Geometric & Negative Binomial

The number of trials until the first success, and its generalization to the $r$-th success.

1. The idea

A Geometric $(p)$ random variable counts the number of independent Bernoulli $(p)$ trials up to and including the first success: $P (X = k) = (1 - p)^{k - 1} p$ for $k = 1, 2, \dots$

Its defining property is memorylessness: given that the first $m$ trials failed, the remaining number of trials needed is still Geometric $(p)$ , independent of $m$ . This makes conditioning arguments collapse to restarting the clock.

Negative Binomial $(r, p)$ generalizes this to the number of trials until the $r$ -th success; it is the sum of $r$ independent Geometric $(p)$ variables, so its mean and variance scale by $r$ .

2. Why it matters

Memorylessness is one of the most reused tricks in interviews: any "waiting for the first occurrence" question, once you show it's Geometric, can be answered without ever writing an infinite sum.

3. Core mathematics

Parameters

p \in (0, 1]

Support

k \in {1, 2, 3, \dots}

trials up to and including the first success

Expectation

E [X] = \frac{1}{p}

Variance

Var (X) = \frac{1 - p}{p ^{2}}

Key property

P (X > m + n ∣ X > m) = P (X > n)

memorylessness

Negative Binomial

E [X] = \frac{r}{p}, Var (X) = \frac{r ( 1 - p )}{p ^{2}}

sum of $r$ iid Geometric $(p)$

4. How to recognize it

Think of this when…

"Number of trials/flips/attempts until the first success"
"Expected number of rolls to see a 6" or similar first-occurrence questions
"Until the $r$ -th success" signals Negative Binomial
Any conditioning on "given it hasn't happened yet" — a memorylessness cue

5. Worked example

Worked example

A die is rolled repeatedly until a 6 appears. Given that the first 6 has not appeared in the first 4 rolls, find the expected number of additional rolls needed.

1
Let $X$ be the roll number of the first 6; $X \sim Geometric (1/6)$ .
2
By memorylessness, conditioning on $X > 4$ resets the count: the additional number of rolls needed is again Geometric $(1/6)$ .
3
So the expected additional rolls is just $E [X]$ itself.
$E [X] = \frac{1}{1/6} = 6$
Answer: $6$ additional rolls are expected, regardless of how many rolls have already failed.

6. Common mistakes

Writing the PMF as $(1 - p)^{k} p$ instead of $(1 - p)^{k - 1} p$ (off-by-one in the exponent)
Confusing the two common conventions: number of trials until success (support starts at 1) vs. number of failures before success (support starts at 0, mean $(1 - p) / p$ )
Forgetting to invoke memorylessness and instead re-deriving a conditional distribution from scratch
Applying memorylessness to a non-Geometric/non-Exponential variable, where it does not hold

7. Quick check

A biased coin has $P (heads) = 0.2$ . What is the expected number of flips until the first heads?

Which property lets you say that a Geometric variable's remaining wait, given $10$ failures so far, has the same distribution as the original?

8. Practice

20 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Memorylessness Conditioning on the First Event

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