Probability of choosing 2 items of the same color from a bucket with k colors I think this is right, just want to double check. Let's say I have a bucket of n items. Those n items have k colors uniformly distributed. Then the probability that I get two items from the same color is:
$$
P = \frac{\binom{k}{1}\binom{n/k}{2}}{\binom{n}{2}}
$$
So, for a bucket of 200 items with 10 colors, this is:
$$
P = \frac{\binom{10}{1}\binom{200/10}{2}}{\binom{200}{2}}
$$
And the intuition being: There's a population n where I want to choose two items (denominator). From this population there are 10 colors and I want to pick 1 (numerator). And from a particular color i want to choose 2. Since there is an equal number of colors, then I can set that up as items/colors, n/k.
Are the formulation and intuition correct?
 A: Your method is a good one if you choose exactly two items. If you chose three items and the question was whether any two were the same (so the alternative, which we don't want, is items of three different colors) then the probability would be higher. If you chose $k+1$ items the probability of a matching pair would be $1.$ That's why you get questions like, "How many items are you choosing?"
Choosing just two items from the bucket, you have (as you say) $\binom{200}{2}$ possible pairs of items that you could choose, and the assumption is that no pair is more likely to be chosen than any other pair, so all $\binom{200}{2}$ pairs are equally likely.
Because of that, you just have to count the "successful" pairs and divide by the total number of possible pairs to get the probability.
Another approach is as follows.
For convenience, let $m = n/k.$
You pick one item from the bucket. It has a color.
Now you pick an item from the $n - 1$ items remaining in the bucket.
That item is either one of the $m - 1$ remaining items of the color you already picked, or one of the $(k-1)m$ items of the other $k-1$ colors.
The chance that you choose the same color twice is therefore
$$ \frac{m - 1}{n - 1} = \frac{\frac nk - 1}{n - 1}.$$
If you have $200$ items, of which $20$ are in each of $10$ colors,
the answer comes out to $\frac{20-1}{200-1} = \frac{19}{199}.$
Now, recalling that 
$$\binom xy = \frac{x!}{(x-y)! y!} = \frac{x (x-1)(x-2) \cdots (x - y + 1)}{y!},$$
in particular $\binom x1 = x$ and $\binom x2 = \frac{x(x-1)}{2},$
your approach gives us
$$
\frac{\binom k1\binom{n/k}{2}}{\binom n2}
= \frac{\binom k1\binom{m}{2}}{\binom n2}
= \frac{k \left(\frac{m(m-1)}{2}\right)}{\left(\frac{n(n-1)}{2}\right)}
= \frac{km\left(m - 1\right)}{n(n-1)}
= \frac{m - 1}{n - 1}
= \frac{\frac nk - 1}{n - 1}.
$$
So we get the same answer either way.
A: Yes, this is the right expression. We can also say that we firstly choose one item with color $k=1$. The probability is $\frac{\frac{n}{k}}{n}=\frac1k$. Then we choose the next item with color $k=1$. The probability is $\frac{\frac{n}{k}-1}{n-1}$. Since we have k colors the probability to choose two items with same colors is $\frac1k\cdot \frac{\frac{n}{k}-1}{n-1}\cdot k=\frac{\frac{n}{k}-1}{n-1}$. This term is equal to $$\frac{\binom{k}{1}\binom{n/k}{2}}{\binom{n}{2}}=\frac{\frac{k}{1}\cdot \frac{n/k\cdot (n/k-1)}{2\cdot 1} }{\frac{n\cdot (n-1)}{1\cdot 2}}=\frac{k\cdot n/k\cdot (n/k-1)}{n\cdot (n-1)}=...$$
