The simple version: one divided by the chance
Suppose an item drops with a fixed 1 percent chance on each attempt, and every attempt is independent. The number of attempts until the first success follows what statisticians call a geometric distribution. Its average is easy to compute: divide 1 by the chance. Here that is 1 / 0.01 = 100 attempts.
That figure is useful as a budgeting tool, but it is easy to misread. It is an average over many players, not a prediction for you, and not a point after which the drop is “due”.
Why the average is not the typical result
The probability of failing every one of n attempts at a 1 percent rate is 0.99 raised to the power n. So the chance of at least one success in n attempts is 1 minus that figure. A few useful points:
| Attempts | Chance of at least one drop |
|---|---|
| 69 | about 50 percent |
| 100 | about 63 percent |
| 230 | about 90 percent |
| 299 | about 95 percent |
Two things stand out. First, after exactly 100 attempts, the “expected” number, about 37 percent of players still have nothing. Second, the median is lower than the mean: half of players finish within roughly 69 attempts, but a minority take far longer, and those long runs pull the average up to 100.
That asymmetry is the signature of the geometric distribution. It has a hard floor (you cannot succeed in fewer than one attempt) and a long right tail with no upper limit.
How wide is the spread?
The standard deviation of a geometric distribution with success chance p is the square root of (1 - p) divided by p. For p = 0.01 this is about 99.5, nearly the same size as the mean. A spread that large means results vary enormously from player to player.
One consequence is that being “unlucky” is not rare. The chance of needing more than 200 attempts is 0.99 to the power 200, about 13 percent, or roughly one player in eight. The chance of needing more than 300 is about 5 percent. If a thousand players each attempt the same farm, around 50 of them will reach 300 failures, with no flaw in the system at all.
The same applies to luck on the good side. Roughly 1 player in 100 will get the item on the very first attempt, which is the same 1 percent as any other attempt.
Planning with percentiles instead of averages
If you want to plan how long a farm will take, a percentile is often more helpful than a mean. To find the number of attempts for a target confidence, solve 1 - 0.99^n = target:
- For 50 percent: n is the natural log of 0.5 divided by the natural log of 0.99, about 69.
- For 90 percent: the same with 0.1, about 229, which rounds up to 230.
- For 95 percent: about 298, rounded up to 299.
So a sensible plan for a 1 percent drop is “about 70 attempts if things go averagely well, a couple of hundred to be fairly safe, and be prepared for more”. Even at 95 percent confidence, there is a one in twenty chance of exceeding the estimate.
Scaling to other rates
The same method works for any fixed rate. Halving the chance doubles the average: a 0.5 percent drop averages 200 attempts. Doubling it halves the average: 2 percent averages 50. The percentile figures scale in roughly the same proportion, so the 90 percent mark for a 2 percent drop is about 114 attempts, and for 0.5 percent it is about 459. These are approximations from the formula, and the exact numbers shift slightly with rounding.
It also works when several steps stack. If an enemy has a 20 percent chance to drop a pouch and the pouch has a 5 percent chance of containing your item, the chance per kill is 1 percent, and the numbers above apply per kill, not per pouch.
Timing: attempts versus hours
An attempt count only becomes useful when you attach a time cost. If one run takes four minutes, 100 attempts is about 400 minutes, or roughly 6.7 hours, while the 90 percent figure of 230 attempts is about 15 hours. Stating the range, rather than a single figure, gives a more honest picture of what you are signing up for. It also helps with deciding whether the time is worth it, a judgment that depends on what else you could do with it. Weighing time against reward in this way is a simple case of expected value. Keep in mind too that your own few runs are a tiny sample size, so a result far from the average says little about whether the stated rate is wrong.
When the formula does not apply
The one-over-p rule assumes the chance stays fixed. Many games add protection: a guaranteed reward after a set number of misses, or a chance that rises after each failure. Those systems shorten the tail, and their effect on cost is covered in the articles on pity timers and pseudo random distribution. Other situations that break the assumption include:
- items that can only drop once per account or per week
- per-player shared loot where you are competing for a single drop
- stages of upgrading where each step has its own chance
In each case you would model the process differently, though the geometric distribution often remains a building block.
A caution about “due” drops
Because the average is 100, it is tempting to feel that attempt 101 is more likely to pay out. In a fixed-rate system it is not: every attempt carries the same 1 percent, whatever has happened before. This is a classic error, discussed further under the gambler fallacy. What the geometric distribution gives you is a way to measure the spread before you start, so a dry streak is less of a surprise when it arrives. Actual outcomes remain uncertain.