Probability of type-specific grouping in sets containing 2 or more types Given a number of items of a number of types randomly arranged, how likely is it that the items are grouped according to type?
Example 1: Given the numbers 1-8, randomly arranged in a line, how likely is it that all of the even numbers or all of the odd numbers are next to each other?
Example 2: Eight people are standing in a circle. Four are wearing red shirts and four are wearing green shirts. Assuming they are standing in a random order (they aren't conferring teams or anything), how likely is it that all the people in red shirts are next to each other?
(Bonus: What about the probability of smaller groups, e.g. three red shirts and three green shirts together? What about if we have blue shirts in the mix too?)
Edit: 
One of the things I tried was multiplying the number of permutations of green shirts by the number of permutations of red shirts, and dividing that by the total number of possible permutations. This number seems to be far too low--i.e. it yields too few favourable cases. (So $4!*4!/8!$ = 576/40320, about 1%. I used the same approach with the case of four shirts, two of each colour--it had few enough cases to draw them out completely--and it was inaccurate.)
I also considered the problem as a case of drawing balls from a bag and lining them up. I calculated the probability of each favourable case, which yielded 1/35 for each favourable case, and added up the probabilities of the favourable cases. However, yet again, applying this method to the simple case (four balls, two of each colour) doesn't seem to work: it indicates a 1/3 probability for each of four favourable cases, which would give me a 133% probability, without even considering the non-favourable cases.
If anyone has insight into why these approaches don't make sense I would appreciate it! Many thanks to those who have helped out already.
Edit 2: The second approach works; I misunderstood the probability calculation I was using. In the approach I used, the first probability I used was for drawing a ball of either colour. This means that I don't have to account for the switched cases (e.g. RRRRGGGG and GGGGRRRR) separately, which I thought I should have to do.
I am still uncertain about why the first approach in the above edit doesn't work, so if anyone understands how to get the total number of permutations of red and green when in a circle I would appreciate the insight. Thanks!
 A: We could say that there are $8!$ equally likely permutations of our numbers, and then count the "favourable" permutations. We do it a slightly different way.
On the $8$ chairs, we will put two sorts of "Reserved" signs, "Evens Only" and "Odds Only." There are $\binom{8}{4}$ equally likely ways to place the "Evens Only"  signs.
How many placements have the "Evens Only" all together? The leftmost such sign can be put in $5$ positions ($1$ to $5$), and then the positions of the rest are determined.
Similarly, there are $5$ placements in which the odds are all together. However, if we calculate $5+5$, we will double-count the arrangements in which the "Evens Only" and "Odds Only" are both all together. Thus the total number of favourables is $8$, and our probability is $\frac{8}{\binom{8}{4}}$. 
A: For the second example.  We have two groups of four members, all arranged in a circle.  
Since we are interested in probabilities we are not interested in rotational distinction.  Treating the situation as equivalent to ways to string beads on a necklace will count equally likely cases as a proportion of the total space.

(1) The probability that either all red shirts are clustered together, or all green shirts are clustered together.
8 of the $\frac{8!}{4!4!}$ possible rearrangements of the string $\{\mathrm{r,r,r,r,g,g,g,g}\}$ will cluster 4 r, or 4 g together when it is looped into a circle.  In fact every arrangement that clusters 4 r together also clusters 4 g together (and vice versa). 
We can also say this as: There are 8 places to start a cluster of 4 red shirts, which fixes the placement of the green shirts, also in a cluster of 4.
Let $R$ measure the size of the largest cluster of red shirts in an arrangement, and $G$ be that of green shirts.
$$\Pr(R=4 \cup G=4) = \frac{\;4}{35}$$

(2) The probability that either red shirts or green shirts are clustered together in clusters of 3 or 4.
Case A: R=4,G=4: eg $\{\mathrm{g,g,g,r,r,r,r,g}\}$  As noted above: There are 8 places to start a cluster of 4 red shirts, which fixes the 4 green shirts, also in a cluster of 4. $\Pr(R=4\cap G=4)=\frac{8}{70}$
Case B: R=3: eg $\{\mathrm{r,r,g,g,r,g,g,r}\}$ There are 8 places to start a cluster of three red shirts, and there are then 3 places to put the fourth red shirt such that it does not form a cluster with the other three.  $\Pr(R=3)=\frac{24}{70}$
Similarly: $\Pr(G=3)=\frac{24}{70}$
Case C: R=3, G=3: eg $\{\mathrm{r,r,g,g,g,r,g,r}\}$ There are 8 places to start a cluster of three red shirts, then only 2 places to start a cluster of three green shirts, such that the remaining red and green shirt can be placed in the remaining positions without forming a cluster of four. $\Pr(R=3\cap G=3)=\frac{16}{70}$
Putting it together, using the Principle of Inclusion-Exclusion (PIE).
$$\begin{align}\Pr(R\geq 3\cup G\geq 3) & = \Pr(R=3)+\Pr(G=3)-\Pr(R=3\cap G=3)+\Pr(R=4\cap G=4) \\ & =\frac{24+24-16+8}{70} \\&=\frac{4}{7}\end{align}$$

(3) The probability that either red shirts or green shirts occur in clusters of 2 or more.
It's easier here to count the ways this does not happen; which is when red and green shirts alternate one by one.  That is the two arrangements of $\{\mathrm{r,g,r,g,r,g,r,g}\}$ and $\{\mathrm{g,r,g,r,g,r,g,r}\}$.
$$\begin{align}\Pr(R\geq 2\cup G\geq 2) & = 1-\Pr(R=1\cap G=1) \\ & = 1-\frac{1}{35} \\ &= \frac{34}{35}\end{align}$$
