Combinatorics rolling a die six times How many ways are there to roll a die six times such that there are more ones than twos?
I broke this up into six cases: 
$\textbf{EDITED!!!!!}$
$\textbf{Case 1:}$ One 1 and NO 2s --> 1x4x4x4x4x4 = $4^5$. This can be arranged in six ways: $\dfrac{6!}{5!}$. So there are $\dfrac{6!}{5!}$$4^5$ ways for this case. 
$\textbf{Case 2:}$ Two 1s and One 2 OR NO 2s --> 1x1x5x4x4x4 = $5x4^3$. This can be arranged in $\dfrac{6!}{2!3!}$ ways. So there are $(6x4^3)$$\dfrac{6!}{2!3!}$ ways for this case. 
$\textbf{Case 3:}$ Three 1s and Two, One or NO 2s --> 1x1x1x5x5x4 = 4x$5^2$. This can be arranged in $\dfrac{6!}{2!3!}$ ways. So there are $(4x5^2)$$\dfrac{6!}{2!3!}$ ways for this case. 
$\textbf{Case 4:}$ Four 1s and Two, One or NO 2s --> 1x1x1x1x5x5 = $5^2$. This can be arranged in $\dfrac{6!}{4!2!}$ ways. So there are $(5^2)$$\dfrac{6!}{4!2!}$ ways for this case. 
$\textbf{Case 5:}$ Five 1s and One or NO 2s --> 1x1x1x1x1x5 = 5. This can be arranged in $\dfrac{6!}{5!}$ ways = 6 ways. So there are $5^3$ ways for this case. 
$\textbf{Case 6:}$ Six 1s and NO 2s --> 1x1x1x1x1x1 = 1. There is only one way to arrange this so there is only 1 way for this case. 
With this logic...I would add the number of ways from each case to get my answer. 
 A: Case 1 is wrong because there are $4$ numbers that are neither $1$ nor $2$, so there are $6 \cdot 4^5$ ways to roll one $1$ and no $2$s.  That error repeats.  
Added:  Case 2 is wrong because the number of ways to arrange the numbers is dependent on whether there is a $2$ or not.  The $1 \cdot 1 \cdot 5 \cdot 4^3$ is really $1 \cdot 1 \cdot (4+1) \cdot 4^3$ where the $4+1$ is four ways to not get a $2$ and one way to get one.  If you don't get one there are $\frac {6!}{2!4!}={6 \choose 2}=15$ ways to arrange them.
A: If the question is "where is my error", than plese ignore this answer, which is giving an other way to count. The idea is that there are either more $1$'s, case (1), or more $2$'s, case (2), or they occur equaly often, case $(=)$. Of course, the count of possibilities for case (1) is the same as for case (2), so we simply count the possibilities for $(=)$, subtract them from the total, divide by $2$, so the number we are searching for is:
$$
N = \frac 12\left(\ 6^6 
-\binom 60\binom 60 4^{6-0}
-\binom 61\binom 51 4^{6-2}
-\binom 62\binom 42 4^{6-4}
-\binom 63\binom 33 4^{6-6}
\ \right) \ .
$$
The products of binomial coefficients of the shape $\binom 6k\binom{6-k}k$, $k=0,1,2,3$ count the possibilities to fix the $k$ places among $6$ places for the $1$'s, then the $k$ places for the $2$'s from the remaining $6-k$.
We get then
$$
N = \frac 12\left(\ 6^6 
- 4^6 
- 6\cdot 5\cdot 4^4 
- 15\cdot 6\cdot 4^2
- 20\cdot 1\cdot 4^0
\ \right) 
=
16710\ .
$$
We can also check this answer with sage, by enumerating all possibilities.
sage: len( [ x for x in cartesian_product( [[1..6] for _ in [1..6]] )
....:        if list(x).count(1) > list(x).count(2) ] )
16710

