The question most players actually ask
When an item drops with some stated chance, the useful question is rarely “what is the chance on one attempt?” It is “what is the chance I get it at least once within n attempts?” Answering that takes one idea and one formula.
The setup assumes each attempt is independent and has the same chance p. Many games work this way, but not all do. Some use pity systems or other adjustments, which change the maths and are noted at the end.
The complement rule
Counting every way to succeed at least once is awkward: you could succeed on the first attempt, the second, the fifth, or several times. It is much easier to count the single case you want to exclude: failing every time. The complement rule says the chance of something happening is one minus the chance of it not happening.
For one attempt, the chance of failing is 1 − p. For n independent attempts, the chance of failing every time is (1 − p) multiplied by itself n times, written (1 − p)^n. So:
P(at least one drop in n attempts) = 1 − (1 − p)^n
Worked example: a 5% drop
Here p = 0.05, so the chance of failing one attempt is 0.95.
- 10 attempts: 0.95^10 is about 0.599, so the chance of at least one drop is about 1 − 0.599 = 0.401, roughly 40%.
- 20 attempts: 0.95^20 is about 0.358, so the chance is about 64%.
- 60 attempts: 0.95^60 is about 0.046, so the chance is about 95%.
Notice that 20 attempts, the number you might naively expect to be enough for a 1-in-20 chance, gives about 64%, not certainty. About 36% of players would still be waiting.
Worked example: a 1% drop
With p = 0.01, the chance of failing once is 0.99.
- 100 attempts: 0.99^100 is about 0.366, so the chance of at least one drop is about 63%.
This is a common pattern. For a 1-in-N event, N attempts give about 63% for large N, so an item with a 1-in-100 chance still leaves roughly a third of players empty-handed after 100 tries. This is the same pattern that makes low-probability events feel so unfair.
Average wait versus typical wait
Two different questions are often mixed up.
The average number of attempts for a drop with chance p is 1/p. For 5%, that is 20 attempts. This is an average over many players, so it is a useful planning figure for a population, not a prediction for you.
The median is the number of attempts by which half of players have succeeded. For 5%, the chance of at least one drop reaches 50% a little before the average. After 13 attempts it is about 48.7% (0.95^13 is about 0.513) and after 14 attempts it is about 51.2% (0.95^14 is about 0.488). So the median is 14 attempts, still below the 20 you might expect.
The distribution is skewed: many players succeed early, and a smaller group waits much longer, pulling the average up.
What the numbers cannot tell you
A few misunderstandings come up repeatedly.
- Failing many times does not make success more likely. If 30 attempts have failed, the next attempt still has chance p. The 95% figure above describes the chance before you start, not after a run of misses. Thinking otherwise is the gambler’s fallacy.
- A drop rate is a long-run description. It tells you what happens across many players or many attempts, and small sample sizes can look far from it.
- A stated rate may be simplified. The true probability might be rounded, or depend on conditions you cannot see.
When attempts are not independent
If a game guarantees a drop after a fixed number of failures, the attempts are dependent events: each failure changes the next chance, so the formula above no longer applies directly. You would need to model the specific rule. A simple way to check such a rule when you cannot work it out by hand is a probability simulation, where you run the process many times and count outcomes.
Calculating several items at once
For a single attempt with different outcomes, such as a common, uncommon and rare drop, the stated chances should add to 100% in total across the table. For example, 70%, 25% and 5% sum to 100%, and 0.70 + 0.25 + 0.05 = 1.00. Within a table like that, outcomes are mutually exclusive, so adding is correct. Across separate attempts, they are independent, so multiplying is correct. Mixing the two up is one of the most common errors in adding probabilities wrongly.
A quick procedure
- Write the chance p of one attempt as a decimal.
- Compute 1 − p, the chance of failing once.
- Raise it to the power n for n attempts.
- Subtract the result from 1.
- Remember it describes likelihood, not a promise.
Whenever real money is attached to randomised rewards, the same maths applies but the stakes are different. Outcomes stay uncertain no matter how the calculation looks, so it is wise to decide spending limits in advance.