Dice Problem

The number of possible outcomes is $6^4$ because there are 6 possible numbers on each die and there are 4 dice so 6 x 6 x 6 x 6 possible outcomes. The number of outcomes without a three on any of the dice is $5^4$ because there are five possible outcomes for each die if 3 is not on any of them. So subtracting this from the total number of outcomes leaves $6^4 - 5^4 =\boxed{671}$ outcomes where a three is on at least one of the dice.

Here is a way by PIE.

Number of ways you can have one die show up $3$ is $\binom{4}{1}\times 6^{3}$ because you have $4$ options for which die is $3$ and each of the other $3$ die have $6$ possible numbers. Similarly, the number of ways two dice show up $3$ is $\binom{4}{2}\times6^{2}$ , three dice is $\binom{4}{3}\times6$ , and four dice is just $1$ .

The number of ways for two dice is counted $\binom{2}{1}=2$ times, so subtract it once. The number of ways for three dice is counted $\binom{3}{1}-\binom{3}{2}=0$ times, so add it once. The number of ways for four dice is counted $\binom{4}{1}-\binom{4}{2}+\binom{4}{3}=2$ times, so subtract it once.

Total: $\binom{4}{1}6^{3}-\binom{4}{2}6^{2}+\binom{4}{3}6-\binom{4}{4}=\boxed{671}$