Problem 26 - A First Course in Probability

Question

One of the numbers 1 through 10 is randomly chosen. You are to try to guess the number chosen by asking questions with "yes-no" answers. Compute the expected number of questions you will need to ask in each of the following two cases: (a) Your \(i\) th question is to be "Is it i?" \(i=\) 1,2,3,4,5,6,7,8,9,10 (b) With each question you try to eliminate one-half of the remaining numbers, as nearly as possible.

Step-by-Step Solution

Answer

" for each number between 1 and 10, we expect to need an average of \(5.5\) questions to guess the randomly chosen number. In case (b), where we aim to eliminate half of the remaining numbers with each question, we expect to need an average of \(1.75\) questions to guess the number.

Step 1: Case (a) - Asking "Is it i?" for each i between 1 and 10

In this case, the probability of needing to ask exactly k questions before finding the number is the same for all k values ranging from 1 to 10. Each guess has a probability of being correct of \(\frac{1}{10}\) and a probability of being wrong of \(\frac{9}{10}\). Since we only have 10 numbers to guess, the probability distribution for the number of questions needed can be represented as: \( P(X=k) = \frac{1}{10}, \hspace{5mm} k = 1,2,3, \dots, 10.\) The expected number of questions needed can be calculated as: \(E(X) = \sum_{k=1}^{10} k \cdot P(X=k) = \sum_{k=1}^{10} k \cdot \frac{1}{10}\).

Step 2: Calculation of Case (a)

Calculate the expected number of questions using the formula derived earlier: \(E(X) = \sum_{k=1}^{10} k \cdot \frac{1}{10} = 5.5.\) So in case (a), we expect to need an average of 5.5 questions to guess the number.

Step 3: Case (b) - Eliminating half of the remaining numbers with each question

In this case, the aim is to minimize the number of questions needed on average, and the best approach is using a binary search-like method. With each question, we eliminate half of the remaining possibilities. Initially, we have 10 possible numbers. After the first question, we will have 5 remaining possibilities; after the second question, we'll have 3 remaining possibilities (we cannot split the possibilities perfectly in half); after the third question, only one possibility remains (half of 3, rounded up).

Step 4: Calculation of Case (b)

Now we can calculate the probability of needing each number of questions in case (b) as follows: - 1 question needed: \(\frac{10}{2} = 5\) possibilities eliminated, probability \(\frac{5}{10}\). - 2 questions needed: \(\frac{10}{2} - \frac{10}{4} = 2.5\) possibilities eliminated, probability \(\frac{2.5}{10}\). - 3 questions needed: \(\frac{10}{4} = 2.5\) possibilities eliminated, probability \(\frac{2.5}{10}\). Using these probabilities, we can now calculate the expected number of questions needed in case (b): \(E(X) = 1 \cdot \frac{5}{10} + 2 \cdot \frac{2.5}{10} + 3 \cdot \frac{2.5}{10} = 1.75.\) So in case (b), we expect to need an average of 1.75 questions to guess the number. In conclusion, the expected number of questions needed to guess the randomly chosen number between 1 and 10 in case (a) is 5.5 and in case (b) is 1.75.

Key Concepts

Probability Distribution

When dealing with probability distributions, we are essentially working with a mapping that associates each possible outcome of a random experiment with its probability of occurrence. In our context, if we consider case (a) of the exercise where we ask "Is it i?" for each number from 1 to 10, we have a very straightforward probability distribution.

Here, each number from 1 to 10 has an equal chance of being the correct choice. This means that the probability of guessing the correct number on any single try is consistently \(P(X=k) = \frac{1}{10}\) for all \(k\) values from 1 to 10.

To calculate the expected number of questions, we are utilizing the concept of expected value in probability, which is like finding the 'mean' of a distribution. We sum the products of each outcome's value and its probability, which in this simple setup, boils down to the arithmetic mean of the first 10 integers:\[E(X) = \sum_{k=1}^{10} k \cdot \frac{1}{10} = 5.5\]

In essence, probability distribution helps us to quantitatively describe the randomness in this guessing game.

Binary Search

Binary search is a classic algorithmic technique, primarily used in computer science to efficiently locate an element within a sorted array by dividing the search interval in half with each step.

In the context of our exercise, binary search is similar to the method used in case (b), where each question aims to eliminate about half of the remaining possibilities. This approach employs the strategy of dividing the problem repeatedly until it’s reduced to a simple decision.

With 10 numbers, we start by asking a question that divides these numbers roughly in half. If the number of possibilities cannot be perfectly split, we choose one section slightly larger. Through each question, the number of potential choices is gradually reduced:

• Begin with 10 numbers; divide into groups of 5.
• After 1st question, reduce to 5.
• After 2nd question, further reduce to 3.
• After 3rd question, zero in on 1 remaining possibility.

This binary search-like method greatly enhances efficiency. Thus, it results in an average need of roughly 1.75 questions, much fewer compared to direct guessing, as each question maximally informs us about where exactly the correct option lies.

Probability Theory

Probability theory is the field of mathematics that deals with the analysis of random phenomena. It provides the foundational language to adequately describe random events and their likelihoods, essential for understanding problems involving chance, such as guessing game exercises.

At the heart of probability theory is the notion of outcomes and events, linked with their respective probabilities. For any random experiment, probability is the measure of the chance that one outcome will occur over another.

In the exercise, probability theory allows us to calculate expected values or average outcomes by accounting for each potential scenario—like guessing a number in case (a) or systematically eliminating possibilities in case (b). Using the laws and principles embedded in probability theory, we quantify uncertainty and make more informed predictions about random events, employing formulas like:\[E(X) = \sum_{i} x_i \cdot P(x_i)\]

Utilizing probability theory isn't just about calculating; it helps identify the most efficient approaches to solve problems, such as choosing between random guessing versus using a systematic search strategy like binary search.