Simple balls in bins clarification? Suppose you throw 4 balls into 5 bins. What's the probability that you end up with 1 bin with 1 ball and 1 bin with 3 balls?
Since this does not have an infinite number of balls (like the $ith$ bin), would the answer be just be
$$\frac{1}{5}^4?$$ 
 A: For the analysis, think of the balls as being thrown towards the bins one at a time.  We can then think of the outcomes as the set of quadruples that record where the first ball goes, the second, and so on. There are $5^4$ equally likely possibilities. Now we count the "favourables."
The lucky bin that will get $3$ balls can be chosen in $\binom{5}{1}$ ways. For each such way, the balls it gets can be chosen in $\binom{4}{3}$ ways. And then the bin that  will get the remaining ball can be chosen in $\binom{4}{1}$ ways, for a total of $\binom{5}{1}\binom{4}{3}\binom{4}{1}$.
For the probability, divide by $5^4$.
A: On behalf of the bins, here is their perspective:
$5\choose {2}$ ways to pick two of the five bins to put the four balls into. We want one of the bins to have 1 ball, the other to have three. So picking bin 1 to have 1 ball and bin 2 to have 3 balls is different from bin 1 having 3 balls and bin 2 having 1 ball. So in fact there are $2{5\choose{2}}$ ways to pick one bin having 1 ball, and another bin having 3 balls. 
Assuming each bin is equally likely to receive each ball, the probability of a particular arrangement is $(\frac{1}{5})^4$. So the probability of the desired event is $2{5\choose{2}}(\frac{1}{5})^4$.
