4.3 Geometric Distributions

Consider the board game Monopoly. In order to get out of jail a player must roll doubles. What is the probability of getting doubles for the first time on your fourth roll of the dice?

First let’s try to define the problem in terms of a probability experiment:

Let X be the number of rolls before rolling doubles.

Let p be the probability of rolling doubles on any particular roll (success)

Let q be the probability of not rolling doubles on any particular roll (failure)

Then p = 6/36 = 1/6

and q = 1-p = 5/6.

In terms of successes and failures, getting doubles on the fourth try looks like:

F, F, F and S

Since the trials are independent we the probability of this event is:

P(X=4) = q³p

= (5/6)³(1/6)

= 125 / 1296

= 0.096

So the probability of getting doubles for the first time on your fourth roll is 9.6 %

Defn: Consider an experiment involving successive independent trials where the outcomes can be measured as success/failure. Let X be the number of trials before success occurs. Then X follows a geometric distribution.

The Monopoly scenario above is example of a geometric distribution.

Note the key features in identifying a geometric distribution are:

• the idea of waiting for something to happen

• independent trials

Probability in a Geometric Distribution

Let’s now generalize this process. Let the random variable X represent the number of trials before success occurs. Let p be the probability of success and q the probability of failure. Then the probability of having x trials before success is

Expected Value for a Geometric Distribution

The expected value of a random variable X that follows a geometric distribution is given by:

Example: If 2% of the transactions at a bank are applications for loans, what is the probability that the first loan request on a given day occurs with the 10^th customer? How many customers should we expect to see before we see one with a loan request?

Solution: Let X represent the number customers before a customer with a loan request.

p = 0.02, q= 0.98 and the quantity we’re looking for is: P(X=9)

P(X=9) = q⁹p

= (0.98)⁹(0.02)

= (0.0166)

So the probability of the first loan request being from the 10^th customer is 1.66%

To calculate the expected value:

E[X] = q/p

= 0.98/0.02

= 49

So we would expect 49 customers before getting a loan request.

Homework: pg 394 #1-4, 7-13