Drop Rate Calculator

Frequently Asked Questions

What is the chance of at least one drop in N tries?

P(≥1) = 1 − (1 − p)^N. At a 2% drop over 50 kills: 1 − 0.98^50 ≈ 64%. It approaches but never reaches 100% with independent rolls.

How many attempts for a given confidence?

N = ln(1 − confidence) ÷ ln(1 − p). For 95% confidence at a 1% rate: ln(0.05) ÷ ln(0.99) ≈ 298 attempts. Rare drops have surprisingly long tails.

What does "expected" number of attempts mean?

Expected attempts = 1 ÷ p (a 1% drop averages 100 tries), but this is a mean, not a guarantee. Only about 63% of players get it by the expected count due to variance.

Why did I get nothing in way more than the drop rate suggests?

Each attempt is independent - past failures do not raise future odds (unless the game has bad-luck protection). Long dry streaks are statistically normal for low drop rates.

Provided by AllCalculators.io
Free online calculators for everyday. No registration required.

Estimates for informational purposes only.

Important Disclaimer: Estimates for informational purposes only.

This calculator provides estimates for informational purposes only. Results are based on assumptions and may not reflect actual outcomes. Consult qualified professionals in relevant fields before making important decisions based on these results.

JavaScript is required to use the interactive calculator above. The questions and answers below remain readable without JavaScript.

How This Calculator Works

A boss has a 2% chance to drop the mount. You have run it 50 times with nothing. Are you unlucky, or is 50 simply not enough runs? This calculator answers that with the at-least-one probability for repeated independent attempts. Each kill is a Bernoulli trial with success probability p, so the chance of failing every time across N attempts is (1 − p)^N, and the chance of at least one drop is the complement. It also reports the expected number of attempts for the first drop (1 ÷ p) and inverts the equation to tell you how many runs you need to be, say, 95% confident - the realistic farming budget rather than the wishful one.

Formula / Method

P(at least one in N) = 1 − (1 − p)^N. Expected attempts to first drop = 1 ÷ p (the mean of a geometric distribution). Attempts for confidence c = ⌈ln(1 − c) ÷ ln(1 − p)⌉, found by solving 1 − (1 − p)^N ≥ c. Expected number of drops in N attempts = N × p. These assume every attempt is independent with a fixed rate, which holds for most loot tables but not for systems with bad-luck protection or guaranteed pity. The rate is entered as a percentage and must be above 0 and at most 100.

Worked Example

Rate 2% (p = 0.02), 50 attempts. Chance of nothing = 0.98⁵⁰ ≈ 0.364, so P(at least one) ≈ 1 − 0.364 ≈ 63.6% - meaning a 36% chance of an empty 50-run streak is completely normal, not a bug. The expected number of runs for the first drop is 1 ÷ 0.02 = 50, so you are right at the average. To be 95% confident you would need ⌈ln(0.05) ÷ ln(0.98)⌉ = ⌈148.3⌉ = 149 runs. Over 50 runs you expect 50 × 0.02 = 1 drop on average, but "on average one" still leaves a real chance of zero.

Common Mistakes

Multiplying rate by attempts (2% × 50 = 100%) - that wrongly implies a guarantee and exceeds 100% beyond 50 runs.

Believing you are "due" after a dry streak - independent drops have no memory unless the game adds explicit pity.

Treating the expected value as a cap - it is the mean; roughly a third of players need more than 1 ÷ p attempts.

Using a shared-table rate as the specific-item rate when several rares roll on the same roll.

Ignoring per-run modifiers (magic find, difficulty, group size) that change p between attempts.

Tips & FAQ

How many runs should I plan for? Budget the attempts-for-confidence figure at a confidence you can accept (90–95% is common) rather than the expected value, which falls short more than a third of the time; planning to the mean is the single most common reason farmers feel "robbed" by a drop table that is behaving exactly as designed. Am I really unlucky? Compare your run count to the P(none) figure: if there is still a sizeable chance of zero at your count, you are within normal variance, not cursed - for a 1% drop you need roughly 230 attempts before P(none) falls under 10%, so a hundred empty runs is unremarkable. Does a longer dry streak improve future odds? Not for pure independent drops - the coin has no memory, so each run is the same flat chance; only bad-luck protection or a pity ceiling changes that, in which case your real odds are better than this model says late in the streak. What about magic find or difficulty bonuses? Enter the effective boosted rate; because the math is non-linear, even a small absolute increase to a very low rate can sharply cut the attempts needed for high confidence - doubling a 0.5% rate to 1% roughly halves the runs to 95% confidence. Should I farm a higher-rate alternative? Often yes - if a different source has triple the rate, the attempts-for-confidence drop by about the same factor, so compare the total time per attempt across sources, not just the headline percentage.

Frequently Asked Questions

Related Gaming Calculators