Problem 7 - Active Calculus

Question

Suppose you play a game with a friend that involves rolling a standard six- sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six. a. Explain why the probability of rolling a six on any single roll (including your first turn) is \(\frac{1}{6}\) b. If you don't roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right) $$ c. Now suppose you fail to roll the first six on your second turn. Explain why the probability is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right) $$ that you to roll the first six on your third turn. d. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is $$ \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right)+\cdots $$ Find the sum of this series and determine the probability that you roll the first six.

Step-by-Step Solution

Answer

The probability that you roll the first six is \( \frac{6}{11} \).

Step 1 Title - Understanding the Probability of Rolling a Six

The probability of rolling a six on any single roll of a six-sided die is determined by the fact that there are six possible outcomes (1, 2, 3, 4, 5, and 6), and only one of these outcomes is a six. Therefore, the probability is \(\frac{1}{6}\).

Step 2 Title - Calculating the Probability for the Second Turn

To determine the probability that you roll the first six on your second turn, both you and your friend must fail to roll a six on your first turns. The probability of not rolling a six on any single roll is \(\frac{5}{6}\). Therefore, the probability of both you and your friend not rolling a six is: \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\. If you succeed in rolling a six on your second turn, the combined probability is: \[ \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right) \]

Step 3 Title - Extending to the Third Turn

If you fail to roll a six on your second turn, we must consider the events leading to your third turn. The probability of not rolling a six on each of the first two turns is \left( \frac{5}{6} \right) \, and similarly for your friend. The combined probability of all these failures and finally succeeding on your third turn is: \[ \left( \frac{5}{6} \right)^4 \left( \frac{1}{6} \right) \]

Step 4 Title - Summing the Infinite Series

The probability of you rolling the first six can be modeled by an infinite series, where each term represents the scenario of succeeding on the nth roll after repeatedly failing. This series is given by: \[ \frac{1}{6} + \left(\frac{5}{6}\right)^2 \left(\frac{1}{6}\right) + \left( \frac{5}{6} \right)^4 \left( \frac{1}{6} \right) + \cdots \] This is a geometric series with the first term \(a = \frac{1}{6}\) and common ratio \(r = \left( \frac{5}{6} \right)^2 \). Using the formula for the sum of an infinite geometric series, \[ S = \frac{a}{1 - r} \] we get: \[ S = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^2} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11} \]

Key Concepts

Rolling a Six-Sided Die

Rolling a six-sided die is a classic example in probability theory. A six-sided die, also known as a d6, has six faces numbered 1 through 6. When you roll the die, each face has an equal probability of landing face-up. That means each number, including the number six, has a probability of \( \frac{1}{6} \). This is because there are six possible outcomes and only one desired outcome. Understanding these basic probabilities is the foundation for solving more complex problems.

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio. In our context, consider the series: \[ \frac{1}{6} + \frac{5}{6} ^ 2 \frac{1}{6} + \frac{5}{6} ^ 4 \frac{1}{6} + \cdots \] Here, the common ratio is \( \frac{5}{6} ^ 2 \) and the first term is \( \frac{1}{6} \). This series represents the decreasing probabilities of rolling a six for the first time on successive turns. Geometric series are particularly interesting because their infinite sum can be calculated using the formula: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio.

Probability Calculations

Probability calculations help us determine the likelihood of different outcomes. For instance, the probability of not rolling a six on any roll is \( \frac{5}{6} \). Combining several independent events, such as not rolling a six on consecutive rolls, involves multiplying their probabilities. In our problem: \[ \text{Probability of not rolling six twice} = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36} \] To find the probability of rolling a six on the second turn after both you and your friend fail the first time: \[ \frac{25}{36} \times \frac{1}{6} = \frac{25}{216} \] This kind of step-by-step multiplication and addition is key to understanding and solving probability puzzles.

Infinite Series Sum

The sum of an infinite series might sound daunting, but using geometric series formulas simplifies the task. Our series is: \[ \frac{1}{6} + \frac{25}{216} + \frac{625}{7776} + \cdots \] This represents a geometric series with the first term \( a = \frac{1}{6} \) and the common ratio \( r = \frac{25}{36} \). To find the sum: \[ S = \frac{a}{1 - r} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11} \] So, the probability that you roll the first six is \( \frac{6}{11} \). Summing an infinite series in this way provides an exact probability, revealing the power of geometric series in probability theory.