random independent drops. Guaranteed or pity-based drops do not follow this formula; for those use a pity calculator instead, since their odds change as your counter rises.

Final Fantasy XIV Drop Rate & Probability Calculator

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A 5 percent drop does not mean you are owed the item after 20 runs — random drops in Final Fantasy XIV reset every attempt. This calculator gives you the honest cumulative odds: the chance of at least one drop over your planned runs, the expected count, and how many runs you need to be reasonably confident.

How it works

Each run is an independent trial. The probability of getting nothing in N runs is the failure chance compounded, so the at-least-once probability is its complement:

P(at least one) = 1 − (1 − p)^N
expected drops  = p × N
runs for C%     = ln(1 − C) / ln(1 − p)

The last formula inverts the cumulative odds to tell you how many runs reach a chosen confidence level C (for example 0.90 for 90 percent).

Example and tips

At a 5 percent drop rate over 40 runs, the at-least-once chance is 1 − 0.95^40 ≈ 87 percent, and you expect 0.05 × 40 = 2 drops on average. To be 90 percent sure of at least one, you need about ln(0.10) / ln(0.95) ≈ 45 runs. Remember the gambler’s-fallacy trap: a long dry streak does not raise your next-run odds, because each attempt is independent.