2014 AMC 10B Problem 16

There are $6^4 = 1296$ equally likely rolls. "At least three alike" means exactly three alike or all four alike; count each.

All four alike. The common value can be any of $6$ numbers: $6$ outcomes.

Exactly three alike. Choose the repeated value ($6$ ways), choose which of the four dice is the odd one out ($4$ ways), and choose its value, which must differ from the repeated one ($5$ ways):
$$6 \cdot 4 \cdot 5 = 120.$$

Favorable outcomes: $120 + 6 = 126$. The probability is
$$\frac{126}{1296} = \frac{7}{72}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{7}{72}}$.

"At least $k$" on a small sample space is best handled by splitting into exact counts that do not overlap. For "exactly three alike," the three choices (which value, which die is different, what that die shows) are independent, so they multiply; insisting the odd die shows a different value is what keeps the two cases disjoint.

Think sequentially: the first die is anything. For all four to match, the next three each match with probability $\frac16$: $\frac{1}{216}$. For exactly three to match, pick the odd die ($4$ ways); if it is the first die, the other three match each other ($\frac{1}{36}$) and the first differs ($\frac56$); the same product holds whichever die is odd. Total $\frac{1}{216} + 4 \cdot \frac{1}{36} \cdot \frac{5}{6} = \frac{1}{216} + \frac{20}{216} = \frac{21}{216} = \frac{7}{72}$.

• Counting exactly-three-alike as 646 = 144 by letting the odd die repeat the common value, which double counts the four-alike outcomes and gives $\frac{150}{1296}$.
• Forgetting the $4$ ways to choose which die is different, giving $\frac{30 + 6}{1296} = \frac{1}{36}$, choice (A).