Problem 42 - A First Course in Probability

Question

Two dice are thrown \(n\) times in succession. Compute the probability that double 6 appears at least once. How large need \(n\) be to make this probability at least \(\frac{1}{2} ?\)

Step-by-Step Solution

Answer

The probability of getting at least one double 6 in n trials is \(1 - \left(\frac{35}{36}\right)^n\). The smallest value of n for which this probability is at least \(\frac{1}{2}\) is 24.

Step 1: Find the probability of not getting a double 6 in a single trial

We know that a single die has 6 faces, numbered from 1 to 6. When two dice are thrown, there are 6 x 6 = 36 possible outcomes. We are interested in the probability of not getting a double 6, which is (6, 6). There is only 1 double 6, thus there are 36 - 1 = 35 favorable outcomes for not getting a double 6. Then, the probability of not getting a double 6 in a single trial can be written as: \(P(\text{not } \text{double} \: 6\) (Single Trial)) = \(\frac{35}{36}\)

Step 2: Find the probability of not getting a double 6 in n trials

We assume that each throw is independent. Therefore, the probability of not getting a double 6 in n trials is given by: \(P(\text{not } \text{double} \: 6 \: \text{in} \: n \: \text{trials}) = P(\text{not } \text{double} \: 6 \: \text{(Single Trial)})^n = \left(\frac{35}{36}\right)^n\)

Step 3: Find the probability of getting at least one double 6 in n trials

Now, we use the complementary probability to find the probability of getting at least one double 6 in n trials: \(P(\text{at least one double } 6 \: \text{in} \: n \: \text{trials}) = 1 - P(\text{not } \text{double} \: 6 \: \text{in} \: n \: \text{trials})\) = \(1 - \left(\frac{35}{36}\right)^n\)

Step 4: Find the smallest value of n for which the probability is at least 1/2

We want to find the smallest n such that \(P(\text{at least one double } 6 \: \text{in} \: n \: \text{trials})\) \(\geq \frac{1}{2}\). We can substitute the expression for this probability from Step 3 and solve for n: \(1 - \left(\frac{35}{36}\right)^n \geq \frac{1}{2}\) \(\left(\frac{35}{36}\right)^n \leq \frac{1}{2}\) To solve for n, we can use the logarithm. We have: \(n \log{\left(\frac{35}{36}\right)} \leq \log{\left(\frac{1}{2}\right)}\) Divide by the negative value of the logarithm of \(\frac{35}{36}\): \(n \geq \frac{\log{\left(\frac{1}{2}\right)}}{\log{\left(\frac{35}{36}\right)}}\) Using a calculator, we find that the smallest integer value of n that satisfies this inequality is: n = 24 Thus, at least 24 trials are needed to make the probability of getting at least one double 6 greater than or equal to 1/2.

Key Concepts

Complementary Probability

Complementary probability is a powerful concept in probability theory. It helps us find the likelihood of an event happening by first considering the event not happening. For instance, when calculating the odds of getting at least one double 6 when two dice are rolled multiple times, it is often easier to first find the probability of not getting a double 6. This non-occurring scenario is essentially the complement of your desired event.

In the exercise you see the complement concept in action:

• The probability of not rolling a double 6 with two dice is \( \frac{35}{36} \).
• This probablity represents all other outcomes when two dice are rolled.

To find the probability of rolling at least one double 6 in several trials, we use: \[ P(\text{at least one double 6}) = 1 - P(\text{not double 6 in } n \text{ trials}) \]

This forecasts how probable it is to achieve at least one double 6 through the complement of continuously failing to roll one.

Probability of Independent Events

Independent events are those whose outcomes do not affect each other. In our case, each time you roll two dice, the result is independent of previous rolls.

Calculating the probability for independent events involves simply multiplying the probabilities of each event.
In the problem presented, the probability that you do not roll a double 6 with each pair of dice rolled is consistently \( \frac{35}{36} \).

Because these events are independent:

• The probability of not seeing a double 6 in each roll, across \( n \) trials, becomes \( \left( \frac{35}{36} \right)^n \).

This product accounts for each roll independently resulting in an outcome other than double 6.

Logarithms in Probability

Logarithms present a useful approach in probability, particularly when exponential growth or decay functions are involved.

When dealing with exponential probabilities, such as repeated independent events, logarithms help unravel the exponential functions for direct solutions.In the dice example, solving for how many rolls are needed where probability dictates you will roll at least one double 6 force us into exponential territory:

• You start with \( \left( \frac{35}{36} \right)^n = \frac{1}{2} \) because you want this outcome to be under half, ensuring over half is achieving a double 6 at some point.
• By taking the log of both sides, you convert the power of \( n \) from an exponent back into a solvable algebraic variable.
• This gives us \( n \log\left( \frac{35}{36} \right) \leq \log\left( \frac{1}{2} \right) \), which then lets you solve for the smallest \( n \) using handy logarithmic rules and a calculator.

Logarithms show their strength dealing with these kinds of inequalities and transitions from exponential expressions into manageable numbers.