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?