Work out how unlikely your dry streak is at any drop rate, and how many more attempts a 50%, 90%, or 99% chance actually takes.
Change the inputs that matter to your decision. Optional assumptions stay collapsed until you need them.
At 1 in 512 over 500 attempts, 37.62% of players would still be dry.
About 37.62% of players would still be without the drop after this many attempts. You are 1.0× the expected attempt count.
| Chance you would still be dry | 37.62%After 500 attempts |
|---|---|
| Chance you would have it | 62.38% |
| Expected attempts | 512The average, not a guarantee |
| Half of players by | 355Already past |
| Nine in ten by | 1,178678 more from here |
| Ninety-nine in a hundred by | 2,3561,856 more from here |
Every attempt is an independent trial at a constant rate, so this models a plain drop table. Games with a pity timer, a threshold, or bad-luck protection change the distribution and are not described here. The expected count is an average — the confidence figures are the more useful number when you are planning your own time.
Practical notes about the inputs, assumptions, and result.
No — it is the single most common place to be. At the expected attempt count only about 63% of players have the drop, so more than a third are exactly where you are. The median is well below the expected count, and the tail above it is long. Being at the rate with nothing is unremarkable.
That depends on what you mean by probably. The result gives three answers: the point half of players are done by, the point nine in ten are done by, and the point ninety-nine in a hundred are. For planning your own time the 90% figure is usually the honest one to work from.
No. Each attempt is independent, so the next roll has exactly the same chance as the first one you ever made. There is no accumulating pressure and nothing owed. The exception is a game with an explicit pity timer or threshold mechanic, where the rate genuinely rises — this model does not describe those.
Divide one by N. A 1 in 512 is 0.195%, a 1 in 128 is 0.781%, a 1 in 5,000 is 0.02%. You do not have to do it here though — the rate field accepts 1/512, 3/1000, 0.2%, and 0.002 directly, including the n-over-N form many rates are published in.
Switch to several and give the count. Getting three of a 1% drop takes about 300 attempts on average, but the confidence points spread much wider than for one — the variance compounds. The tool sums the exact distribution rather than multiplying the single-drop answer, which would understate it.
Yes. The mathematics is the same for any independent roll at a fixed rate — loot boxes, gacha pulls, card packs, raid drops. Enter your game's rate in whatever form it is published and the answer holds, provided there is no pity system layered on top.
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