2014 AMC 10B Problem 16

1. All possible rolls. Each die has $6$ faces, so there are $6 \times 6 \times 6 \times 6 = 1296$ equally likely rolls.
2. "At least three the same" means exactly three the same or all four the same. These two cases do not overlap, so count each and add.
3. All four the same. The rolls $1111, 2222, \ldots, 6666$: that is $6$ rolls.
4. Exactly three the same. Build such a roll in three steps:
- choose the value that appears three times: $6$ ways;
- choose which of the four dice is the odd one out: $4$ ways;
- choose what the odd die shows. It must be different from the other three, or we would be back in the all-four case: $5$ ways.

That gives $6 \times 4 \times 5 = 120$ rolls. (For example, value $2$, odd die second, showing $5$: the roll $2522$.)
5. Total good rolls: $120 + 6 = 126$.
6. Probability: $\dfrac{126}{1296}$. Divide top and bottom by $18$: $\dfrac{7}{72}$.

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

"At least three" is easiest to count by splitting it into exact cases that do not overlap. Within the "exactly three" case, the choices (which value, which die is different, what it shows) do not affect one another, so they multiply. Requiring the odd die to be different is what keeps the two cases apart.

Use probabilities directly. All four match when dice two, three and four each match the first: $\frac16 \times \frac16 \times \frac16 = \frac{1}{216}$. For exactly three matching, pick the odd die ($4$ ways). The three others must show the same value (the second and third each match the first of them: $\frac{1}{36}$), and the odd die must be different ($\frac56$). That gives $4 \times \frac{1}{36} \times \frac56 = \frac{20}{216}$. Total $\frac{21}{216} = \frac{7}{72}$.

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