A tempting wrong answer
You roll a die six times. Each roll has a 1/6 chance of showing a six. A natural thought is that six tries at 1/6 each should add up to 6/6, so a six is certain. That cannot be right, since it is quite possible to roll six times and see no sixes at all.
The mistake is adding probabilities of events that can overlap. Two sixes in six rolls is a possible result, and adding counts that outcome twice. There is a cleaner way to handle “at least one”, and it works almost every time.
The complement trick
“At least one success” is the opposite of “no successes at all”. Those two outcomes cover every possibility between them, so their probabilities add to 1. That gives:
P(at least one) = 1 − P(none)
The probability of none is usually easy: for independent tries, multiply the chance of failing on each one. This is the multiplication rule applied to failures.
Example: dice
The chance of at least one six in six rolls:
- Chance of no six on one roll: 5/6.
- Chance of no six on all six rolls: (5/6)^6 = 15,625/46,656, about 0.335.
- Chance of at least one six: 1 − 0.335 = 0.665, or roughly 66.5 percent.
So rolling six times gives about a two-in-three chance of seeing a six, not a certainty. For four rolls, the answer is 1 − (5/6)^4 = 1 − 625/1,296 = 671/1,296, about 51.8 percent. Four tries are enough to make it slightly better than even.
Example: a drop chance
An item drops with a 5 percent chance on each run, and you do 20 independent runs. The chance of no drop in a single run is 0.95. After 20 runs:
0.95^20 is about 0.3585, so the chance of at least one drop is 1 − 0.3585 = 0.6415, roughly 64 percent.
Note what this says. Twenty runs, which is the “expected” number of attempts for a 1-in-20 event, gives only about a 64 percent chance of at least one success. More than a third of players would still be empty-handed. The expected wait is a different measure from the chance of success within that many tries, and the gap is worth knowing about when planning how many attempts something will take. The same figure describes the chance of success under any published drop rates, as long as the tries are independent.
Example: cards, without replacement
Suppose you are dealt five cards from a standard deck. What is the chance of holding at least one ace?
The complement is a hand with no aces, drawn entirely from the 48 non-aces. Count the hands:
- Hands with no aces: 48 choose 5 = 1,712,304.
- All hands: 52 choose 5 = 2,598,960.
P(no ace) = 1,712,304 ÷ 2,598,960, about 0.6588. So P(at least one ace) is about 1 − 0.6588 = 0.3412, or roughly 34.1 percent.
You can reach the same answer by multiplying the chances for each card in turn: (48/52) × (47/51) × (46/50) × (45/49) × (44/48). This product also equals about 0.6588. Because draws without replacement are dependent, each factor has a different numerator and denominator. This is the setting for the hypergeometric distribution.
A formula you can reuse
For n independent tries each with success chance p:
P(at least one success) = 1 − (1 − p)^n
This formula is the same quantity that appears in waiting-time problems: it is the chance that the first success arrives within n tries.
A rough rule can help for small p. If np is small, then 1 − (1 − p)^n is a bit less than np. For example, 10 tries at 1 percent gives about 9.6 percent (1 − 0.99^10 ≈ 0.0956), slightly under the 10 percent you would get by adding. When np reaches 1 the shortfall is large: 100 tries at 1 percent gives about 63.4 percent, not 100.
Solving for the number of tries
You can also run the formula backwards. To find how many tries you need for at least a 50 percent chance of one success with p = 0.01, solve 0.99^n ≤ 0.5. Trying values shows that n = 69 gives about 0.4999, so around 69 tries are needed. For a 90 percent chance, you need 0.99^n ≤ 0.1, which takes about 230 tries. Check: 0.99^229 is about 0.100, so 230 tries just clears the bar.
These counts only describe the odds before you start. Any single run of tries can succeed on the first or take several hundred; the probability does not change as tries pass without a hit.
When to be careful
- Dependent events. If tries are not independent, as with cards or a pity timer that raises odds after failures, you cannot just raise (1 − p) to the power n. Use the actual conditional probabilities.
- Unequal chances. If each try has a different success chance, multiply the individual failure chances instead of using a single power.
- “Exactly one” is a different question. The complement trick gives “one or more”, not “exactly one”. For exactly one, you need to count the number of ways, as in the binomial approach.
- Overlapping goals. If you want at least one of several different things, such as any of three rare items, treat each carefully, since failure on one does not tell you about the others unless they are independent.
The trick turns many hard problems into a one-line subtraction. Whenever a question says “at least”, “one or more” or “any”, it is worth asking first whether it is easier to calculate the chance of none.