Combinatorial Probability Proof for Dice Rolls Here is my solution:
There are $\binom{6}{2}$ ways for two numbers to appear. On each of the 4 rolls of the die, there are two possible outcomes giving $\binom{6}{2}$$\cdot$$2^4$ possible outcomes which two of the six numbers appear. This probability is $\binom{6}{2}(\cdot2^4-2)\over 6^4$ 
Is this correct?
 A: We can work out the probability that 2 fives appear exactly 2 times and then take the compliment.
The probability that 2 chosen roles of the 4 roles are 5 and the other 2 are not is:
$$ \left (\frac{1}{6} \right ) ^2  \left(\frac{5}{6}\right) ^2  $$
Then there are $ {4 \choose 2} $ ways of choosing which are the 2 chosen roles. from this we deduce your final answer to be:
$$ 1- {4 \choose 2} \left (\frac{1}{6} \right ) ^2  \left(\frac{5}{6}\right) ^2 $$
This does give a different value than your answer, but I'm not really sure where your method went wrong. I didn't really understand the wording at the beginning about 2 numbers appearing. Hope this helps and feel free to help me understand the wording! :) 
A: Your denominator is correct: six outcomes and four trials gives $6^4$ possible sequences.
For the numerator, ask how many sequences do contain exactly two 5's. We can places the 5's in the sequence in $\binom{4}{2}$ ways. After that, we can fill in the remaining two positions with any non-5 roll. This gives a total of $\binom{4}{2} \cdot 5^2$ sequences containing exactly two 5's.
Putting it all together, we have a probability of
$$
\frac{6^4 - \binom{4}{2} \cdot 5^2}{6^4},
$$
where we are subtracting the number of "bad" outcomes from the total number of outcomes in the numerator.
