Question

The expected number of times that a fair coin will come up heads is defined as the sum over \(n=1,2, \ldots\) of \(n\) times the probability that the coin will come up heads exactly \(n\) times in a row, or \(n / 2^{n+1} .\) Compute the expected number of consecutive times that a fair coin will come up heads.

Short answer

The expected number of consecutive heads is 1.

Step-by-step solution

Understand the Formula

We need to calculate the expected value of a fair coin showing heads consecutively. The expression for this is given as \(\sum_{n=1}^{\infty} \frac{n}{2^{n+1}}\). This formula represents the expected number of heads occurring consecutively.

Recognize It as a Series

This sum is an infinite series where each term is \(\frac{n}{2^{n+1}}\). Recognize that this series sums infinitely from \(n = 1\) to \(\infty\). The challenge lies in simplifying and then calculating this series.

Simplify the Series

Notice that \(\frac{n}{2^{n+1}} = \frac{n}{2} \cdot \frac{1}{2^n}\). Let's focus on simplifying this further into a form that is recognizable or easier to calculate.

Differentiate the Geometric Series

The geometric series sum \(\sum_{n=0}^{\infty} x^n\) for \(|x| < 1\) is \(\frac{1}{1 - x}\). Differentiate this sum with respect to \(x\), yielding \(\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1 - x)^2}\).

Calculate the Original Sum

Replace \(x\) with \(\frac{1}{2}\) in the differentiated series: \(\sum_{n=1}^{\infty} n \left(\frac{1}{2}\right)^{n-1} = \frac{1}{(1 - \frac{1}{2})^2} = 4\). However, our series has a \(\frac{1}{2}\) multiplied initially, so our series \(\frac{1}{2} \sum_{n=1}^{\infty} n \left(\frac{1}{2}\right)^{n} = \frac{1}{2} \cdot 2\) which equals 1.

Key concepts

Infinite Series

An infinite series is a summation of infinitely many terms. These terms typically follow a particular pattern or function. When you sum all these terms, you get a series, and if it goes on indefinitely, it's called an infinite series.
Infinite series are crucial in mathematical analysis, as they help approximate values and functions that are otherwise complex. The sum of an infinite series can converge, meaning it approaches a specific value, even though it has an infinite number of terms.
For instance, consider the series: \(\sum_{n=1}^{\infty} \frac{1}{2^n}\). This series is a classic example of a convergent infinite series. As \(n\) increases, the terms become progressively smaller, making the total sum approach a finite value.
Understanding infinite series is vital to solving problems related to the expected value, as they help calculate scenarios with numerous possible outcomes. In the original exercise, the task was to find the expected number of consecutive heads, which utilized such a series.

Geometric Series

A geometric series is a type of infinite series where each term after the first is found by multiplying the previous term by a constant called the "common ratio."
Mathematically, a geometric series can be expressed as:

• \(a + ar + ar^2 + ar^3 + \ldots \)
• Where \(a\) is the first term, and \(r\) is the common ratio.

If \(|r| < 1\), the series converges and the sum can be calculated using the formula \(\frac{a}{1 - r}\).
In our context, consider the geometric series sum \(\sum_{n=0}^{\infty} x^n\) for \(|x| < 1\), which is \(\frac{1}{1 - x}\). By differentiating this, we derive another useful series for solving the original exercise.
The differentiation process allows us to find the sum of terms involving \(n\), which is crucial for expectations involving a counting variable like the number of consecutive heads in a coin toss. By strategically substituting and calculating, this leads to finding an expected value.

Probability

Probability represents the likelihood of a particular event happening. In mathematical terms, it's often expressed as a fraction or decimal between 0 and 1, where 0 indicates impossibility, and 1 indicates certainty.
In scenarios involving coins, each flip is an independent event, with a fair coin having a 0.5 probability of landing heads. When considering multiple flips, concepts of probability intertwine with series to calculate outcomes like consecutive heads.

• Probability of an independent event sequence: The joint probability of multiple independent events is the product of their individual probabilities.
• Expected value: The average outcome if an experiment is repeated many times, calculated using weighted probabilities for each potential outcome.

In the discussed exercise, the focus is on understanding how likely it is to encounter several heads consecutively. By leveraging concepts of probability combined with infinite series, one computes the expected number of such occurrences.