Loot Expected Value Calculator

Frequently Asked Questions

How many drops should I expect over N attempts?

Expected drops = N × p, where p is the per-attempt drop chance. At a 2% rate over 100 kills you average 100 × 0.02 = 2 drops - though variance means you might get 0 or 5.

How many attempts until the first drop?

On average 1 ÷ p attempts. A 1% drop takes about 100 tries on average, but this is a geometric distribution - many players get it far sooner or much later.

What are the odds of at least one drop in N tries?

P(at least one) = 1 − (1 − p)ᴺ. At 2% over 50 attempts that is 1 − 0.98⁵⁰ ≈ 64%. Notably this never reaches 100% - there is always a small chance of a dry streak.

What is "pity" and does it change the math?

A pity system guarantees a drop after a set number of failures, capping the worst-case streak. Pure random drops have no such floor, which is why this calculator's "at least k" odds matter for planning grinds.

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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.

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How This Calculator Works

Loot drops, gacha pulls, and rare-mob farming all run on the same probability engine: each attempt is an independent trial with a fixed chance p of success. This calculator answers the three questions every grinder asks. First, the expected number of drops over N attempts - on average, how many of the item you should walk away with. Second, the expected number of attempts to get your first drop - the "average grind length." Third, the probability of getting at least k drops in N attempts, so you can judge whether your bad luck is genuinely unusual or perfectly normal. Enter the drop rate as a percentage, the number of attempts you plan or have made, and how many copies you are hoping for, and the tool returns all four figures.

Formula / Method

Each attempt is a Bernoulli trial with success probability p. Expected drops = N × p by linearity of expectation. The number of attempts until the first success follows a geometric distribution whose mean is 1 ÷ p. The probability of at least one drop is the complement of zero drops: 1 − (1 − p)^N. For at least k drops we use the binomial distribution: P(exactly i in N) = C(N, i) × pⁱ × (1 − p)^N−i, and P(at least k) = 1 − Σ from i = 0 to k−1 of those terms. To stay numerically stable for large N, the binomial terms are computed in log-space and exponentiated. All probabilities are clamped to the valid 0–100% range.

Worked Example

A boss drops a mount at a 1% rate (p = 0.01) and you farm it 100 times. Expected drops = 100 × 0.01 = 1.0 - on average one mount, though average is not guarantee. Expected attempts for the first drop = 1 ÷ 0.01 = 100. Probability of at least one in 100 tries = 1 − (0.99)¹⁰⁰ ≈ 1 − 0.366 = 63.4% - meaning even after the "expected" 100 kills, more than a third of players still have nothing, which is exactly why drop-rate frustration is so common. Probability of at least two drops in 100 tries is about 26.4%, so doubling up is a genuine but uncommon stroke of luck.

Common Mistakes

Believing "the drop is due" after a long dry streak - independent trials have no memory; your odds reset every attempt.

Reading expected drops of 1.0 as a guarantee; it is an average, and you have only a ~63% chance of even one at the expected count.

Confusing "1 in 100" with "100 attempts guarantees it." You need roughly 230 attempts for a 90% chance and 460 for 99%.

Entering the drop rate as a decimal in the percent field (e.g. 0.01 instead of 1) - it expects a percentage.

Ignoring pity systems - some games add a guaranteed drop after a threshold, which breaks the pure independent-trial model this tool assumes.

Tips & FAQ

Why does the expected count feel like a lie? Because the distribution is skewed: at the expected number of attempts you have only about a 63% chance of success, since (1 − p)^1/p tends toward 1/e ≈ 0.368 as p shrinks. How many attempts for near-certainty? For a 90% chance, attempts ≈ ln(0.10) ÷ ln(1 − p); for a 1% drop that is about 230 tries, and roughly 460 for 99%. Do drops have memory? No - absent a pity system, each attempt is independent, so a long unlucky streak does not improve your next roll. What about pity or bad-luck protection? Many modern gacha games raise the rate or guarantee a drop after a set number of failures; this calculator models the pure fixed-rate case, so treat its numbers as a floor when a pity system exists. Why compute "at least k"? When you need multiple copies (duplicates to upgrade an item), the binomial "at least k" figure tells you how realistic that goal is within your planned attempts.

Loot Expected Value Calculator

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