Suppose you have probability p of succeeding on any one try. If you make
independent attempts over and over, then the geometric random variable, denoted by X ~ geo(p), counts the number of attempts needed to obtain the first success.

We often let q = 1 - p be the probability of failure on any one attempt. Then the **probability** of having the first success on the kth attempt, for k >= 1, is given by

(from k-1 failures followed by a success).

The **average** (or **mean**) number of attempts needed to succeed is given by E[X] = 1 / p.

The **variance** of the number of attempts needed is given by Var(X) = (1 - p) / p^2.

The most likely number of attempts needed, or **mode**, is given by the integer k such that P(X = k) is maximized. In this case, it will always be k = 1. Thus, if you are going to succeed, then you are most likely to succeed on the first try.

The cumulative distribution function (or cdf) is given by P(X <= k) = 1 - q^k. This function can be interpreted as the probability of succeeding within k attempts.

We can use the cdf for computing probabilities such as P(j <= X <= k), which is the probability that it will take from j attempts to k attempts to succeed. This value is given by

Lastly, by solving the equation P(X <= k) >= r, (and rounding up to the nearest integer) we can find the number of attempts needed to have at least probability r of success within this number of attempts. Solving 1 - q^k >= r, we obtain k >=
ln(r) / ln(q).

The **GEOMET** program can be used to compute probabilites such as P(j <= X <= k), P(X = k), and P(X <= k). To execute the program, we enter the value of p and the lower and upper bounds of j and k, for 1 <= j <= k. (Enter the same value k for both the lower and upper bound to compute a pdf value P(X = k).) The program also asks if you want a complete distribution to be entered into the **STAT Edit** screen. If so, then enter **1**. If not, then enter **0**. The program then displays P(j <= X <= k) along with the average number of attempts needed to succeed and the standard deviation.

If you enter **1**, then most of the distribution will be entered into the **STAT Edit** screen. Under **L1**, the possible number of attempts 1, . . . , n are listed. Under **L2**, the pdf values of P(X = k), for 1 <= k <= n, are listed. Under** L3**, the cdf values P(X <= k), for 1 <= k <= n, are listed. The upper bound n is chosen so that P(X <= n) exceeds 0.975.

Click here for info on the TI-86 and TI-89 Stat Edit displays.

** Example.** Suppose one die is rolled over and over until a Two is rolled. What is the probability that it takes

(a) from 3 to 6 rolls?

(b) exactly 4 rolls?

(c) at most 5 rolls?

(d) How many rolls are needed so that there is at least 0.95 probability of rolling a Two within this number of rolls?

*Solution.*After calling up the **GEOMET** program, enter **1 / 6** for **PROBABILITY**, enter **3** for** LOWER BOUND**, and enter **6** for **UPPER BOUND**. Also enter **1** for a complete distribution.

(a) We find that P(3 <= X <= 6) = 0.359546, and that the average number of rolls needed is 6 with a standard deviation of 5.47723.

(b) and (c) From looking at the complete distribution in the list editor, we see that P(X = 4) = 0.09645 and P(X <= 5) = 0.59812.

(If you did not enter **1** for a complete distibution, then you can execute the program again with **4** for both **LOWER BOUND** and **UPPER BOUND** to find P(X = 4), and you can reexecute the program with bounds of 1 and 5 to find P(X <= 5).

(d) The number of rolls needed must satisfy k >= ln(.05) / ln(5/6) = 16.431; thus 17 rolls are needed to have a 95% chance of rolling a Two.

1. If two dice are rolled until a Double Six is rolled, what is the probability
that it will take (a) at most 24 rolls (b) at least 10 rolls?

2. When dealing 5 cards from a shuffled deck, the probability of dealing a
hand with exactly one pair is about 0.42257. If you want the probability of dealing a pair within n independent tries to be at least 0.99, then how many deals n should be made?

1. Here, X ~ geo(1 / 36). Since there must always be at least one attempt, we first wish to find P(1 <= X <= 24). (a) In the **GEOMET** program, enter **1 / 36** for **PROBABILITY**, enter **1** for **LOWER BOUND**, and enter **24** for **UPPER BOUND**. We see that P(1 <= X <= 24) = 0.4914. Alternately, P(X <= 24) = 1 - (35/36)^24.

(b) P(X >= 10) = 1 - P(1<= X <= 9) = 1 - (1 - (35/36)^9) = (35/36)^9 = 0.77605 .

2. Here X ~ geo(0.42257), and we wish to find k so that P(X <= k) >= 0.99. Thus, we must solve for k in the inequality 1 - q^k >= 0.99. We obtain, k >= ln(0.01) / ln(q) = ln(.01) / ln(.57743) = 8.3857. Thus, k = 9 deals are needed.

Return to Table of Contents.