A geometric distribution models the number of trials you need to run until you get your first success in a sequence of independent attempts, each with the same probability of success. Picture flipping a coin until you hit heads: it’s a simple waiting game where the odds don’t change, and the question is “how long until I win?”

Let’s build it up. Start with a single trial—say, a coin flip with success probability - $p$ (e.g., $p = 0.5$ for heads). - Failure is $1 - p$, and each trial is independent, like rolling a die unaware of past rolls. - Now, imagine waiting for the first success: you might fail $k-1$ times (probability $(1-p)^{k-1}$), - Then succeed on the $k_{th}$ try (probability $p$).

The geometric distribution’s probability mass function is: $P(X = k) = (1-p)^{k-1} \cdot p$

where $k = 1, 2, 3, \ldots$. It’s a chain of failures capped by a win, derived from multiplying these basic probabilities together—exponential decay until the breakthrough.

Expectation

You're trying to get a new coupon you don't already have.

If you already have 1 out of $N$ coupons, then there are $N - 1$ new ones left.
The chance the next coupon is new = $\frac{N - 1}{N}$
So, the expected number of draws to get a new one is:

\[ \text{Expected tries} = \frac{1}{\text{Probability}} = \frac{N}{N - 1} \]

This is just how probability works: if success chance is $p$, then expected number of trials until success = $\frac{1}{p}$.

Let $X$ denote the number of tries till the first success. $\(P(X=n) = (1-p)^n \times p$\) Now the expectation can be written as: $$\sum^\infty n (1-p)^n p $$ The sum of a Arithmetic-Geometric progression is: $$ S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$$ Hence the expectation is: $\(E(X) = \frac{1}{p}$\)

Expectation Markov Processes

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Expectation