2024 AMC 10A Problems/Problem 21

Problem

A fair six-sided die is repeatedly rolled until the same number is rolled twice in a row. What is the probability that the last number rolled is equal to the first number rolled?

$\textbf{(A)}~\frac{17}{72} \qquad\textbf{(B)}~\frac{4}{15} \qquad\textbf{(C)}~\frac{5}{18} \qquad\textbf{(D)}~\frac{2}{7} \qquad\textbf{(E)}~\frac{3}{10}$

Solution

Let the answer to the problem be $x$ . WLOG, assume the first number rolled is a $1$ . Then, by symmetry, the probability that the last number rolled is $2$ equals $\tfrac{1-x}{5}$ . Consider the following three cases:

If the second number rolled is a $1$ , there is now a probability of $0$ that the last number rolled is $2$ .
If the second number rolled is a $2$ , there is now a probability of $x$ that the last number rolled is also $2$ .
If the second number rolled is a $3$ , $4$ , $5$ or $6$ , then by symmetry there is now a probability of $\tfrac{1-x}{5}$ that the last number rolled is $2$ .

Thus $\frac{1}{6} \cdot 0 + \frac{1}{6} \cdot x + \frac{4}{6} \cdot \frac{1-x}{5} = \frac{1-x}{5} \implies x = \boxed{\textbf{(D)}~\tfrac{2}{7}}.$

~ihatemath123

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2024 AMC 10A Problems/Problem 21

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