Problem 65 - Single Variable Calculus

Question

Peter and Paul take turns tossing a pair of dice. The first person to throw a 7 wins. If Peter starts the game, then it can be shown that his chances of winning are $$ p=\frac{1}{6}+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2}+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{4}+\cdots $$ Find \(p\).

Step-by-Step Solution

Answer

Peter's probability of winning the game is \(p = \frac{6}{11}\).

Step 1: Identify the geometric series

The given sum contains a geometric series: $$ \frac{1}{6}+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{2}+\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{4}+\cdots $$ Here, the first term, \(a_1\), is \(\frac{1}{6}\), and the common ratio, \(r\), is \(\left(\frac{5}{6}\right)^2\).

Step 2: Use the sum formula for a geometric series

Recall the formula for the sum of an infinite geometric series, when \(|r| < 1\): $$ S = \frac{a_1}{1 - r} $$ In this case, we want to find the sum \(p\), which represents Peter's chances of winning. Applying the sum formula, we get: $$ p = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^2} $$

Step 3: Simplify the expression

Simplify the expression in the denominator by squaring \(\frac{5}{6}\) and subtracting the result from 1: $$ 1-\left(\frac{5}{6}\right)^2 =1- \frac{25}{36} = \frac{11}{36} $$ Now, substitute this result back into the fraction: $$ p = \frac{\frac{1}{6}}{\frac{11}{36}} $$

Step 4: Find Peter's probability of winning

To find \(p\), divide the fractions by multiplying the numerator by the reciprocal of the denominator: $$ p = \frac{1}{6} \times \frac{36}{11} = \frac{1 \times 36}{6 \times 11} $$ Simplify the fraction: $$ p = \frac{36}{66} = \frac{6}{11} $$ Peter's probability of winning the game is \(\frac{6}{11}\).

Key Concepts

Geometric Series

Imagine repeating an action and each time, you do it with a fraction of the intensity of the previous time. This is the essence of a geometric series. It's a mathematical sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

In our example with Peter and Paul, the common ratio is \(\left(\frac{5}{6}\right)^2\) because each term represents a situation where two rounds go by without winning, resulting in two multiplications of \(\frac{5}{6}\), the chance of not rolling a 7 on the two dice. Since the ratio is less than one, the amounts keep shrinking, making up a decreasing geometric series.

Infinite Series Sum Formula

Some series continue forever—or as mathematicians like to say, they are infinite. You might wonder how we could ever sum up infinitely many things, but surprisingly, if the terms get smaller and smaller quickly enough, there's a sweet and finite number they all add up to. That's what the infinite series sum formula is for.

If we have an infinite geometric series, and the absolute value of the common ratio is less than 1, the sum can be found using the formula \[ S = \frac{a_1}{1 - r} \] where \( a_1 \) is the first term and \( r \) is the common ratio. This is the formula used to find the probability of Peter winning the dice game.

Simplifying Expressions

To simplify an expression means to make it as straightforward as possible. This usually involves combining like terms, reducing fractions, and performing arithmetic operations like addition, subtraction, multiplication, and division. In this example, simplifying the expression is a key step in calculating the probability. We start with a complex-looking fraction and end with the clear chance of winning: \(\frac{6}{11}\).

Probability Calculation

Finally, all this talk of series comes down to something quite practical: probability calculation. Probability is the math of chance, of predicting what's likely to happen. In dice games, we calculate it by considering the total number of outcomes and how many of those outcomes are winners.

For Peter, calculating his chance of winning the dice game involved understanding an infinite geometric series because he might, in theory, have to wait infinitely many turns to win—though luckily for him (and us!), the sum of that series is a graspable number. After using the series sum formula and simplifying the expression, we find his probability is a neat \(\frac{6}{11}\). This represents a little more than half, so if you were betting on the game, you might just put your money on Peter!