Probability of Rolling Exactly 4 of a kind on 6 Dice Probability of 3 of a kind with 7 dice
I've looked at the link above - along with several others - and dusted off my combinatorics notebook to try to figure out the answer to the problem: If you roll six 6-sided dice, what is the probability of rolling exactly four of a kind?
Using combinations, I came up with: 
6C1 ways to choose the side of a die 
6C4 ways to choose four of a kind
5 ways to chose the 5th die
5 ways to chose the 6th die.
Then, there are 6^6 possible outcomes.
Putting everything together, I get (15*5*5*6)/6^6 = 2250/
46656
This is the same answer as on this Wolfram site http://www.wolframalpha.com/input/?i=6+dice
What I would like to know is if there is a way to arrive at this answer using counting principles? 
I attempted this by considering the chance of four of a kind of each number as mutually exclusive events (see images). 
For the 1 side of a die, for example: the chance rolling a 1 is 1/6, the chance of rolling a second 1 is 1/6, the chance of rolling a third 1 is 1/6, the chance of rolling a fourth 1 is 1/6, the chance of rolling anything but a 1 for the 5th die is 5/6, and the chance of rolling anything but a 1 for the sixth is 5/6. Then 4(1/6) * 2(5/6) = 25/6^6 chance of four of a kind of 1s. I then repeated this for sets of four of a kind of 2s, 3s, 4s, 5s, and 6s. 
Since these are mutually exclusive events, I then took the sum of the probability of each of the six events. The result of this method was 25/6^6 + 25/6^6 + 25/6^6 + 25/6^6 + 25/6^6 + 25/6^6 = 150/6^6, which is very different from 2250/6^6. What am I missing with this approach? Do I need to consider the different arrangements of the dice?
 A: Your first approach is correct if you permit $4$ of a kind and a pair as well as $4$ of a kind and two different values for the other two dice.  In your second approach, you do indeed need to account for the different arrangements of the dice.  There are $\binom{6}{4} = 15$ orders in which the same number could occur on four of the six rolls.  Multiplying your result by $15$ yields the same probability that you obtained using the first method.
Four of a kind without a pair: The total number of outcomes is $6^6$.  The number of favorable outcomes is 
$$\binom{6}{1}\binom{6}{4}\binom{5}{2}\binom{2}{1}$$
since there are $\binom{6}{1}$ choices for the number that occurs four times, $\binom{6}{4}$ ways for that number to appear in four of the six rolls, $\binom{5}{2}$ choices numbers that could occur once each during the six rolls, and $\binom{2}{1}$ choices for which those of numbers appears first.  Hence, the probability that four of a kind occurs with two different numbers on the remaining rolls is 
$$\frac{\dbinom{6}{1}\dbinom{6}{4}\dbinom{5}{2}\dbinom{2}{1}}{6^6}$$
Four of a kind with a pair:  The total number of outcomes is again $6^6$.  The number of favorable outcomes is 
$$\binom{6}{1}\binom{6}{4}\binom{5}{1}$$
since there are $\binom{6}{1}$ choices for the number that occurs four times, $\binom{6}{4}$ ways for that number to appear in four of the six rolls, and $\binom{5}{1}$ choices for the number that appears in both of the other two rolls.  Hence, the probability that four of a kind and a pair occur is 
$$\frac{\dbinom{6}{1}\dbinom{6}{4}\dbinom{5}{1}}{6^6}$$
Total:  Adding these two disjoint cases yields your result.
