Question

Random digits, each of which is equally likely to be any of the digits 0 through 9 , are observed in sequence. (a) Find the expected time until a run of 10 distinct values occurs. (b) Find the expected time until a run of 5 distinct values occurs.

Short answer

The expected time until a run of 10 distinct values occurs is approximately 29.29 steps, and the expected time until a run of 5 distinct values occurs is approximately 4.08 steps.

Step-by-step solution

(a) Find the expected time until a run of 10 distinct values occurs.

For this part, let's consider each digit as an equally likely event. Since there are 10 distinct digits (0 through 9), the probability of selecting any digit at random is \(1/10\). Let \(T\) be the expected time until a run of 10 distinct values occurs. We can define a recurrence relation for the expected time as follows: \(T = \frac{1}{10}(1+T) + \frac{9}{10}(1+T_9)\) Here, \(T\) and \(T_9\) represent the expected time until a run of 10 and 9 distinct values occurs, respectively. The term \(1/10\) corresponds to the probability of selecting a new unique digit, and the term \(9/10\) accounts for the probability of selecting a digit that has already appeared in the sequence. Now let's solve the recurrence relation recursively for \(T\): \(T = \frac{1}{10}(1 + T) + \frac{9}{10}(1 + \frac{1}{9}(1 + T) + \frac{8}{9}(1 + T_8))\) Continuing recursively until we have all the terms from \(T_1\) to \(T_9\), we will have: \(T=1+\frac{1}{10}(1+\frac{1}{9}(1+\frac{1}{8}(1+\frac{1}{7}(1+\frac{1}{6}(1+\frac{1}{5}(1+\frac{1}{4}(1+\frac{1}{3}(1+\frac{1}{2}(1+1)))))))))\) Now, we can sum all the terms to find the expected time: \(T=\sum_{k=10}^{2} (1+\frac{1}{k}) = 1+ 1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9}\) \(T ≈ 29.28968\) So the expected time until a run of 10 distinct values occurs is approximately 29.29 steps.

(b) Find the expected time until a run of 5 distinct values occurs.

Similar to part (a), let's define a recurrence relation for the expected time until a run of 5 distinct values occurs. Let \(S\) be the expected time and \(S_k\) be the expected time when there are already \(k\) distinct values in the run. \(S = \frac{1}{2}(1+S_4) + \frac{1}{2}(1+S)\) Now let's solve the recurrence relation recursively for \(S\): \(S = \frac{1}{2}(1 + S_4) + \frac{1}{2}(1 + \frac{1}{4}(1 + S_3) + \frac{3}{4}(1 + S))\) Continuing recursively until we have all the terms from \(S_1\) to \(S_4\), we will have: \(S=1+\frac{1}{4}(1+\frac{1}{3}(1+\frac{1}{2}(1+1)))\) Now, we can sum all the terms to find the expected time: \(S=\sum_{k=5}^{2} (1+\frac{1}{k}) = 1+ 1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\) \(S=4.0833\) So the expected time until a run of 5 distinct values occurs is approximately 4.08 steps.

Key concepts

Probability Models

When dealing with random events, such as the appearance of distinct digits, probability models come into play. These models are mathematical constructs that represent the various outcomes of a random process and their likelihood of occurrence.

In our problem, we’re looking at a uniform probability model, where each digit from 0 to 9 is equally likely to appear in a sequence. This equal probability is expressed numerically as \(1/10\) for each digit. Understanding this basic probability model is crucial for predicting the expected time until a sequence of distinct values occurs because it determines the likelihood of any given digit appearing at each step.

When calculating expected time, we make consistent use of this uniform model to establish the foundational probabilities that are then incorporated into more complex calculations such as those involving recurrence relations.

Recurrence Relation

Recurrence relations are equations that recursively define sequences: each term is defined as a function of its preceding terms.

For the expected time until a run of distinct digits appears, we create a recurrence relation to represent this expected time. In the context of our problem, the recurrence relation allows us to express the expected time to observe 10 distinct digits (\(T\)) as a combination of the expected times to observe fewer distinct digits (\(T_9\), \(T_8\), ..., \(T_1\)).

The relation includes terms that account for the event of picking a new digit and terms for picking a digit that was already seen. This method is powerful because it breaks down a complex problem into smaller, more manageable pieces, thereby simplifying the calculation of the expected time.

Run of Distinct Values

A run of distinct values refers to a consecutive sequence in which all elements are different. Understanding the concept of runs is key to solving problems involving sequences of random events.

In probability theory and applications, we are often interested in determining how long it takes for such a run to occur. The approach to these problems typically involves combinatorial methods and probability calculations.

In the given example, we calculate the expected time for runs of 10 and 5 distinct digits, which involves careful consideration of the conditions under which each additional distinct digit can appear in the sequence. By constructing and solving a recurrence relation for the runs, we can find the expected length of time (or steps) needed for such sequences to arise in a random process.