A box contains four tickets, labeled 0,2,4,10. What is the chance that, if you pull 3 tickets out of this box independently, the sum of them is 14?  What I was thinking is
$_4C_3 \cdot (1/4)^3 \cdot 2 = 0.125$.
My logic is we can get a 0, 4, and 10, and the options to choose these is $_4C_3 \cdot (1/4)^3$, but we can also do 2, 2, 10, which is the same probability, thus the times $2$.
Does this look correct?
EDIT: You can replace the tickets.
Thanks!
Bliebervik
 A: If there are only $4$ tickets how can you get $0,2,2,10$? 
It looks like the only possibility of getting a sum $14$ is to pull $0, 4, 10$.
In total, you can get $^4C_3=4$ equally probable results, so the probability of pulling a sum $14$ is $\frac{1}{4}$.
Remember that in this example we're counting subsets, not sequences.
Maybe I misunderstood something you wrote though.
EDIT: If tickets are replaced after each draw, than:
The possible "14-good" outcomes are permutations of $(0, 4, 10)$ and $(2, 2, 10)$. There are $3!$ permutations of the first sequence and $3$ permutations of the second sequence.
There are $4^3$ possible sequences, because on each place (first, second, third) you can choose one of four tickets.
So the probability is $\frac{3!+3}{4^3}=\frac{9}{64}$.
Remember - here we're counting sequences, not subsets. This is why we have to take all permutations of $(0,4,10)$ and $(2,2,10)$ into consideration and this is why the denominator is the total number of pairwise distinct 3-elements-long sequences with element from the set $\{0, 2, 4, 10\}$
