Info
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:
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:
\(\(E(X) = \frac{1}{p}\)\)