Erin rolls 4 four-sided dice all at once, then can roll a subset of her choosing a 2nd time. What is the probability of getting all the same number? Here's what I have so far:
All 4 same on first try = (1/4)^4 * 4
3 same, then get 4th on 2nd roll = 4 * (1/4)^3 * (3/4) * (4!/3!)
Here's where I'm confused:
2 same = 4 * (1/4)^2 * (3/4)(2/4 :to avoid counting double couplets twice) * (4!/2! :the order they can be arranged) + 4*3/4^4
1 same = ???
(I get easily stuck on combinatorics problems like this, and would be so grateful if you could thoroughly explain a simple strategy for solving these kind of problems with combinatorics; while I know what combinatorics are, I don't think I've fully understood how to apply them in different kinds of contexts.)
 A: Let the dice be coloured with four colours, blue, white, red, green. There are $4^4$ possible outcomes, of which $4!$ have all numbers different. So the probability they are all different is $\frac{4!}{4^4}$. In that case she might as well roll all again. That gives a contribution of $\frac{4!}{4^4}\cdot \frac{1}{4^4}$ to our required probability.
Now let us look at the probability that there is at least one pair, possibly $2$.  It is useful to divide into two cases (i) one pair, with the other two different from each other and (ii) two pairs. The calculation is done in the remark at the end. But I would rather cheat.
The probability all are the same is $\frac{4}{4^4}$. The probability $3$ are the same and one different is $\frac{\binom{4}{3}\cdot 4\cdot 3}{4^4}$. And as we saw the probability  all are different is $\frac{4!}{4^4}$. So the probability of one or two pairs is
$$1-\frac{4}{4^4}-\frac{\binom{4}{3}\cdot 4\cdot 3}{4^4}-\frac{4!}{4^4}.\tag{1}$$
If she get one or two pairs she will keep one pair and toss the remaining two dice. The contribution to the probability is $\frac{1}{4^2}$ times the number in (1). 
Now you can put things together. 
Remark: The probability of one pair, with the other two numbers different from each other, is $\frac{\binom{4}{2}\cdot 4\cdot 3\cdot 2}{4^4}$.
Two pairs is a little trickier. The numbers we have pairs in can be chosen in $\binom{4}{2}$ ways. For each choice, the dice that show the larger of the numbers can be chosen in $\binom{4}{2}$ ways. So the probability of two pairs is $\frac{\binom{4}{2}\binom{4}{2}}{4^4}$.
A: If you get confused with this type of problems, a simple stepwise approach with an inbuilt check to compute  # of ways  on the first roll is as under: 
4 of a kind: ${4\choose 1}\cdot 4!/4! = 4$
3-1 of a kind:${4\choose 1}\cdot{3\choose 1}\cdot4!/3! = 48$
2-2 of a kind${4\choose 2}\cdot4!/(2!2!) = 36$
2-1-1 of a kind:${4\choose 1}\cdot {3\choose 2}\cdot 4!/2!) = 144$
1-1-1-1 of a kind: ${4\choose 4}\cdot 4! = 24$
The inbuilt check shows that it adds up to 256, $(4^4)$, as it should.
Now  leave the "maximum of a kind" rolled, and roll the rest in an attempt to get 4 of a kind.
As an example, if 2-1-1 of a kind was rolled on first try,
probability of success = $(144/4^4)$$\times$ (1/4^2)
Do the same for other cases and add up to get the final probability.
