Geometric Distributions

A geometric distribution is a type of probability distribution that tells us how many failures happen before the first duccess in a series of independent trials. Each trial has only two possible outcomes: success or failure, and the probability of success stays the same for each trial.

Here are few keys things to remember when working with geometric distributions:

It counts the number of failures before the first success
Each trial is independent, meaning the outcome of one does not affect the next
The probability of success is constant in each trial

Probability in a Geometric Distribution

The formula for the geometric distribution comes from the way we model the number of trials, which is the Failures that are needed before a success occurs:

\(P(x) = q^{x} \cdot p \)

Where:

\(p\) is the probability of success in each single Trial
\(q\) is the probability of failure
\(x\) is the number of failures before the first success

Expectation for a Geometric Distribution

The expectation of a geometric distribution is the sum of an infinite series. In a geometric distribution, the number of trials until the first success can theoretically extend infinitely, meaning each additional trial adds more possible outcomes to the Failures before the first success occurs.

The expected number of failures is given by:

\(E(X) = \sum_{x=0}^{\infty} x \cdot P(x)\)

\(E(X) = \dfrac{1}{p}\)

Example

A player is rolling a fair 8-sided die, where each face has an equal Probability of appearing. The player continues rolling until they land on a 7 for the first time. What is the Probability that the player rolls the die and does not roll a 7 exactly four times before finally rolling a 7?

First, we can determine the probability of rolling a 7 as such:

\(p = \dfrac{1}{8}\)

This is due to the fact that a die has 8 sides and only one of them is a 7.

Next, we can determine the probablity of not rolling a 7:

\(q = 1 - p\)

\(q = 1 - \cfrac{1}{8}\)

\(q = \cfrac{7}{8} \)

Then, we can use the formula for probability of geometric distribution to determine the probability of not rolling a 7 exactly 4 times before rolling it:

\(P(x) = q^{x} \cdot p\)

Since we need to roll the dice 4 times, we can determine \(x = 4\):

\(P(4) = \left(\cfrac{7}{8} \right)^4 \times \dfrac{1}{8}\)

\(P(4) = \cfrac{2401}{4096} \times \dfrac{1}{8}\)

\(P(4) = \cfrac{2401}{32768} \approx 0.0733 \)

Therefore, the Probability of having exactly 4 failures before the first success is approximately 0.0733.

You are rolling a fair 6-sided die. Let’s define a success as rolling a 6. What is the expected number of rolls required until you roll a 6 for the first time?

Show Answer

First, we can determine the probability of rolling a 6, \(p\), as such:

\(p = \dfrac{1}{6} \)

This is due to the fact that the die has 6 sides and only one of them is a 6.

Next, we can use the formula for Expectation for Geometric Distribution to determine the expected number of rolls until we roll a 6:

\(E(X) = \cfrac{1}{p}\)

\(E(X)= \cfrac{1}{\dfrac{1}{6}}\)

\(E(X) = 6\)

Therefore, the expected number of rolls required until the first 6 is 6.