Combinatorics - objects arranged in a circle Trying to solve a probability question, using indicators I managed to reduce it to what should be two simple combinatoric probability questions (below), yet I seem to get it wrong and I can't figure out why.

Given 25 marbles, 10 green and 15 brown, arranged in a circle. What is the probability that an $i_{th}$ green marble has brown marbles on both his sides along the circle?

This one I think I have gotten right:
$$\frac{\binom{3}{1,2}*\binom{22}{9,13}}{\binom{25}{10,15}}=\frac{21}{46}$$
Also fits nicely with the rest of the computations. The next however I am certain I have gotten wrong, and I can't figure out why:

Given the same situation as above, yet this time we assume that an $i_{th}$ green marble has brown marbles on both his sides along the circle. What is the probability that $j_{th}$ ($j\neq i)$ green marlbe also has brown marbles on both his sides along the circle?

Approaching this I looked at two distinct events: the green marble is in one marble distance than the $i_{th}$ green marble (and thus next to a brown marlbe), and otherwise. Keeping that in mind I have reached:
$$\frac{2*\binom{20}{8,12}+\binom{3}{1,2}*\binom{19}{8,11}}{\binom{22}{9,13}}$$
This leads to a plausible probability that may full down with the rest of the computations, but it is still (logically) way too high. I am obviously missing some basic concept in the counting here, if anyone could point me to my mistake I will appreciate it!
 A: EDIT: The following is for unordered arrangements or marbles; it does not take into account the fact that we are looking at the $i$-th and $j$-th green marbles.
For the first question, we fix the position of the green marble in question. This leaves 9 green and 15 brown marbles unaccounted for. We now pick the neighbors of the green marble. There is a $\frac{15}{24}$ chance of the first neighbor being brown and a $\frac{14}{23}$ chance of the second neighbor being brown given the first neighbor was brown. This gives a total probability of $\frac{15}{24}(\frac{14}{23})=\frac{35}{92}$ of both neighbors being brown.
For the second question, as you have done, we will consider two cases: first where the marble in question is next to the set green marble's brown neighbors, and second where it isn't.
First case: There are two spots next to the set green marble's neighbors, so the probability that the green marble in question is in one of these spots is $\frac{2}{22}$. Given that it is in one of these spots, the probability that its free neighbor is brown is $\frac{13}{21}$ since there are 13 brown marbles unaccounted for and 21 total marbles unaccounted for.
Second case: The probability of the green marble in question not being next to the set green marble's neighbors is just the opposite of the probability that it is, so this probability is $\frac{20}{22}$. From here, we pursue a line of thought similarly to in the first question. We have two spots we want to fill with brown marbles, and there are 13 brown marbles and 8 green marbles unaccounted for. Hence the probability of the green marble in question getting two brown marbles as neighbors in this case is $\frac{13}{21}(\frac{12}{20})=\frac{13}{35}$
Putting this all together, the probability of the green marble in question having brown neighbors given a set green marble has brown neighbors is
$$\frac{2}{22}\bigg(\frac{13}{21}\bigg)+\frac{20}{22}\bigg(\frac{13}{35}\bigg)=\frac{13}{33}$$
A: Here is a solution for the first problem, showing your initial approach is not right. First assume $i$ is neither $1$ nor $10$ (since those are special cases, cf. my comment). Then one can count the configurations in which brown-green-brown occurs with the green in position $i$ among the green marbles as follows: take all arrangements of $10$ green marbles and $15-2=13$ browns ones (there are $\binom{23}{10}$ of those), search the $i$-th green marble and add brown marbles to its left and right. This precisely gives any good configuration once, so for such $i$ the probability that in a random configuration of the problem the $i$-th green mable has brown ones at both sides is (good configurations/total configurations):
$$
  \frac{\binom{23}{10}}{\binom{25}{10}} = \frac{15\times14}{25\times24}
   = \frac7{20}
$$
For $i=1$ we must add to the configurations obtained as above, those good configurations in which the very first marble is green (since these are not obtained by the above method). So our circle starts green-brown and ends brown, with any arrangement of the remaining marbles in between. There are $\binom{22}9$ such arrangements, so we get a total probability of a good configuration for $i=1$ of
$$
\frac{\binom{23}{10}+\binom{22}9}{\binom{25}{10}} = \frac{231}{460}.
$$
The value for $i=10$ is the same as for $i=1$, for symmetry reasons. So we get two different values, depending on$~i$, one significantly smaller than the value $\frac{21}{46}$ you found, the other significantly larger.
If one secretly marks one of the green marbles before shuffling, and lets $i$ be determined by the position of the marked marble after shuffling, then every value of $i$ is equally likely to occur. If you take the average over all$~i$ of the probabilities found, you get
$$
  \left(8\times\frac7{20}+2\times\frac{231}{460}\right)/10 = \frac{35}{92}
$$
as the probability that the marked marble has two brown neighbours. You can calculate this number more directly, as is done the other answer, but this does not answer the first question you asked, for any $i$.
