Question

Probability: tossing for a head The expected (average) number of tosses of a fair coin required to obtain the first head is \(\sum_{i=1}^{\infty} k\left(\frac{1}{2}\right)^{t} .\) Evaluate this series and determine the expected number of tosses. (Hint: Differentiate a geometric series.)

Short answer

Answer: The expected number of tosses to obtain the first head is 2.

Step-by-step solution

Rewrite the series

Given the series \(\sum_{i=1}^{\infty} k\left(\frac{1}{2}\right)^{t}\), observe that it is the sum of terms of the form \(kt\left(\frac{1}{2}\right)^{t}\), where \(k\) is an integer ranging from \(1\) to \(\infty\) and \(t\) is the number of tosses. Now, rewrite the series into a geometric series format that is easier to differentiate. To do this, let \(S\) be the sum of this series: \(S = \sum_{k=1}^{\infty} k\left(\frac{1}{2}\right)^{t} = \left(\frac{1}{2}\right)^{t} + 2\left(\frac{1}{2}\right)^{t} + 3\left(\frac{1}{2}\right)^{t} + \cdots\)

Differentiate the geometric series

In this step, we'll differentiate the geometric series and then determine the value of the sum \(S\). To obtain the sum, we'll differentiate both sides of the equation we got in step 1. So, \(S = \left(\frac{1}{2}\right)^{t} + 2\left(\frac{1}{2}\right)^{t} + 3\left(\frac{1}{2}\right)^{t} + \cdots\) Differentiating both sides with respect to \(t\), we get: \(\frac{dS}{dt} = -\left(\frac{1}{2}\right)^{t} \ln{\left(\frac{1}{2}\right)} - 2\left(\frac{1}{2}\right)^{t} \ln{\left(\frac{1}{2}\right)} - 3\left(\frac{1}{2}\right)^{t} \ln{\left(\frac{1}{2}\right)} - \cdots\) Now simplify the equation: Since \(\ln{\left(\frac{1}{2}\right)} = -\ln{2}\), the equation can be rewritten as: \(\frac{dS}{dt} = \ln{2}\left[\left(\frac{1}{2}\right)^{t} + 2\left(\frac{1}{2}\right)^{t} + 3\left(\frac{1}{2}\right)^{t} + \cdots\right]\) Notice how the expression inside the square brackets is \(S\), the sum we are trying to find: \(\frac{dS}{dt} = \ln{2}S\) Let's solve for S.

Solve for S

Now we have: \(\frac{dS}{dt} = \ln{2}S\) Divide both sides by \(\ln{2}S\): \(\frac{1}{S} \frac{dS}{dt} = \ln{2}\) Integrate both sides with respect to \(t\): \(\int \frac{1}{S} \frac{dS}{dt} dt = \int \ln{2} dt\) This becomes: \(\int \frac{1}{S} dS = \ln{2} \int dt\) Integrate: \(\ln{S} = \ln{2}t + C\) Exponentiate both sides to solve for \(S\): \(S = e^{\ln{2}t + C}\) Now, apply the initial condition when getting the first head (\(t = 1\)) and assume \(C=0\) (since \(e^0 = 1\)): \(S = e^{\ln{2}(1)} = e^{\ln{2}}\) Therefore, the expected number of tosses to get the first head is: \(S = e^{\ln{2}} = 2\) The expected number of tosses to obtain the first head is 2.

Key concepts

Expected Value

Expected value is a fundamental concept in probability, often referred to as the "average" outcome you expect from a random event. It provides a way to predict future results based on probabilities.
For example, if you toss a fair coin, you have a 50% chance (probability of 0.5) to get heads. To find the expected number of tosses required to get the first head, you sum over all possible outcomes, each weighted by its probability. This expected value accounts for all possible scenarios (one toss, two tosses, etc.).

The formula used here is expressed as a series:

• Sum over the outcomes from 1 to infinity
• Each outcome scaled by its probability
  

Simplifying this series helps calculate how many tosses you need on average to see your first head. This concept in probability helps model real-world problems, providing insights into average expected behavior over time.

Geometric Series

A geometric series is a sequence where each term is a constant multiple of the previous one. These are common in probability, especially when dealing with repeated trials.
The basic form of a geometric series is:\[S = a + ar + ar^2 + ar^3 + \cdots\]
Where:

• \(a\) is the first term
• \(r\) is the common ratio

If \(|r| < 1\), the series converges, which means you can find a finite sum even though it continues indefinitely. For the coin tossing problem, the series represents the probability distribution across multiple tosses.

By differentiating and manipulating the series, we can find specific values like expected numbers. Understanding geometric series allows us to handle complex probability sequences efficiently, providing a systematic approach to solve intricate problems.

Coin Tossing

Coin tossing is a simple yet powerful way to explore probability concepts. A fair coin has an equal chance (1/2) to land on heads or tails.
Every toss is independent, meaning prior outcomes don't affect the next one. This independence is crucial when calculating probabilities over multiple tosses.

When dealing with coin tosses:

• A single toss has possible outcomes: heads or tails
• The probability for each is 0.5 if the coin is fair

The exercise explores how many times you need to toss a coin to get the first head. By understanding and applying probability, we find that the expected number is 2 tosses.

Coin tossing serves as an excellent introduction to randomness and chance in statistical studies, laying the groundwork for more complicated probability scenarios.