You answer for "exactly two the same" counts some cases twice - when you get two pairs ($4545$, for example.)
The case of $2$ the same the others different counts to $6\cdot 5\cdot 4\cdot\binom{4}{2}=720$.
The case of two pair is $\binom{6}{2}\cdot\binom{4}{2}=90$.
Those two values add up to $810$, and you over-counted by $90$ - that is, you counted each "two pair" result twice.
This gives a total of $720+90+120+6=936=1296-360$.
This sort of problem is much easier to do by calculating the probability of the opposite (that they are all different) and subtract that from $1$. The probability that they are all different is $\frac{6\cdot5\cdot 4\cdot 3}{6^4} = \frac{360}{1296}$.
You are rolling four dice, and you want to know the chance of at least one three. Because it is at least one there are a few ways this could happen- namely there could be one three, two threes, three threes or four threes. Now, we can work all of these out using something called the binomial coefficients (technically speaking the number of heads is a binomial random variable) - the formula is the probability of a three $(1/6)$ raised to the power of how many, times by the probability of not a three (5/6) times by how many aren't threes, and then multiplied by the number of ways to select the dice (again, binomial coefficients). So the chance of one three is $4*(1/6)(5/6)^3=500/1296$. The chance of two threes is $6*(1/6)^2*(5/6)^2=150/1296$. The chance of three threes is $4*(1/6)^3*(5/6)=20/1296$ and the chance of four threes is $1*(1/6)^4*(5/6)^0=1/1296$.
If we add these all together, we get $(500+150+20+1)/1296=671/1296$ which is the the same as $1-(5/6)^4$ as expected. This is the formal way of working it out, but the short cut is the complement. The way I target these questions is to work out which has the least leg work. You are asked for the chance of at least one which includes the options 1,2,3,4, where as the complement is just 0. If you were asked for the chance of at least two, this is 2,3,4 and the complement is 0 or 1 (which is easier to work out). It is only when you're asked for 3 or more, that it's easier the formal way.
In general, if $X$ is the number of something with maximum value $Y$ (number of heads in Y coin tosses, threes on a dice roll, etc) and you want to calculate the probability that $X>n$ (written $P(X>n)$) it easier to use the complement if n
Best Answer
You should break up the problem into pieces. First let's compute how many ways we can get exactly three of the dice to show the same number. First we choose which three dice we are forcing to have the same value (namely $4\choose 3$) and multiply this by the number of values the three dice can have (there are 6 possibilities here). Then multiply this by all possible values for the fourth dice (this is 5 since we can't have the last die have the same value as the first 3). This gives a total of $5\times 6\times$$4\choose 3 $$=120$ possibilities.
Now we add the above to the number of ways all four dice land on the same number - there are only 6 ways.
There are a total of $6^4 = 1296$ possible configurations of the four dice. The probability you want is $\frac{126}{1296} = \frac{7}{72}$ giving about a $9.7$% chance of this occurring.