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Chapter 3: Problem 73

Suppose that we continually roll a die until the sum of all throws exceeds 100 . What is the most likely value of this total when you stop?

Short Answer

Expert verified
The most likely total when you stop rolling the die is 102, which corresponds to 17 rolls with the highest probability among the sums between 102 and 112. Calculate the probabilities using the recursive formula \(P(n, k) = \frac{1}{6}(P(n-1, k-1) + P(n-1, k-2) + \cdots + P(n-1, k-6))\), and compare them to find the maximum probability.

Step by step solution

01

1. Determine the minimum number of rolls needed to exceed 100.

The minimum sum that can be obtained is 1*numberOfRolls since the smallest number on a die is 1. Therefore, to exceed 100, we require at least 101 rolls. This can be concluded using the following inequality: \[numberOfRolls \geq \frac{100}{6}\] which gives numberOfRolls = 17.
02

2. Calculate the probability of each final total.

Since 6*numberOfRolls is the highest possible sum for a given number of rolls, we need to find the most likely total for 17 rolls, which is 102. To find the probability, we will use a recursive formula. Let \(P(n, k)\) denote the probability of sum \(k\) with \(n\) rolls. Then, the probability of each final total can be computed as follows: \[P(n, k) = \frac{1}{6}(P(n-1, k-1) + P(n-1, k-2) + \cdots + P(n-1, k-6))\] with initial conditions: \[P(1, k) = \frac{1}{6}\] for \(k = 1,2,3,4,5,6\), and \(P(1, k) = 0\) for \(k \ge 7\). Calculate the probabilities for the sums of 17 rolls between 102 and 112 using the given formula.
03

3. Identify the most likely total.

Now that we have the probabilities of each sum for 17 rolls, we compare the probabilities and choose the one with the highest probability. Find the maximum probability and associate it with the corresponding total. The most likely total when you stop rolling the die is the one with the highest probability.

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