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?
Solution
Let the answer to the problem be 
. WLOG, assume the first number rolled is a 
. Then, by symmetry, the probability that the last number rolled is 
equals 
. Consider the following three cases:
- If the second number rolled is a
, there is now a probability of 
that the last number rolled is 
. - If the second number rolled is a , there is now a probability of
that the last number rolled is also . - If the second number rolled is a
, 
, 
or 
, then by symmetry there is now a probability of 
that the last number rolled is .
that the last number rolled is 
, 
, 
or 
, then by symmetry there is now a probability of Thus ![\[\frac{1}{6} \cdot 0 + \frac{1}{6} \cdot x + \frac{4}{6} \cdot \frac{1-x}{5} = \frac{1-x}{5} \implies x = \boxed{\textbf{(D)}~\tfrac{2}{7}}.\]](http://latex.artofproblemsolving.com/3/d/1/3d15e182ec3711605ab3c35bf34a0f41f8b59902.png)
~ihatemath123
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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