The Geometric Distribution

The geometric distribution answers a very natural question: you keep trying the same thing over and over, each attempt succeeds with the same probability, so how many attempts until the first success? Flip a coin until it lands heads, send quotes until one fills, roll a die until you get a six. The count of tries is geometric, and remarkably, a single number, the per-try success probability, tells you everything about it.

Each attempt is an independent Bernoulli trial: it succeeds with probability $p$ and fails with probability $1 - p$. To have your first success land on try number $k$, you need the first $k-1$ tries to all fail and the $k$-th to succeed. Multiply those independent chances together:

$$P(X = k) = (1 - p)^{k-1}\, p, \qquad k = 1, 2, 3, \dots$$

Here $X$ is the trial on which the first success occurs, $p$ is the success probability of a single try, and $(1-p)^{k-1}$ is the chance of failing the first $k-1$ times in a row. The counts run $1, 2, 3, \dots$ with no upper bound: you might, in principle, wait forever.

The average wait is one-over-p

The mean and variance are worth memorising:

$$E[X] = \frac{1}{p}, \qquad \operatorname{Var}(X) = \frac{1 - p}{p^2} .$$

The mean has a beautifully simple reading: if something works one time in $p$, you wait $1/p$ tries on average. A one-in-six event (rolling a six) takes six rolls on average; a coin's heads takes two flips.

Memorylessness

The geometric distribution has a strange and useful property: it forgets the past. Suppose you've already failed ten times. Does that "load the dice" for a success soon? No. The chance you need at least $t$ more tries is exactly the same as it was at the very start:

$$P(X > s + t \mid X > s) = P(X > t) .$$

Coins and dice don't keep score. A run of bad luck doesn't make you "due." The geometric is the only discrete distribution with this no-memory property, which is exactly why it's the discrete twin of the The Exponential Distribution, the continuous distribution that forgets in the same way.

Worked example

You roll a fair die repeatedly until the first six. Here $p = 1/6$.

• Expected number of rolls: $E[X] = 1/p = 6$. On average it takes six rolls.
• Chance the first six lands on roll 3: you need two non-sixes then a six,

$$P(X = 3) = \left(\tfrac{5}{6}\right)^2 \left(\tfrac{1}{6}\right) = \frac{25}{36}\cdot\frac{1}{6} = \frac{25}{216} \approx 0.116 .$$

• Chance it takes more than 6 rolls: all of the first six must miss,

$$P(X > 6) = \left(\tfrac{5}{6}\right)^6 \approx 0.335 .$$

So even though six rolls is the average, you'll still be waiting past six rolls about a third of the time, that fat right tail again.

A note on conventions

Two definitions float around and they differ by one. This page counts the trial of the first success (support $1, 2, 3, \dots$, mean $1/p$). The other convention counts the number of failures before the first success (support $0, 1, 2, \dots$, mean $(1-p)/p$). They describe the same experiment; just check which one a textbook or a library function means before plugging numbers in.

The geometric is the $r = 1$ special case of the The Negative Binomial Distribution (wait for one success instead of $r$), sits among the Common Distributions every quant should know, and its one-over-p mean is a direct application of Expected Value.