Gacha Probability Calculator

Frequently Asked Questions

What is the chance to get an item in N pulls?

Probability of at least one = 1 − (1 − p)^N, where p is the per-pull rate. At a 0.6% rate over 100 pulls: 1 − 0.994^100 ≈ 45% - not the intuitive 60%.

How many pulls does it take on average?

Expected pulls = 1 ÷ p. A 1% banner averages 100 pulls, but "average" hides huge variance - many players need far more or fewer, and 100 pulls is only ~63% likely to succeed.

How many pulls for a target confidence?

N = ln(1 − confidence) ÷ ln(1 − p). For 90% confidence at a 1% rate: ln(0.10) ÷ ln(0.99) ≈ 230 pulls. The tail is long because each pull is independent.

How does pity change the odds?

A pity (hard guarantee at X pulls) caps the worst case and often adds soft-pity rate boosts near the threshold, raising effective probability beyond the base rate as you approach the limit. This is for entertainment planning, not spending advice.

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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

Gacha and loot-box systems advertise a per-pull drop rate, but players really want to know "what are my odds across a whole batch of pulls?" That is a classic at-least-one probability problem. Each pull is an independent trial with success probability p, so the chance of never hitting it in N pulls is (1 − p)^N, and the chance of getting it at least once is the complement, 1 − (1 − p)^N. This tool also reports the expected number of pulls (1 ÷ p, the mean of a geometric distribution) and inverts the formula to tell you how many pulls you need to reach a chosen confidence level - the honest way to budget currency before you spend it.

Formula / Method

P(at least one in N) = 1 − (1 − p)^N. Expected pulls to first success = 1 ÷ p. Pulls needed for confidence c = ⌈ln(1 − c) ÷ ln(1 − p)⌉, derived by setting 1 − (1 − p)^N ≥ c and solving for N. An optional pity counter adds your already-accumulated pulls to N, since those failed attempts still count toward the same independent-trial pool (a true hard-pity guarantee is a separate ceiling this simple model does not enforce). The drop rate is entered as a percentage and must lie strictly above 0 and at most 100.

Worked Example

A banner has a 0.6% rate and you plan 90 pulls. p = 0.006, so the chance of missing every time is (1 − 0.006)⁹⁰ = 0.994⁹⁰ ≈ 0.582, meaning P(at least one) ≈ 1 − 0.582 ≈ 41.8%. The expected number of pulls for the first hit is 1 ÷ 0.006 ≈ 167 - far more than 90, which is why 90 pulls is less than a coin-flip. To be 90% confident you would need ⌈ln(0.10) ÷ ln(0.994)⌉ = ⌈382.6⌉ = 383 pulls. This gap between "advertised rate" and "pulls to be safe" is exactly why pity systems exist.

Common Mistakes

Multiplying the rate by the number of pulls (0.6% × 90 ≈ 54%) - that overcounts and breaks down entirely above 100 pulls.

Believing you are "due" after a long dry streak; independent pulls have no memory - only an explicit pity system changes that.

Treating the expected value (1 ÷ p) as a guarantee - it is the long-run average; many players need far more, some far fewer.

Ignoring rate-up vs off-banner splits: the headline rate may be shared across many items, lowering the chance of the specific one you want.

Forgetting that soft/hard pity makes real probabilities rise near the cap, so this independent model understates odds late in a pity cycle.

Tips & FAQ

How many pulls should I save? Decide a confidence you can live with - 90% is common among players who hate gambling on the mean - and budget the pulls-for-confidence figure rather than the expected value, which leaves you short more than a third of the time; the gap between the two is often several dozen pulls on a 0.6% banner, which is real money or weeks of free currency. Does pity make this wrong? Real games with soft pity raise the effective rate as you approach the cap, so your true odds at high counts are better than this independent model says; treat the calculator as a conservative floor and the hard-pity ceiling as your guaranteed worst case. Why did I hit it on pull 3? Variance: the geometric distribution has a long tail, so early jackpots and brutal dry streaks are both completely expected outcomes when you sum across many banners and many players. Should I do a 10-pull or singles? Mathematically the per-pull odds are identical - ten singles and one ten-pull have the same chance - so 10-pulls only matter if the game guarantees a minimum rarity in each batch, which this tool does not assume. What if the banner shares the rate across many characters? Enter the rate for the specific item you want, not the combined featured rate; a 50/50 split or a shared 5-star pool can halve or quarter the chance the formula reports, and ignoring that is the most expensive mistake new players make.

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