Geometric Distribution: How Long Until the First Success

A different question from “how many successes”

Most probability questions ask for the chance of a certain number of successes in a fixed number of tries. The geometric distribution turns this around. You keep trying until the first success arrives and ask how long that takes. The number of tries is not fixed in advance, it is the thing being measured.

This is the natural model for rolling until you get a particular number, attempting a boss until a rare item drops, or flipping until the first head. It applies whenever each attempt is independent and has the same probability of success, which we will call p.

The basic formula

For the first success to arrive on exactly try number n, you need n − 1 failures followed by one success. Because the tries are independent, the probabilities multiply:

P(first success on try n) = (1 − p)^(n − 1) × p

With a fair die and a target of rolling a six, p = 1/6. The chance the first six arrives on the third roll is (5/6) × (5/6) × (1/6) = 25/216, about 11.6 percent.

A more useful quantity is often the chance of no success in the first n tries, which is just (1 − p)^n. From it you can get the chance of at least one success in n tries by subtracting from 1.

The expected wait

The expected number of tries until the first success is 1 ÷ p. A success that occurs one time in 6 takes 6 rolls on average. A drop with a 10 percent chance takes 10 attempts on average. A 1 percent chance averages 100.

This is an average over many runs, in the sense of expected value, not a promise about any single run. The distribution is lopsided: many runs finish early, and a few run very long, which pulls the average up.

Why “average” and “typical” differ

Take a 1 percent chance per attempt, so p = 0.01 and the mean wait is 100 tries.

• After 100 tries, the chance of still having no success is 0.99^100, which is about 0.366, or 36.6 percent.
• After 69 tries the chance of no success is about 0.50. So the median wait is around 69 tries: half of all runs succeed by then.
• After 300 tries, the chance of no success is 0.99^300, about 0.049, or just under 5 percent.

So an average wait of 100 does not mean most players succeed within 100 tries. Roughly 63 percent do, and more than a third need longer. Some will need several hundred. Mean and median are different summaries of the same distribution, and the gap grows when p is small. The difference between the median outcome and the mean is a general theme in game statistics.

For the die example, p = 1/6: the chance of no six in 6 rolls is (5/6)^6 = 15,625/46,656, about 33.5 percent. A player expecting a six “within six rolls” misses about a third of the time.

The spread of waiting times

The standard deviation of the geometric distribution is √(1 − p) ÷ p. For p = 0.1 this is √0.9 ÷ 0.1, about 9.5 tries. That is nearly as large as the mean of 10. When the standard deviation is about the size of the mean, individual experiences vary enormously. Two players with identical odds can easily differ by a factor of three or more in how long they wait.

This variability is why short-run results rarely match the averages, and why outcomes feel much more unfair than they really are. If you plan around the average wait, expect to be disappointed often.

The memoryless property

The geometric distribution has a surprising and important feature: it has no memory. If you have already failed 50 times at a 1 percent chance, the expected additional wait is still 100 tries, not 50 and not zero. Each attempt is a fresh trial, and the past failures do not make a success any more likely.

Believing otherwise is the gamblers fallacy. The belief that a success is “due” confuses the long-run average with the next independent trial. Many video games deliberately break this rule. A pity timer raises the chance with each failure, so the wait is capped rather than memoryless, and pseudo-random distribution systems adjust the odds after each miss. If a game uses such a mechanism, the geometric model no longer applies and you need the game’s actual rules to work out the waiting time.

Applying it to drop rates

Suppose an item drops with a published rate of 2 percent per kill, and you assume the kills are independent. The mean wait is 1 ÷ 0.02 = 50 kills. The chance of no drop in 50 kills is 0.98^50. Since 0.98^50 is about 0.364, about 36 percent of players will have no drop after the “average” 50 kills.

Two cautions apply. First, a published rate may not mean what you assume; some games describe rates per session, per chest, or after adjustment for other factors. Second, if you are tracking your own results, a small number of attempts is a weak guide to the true rate, so avoid drawing firm conclusions from a short run. Reasoning about drop rates always begins with a clear statement of what one “try” is.

Summary of the useful formulas

These all assume independent tries with a constant probability, and they describe long-run behaviour. Any single run of attempts can be much luckier or unluckier than the numbers suggest.