probability of two teams meeting in the group stage of Champions League in the recent group stage draw for the Champions League, 32 teams were assigned to 4 separate pots, 1, 2, 3, 4 as follows:
Pot 1: Barcelona, Bayern München, Chelsea, Benfica, Paris, Juventus, Zenit and PSV Eindhoven. 
Pot 2: Real Madrid, Atlético, Porto, Arsenal, Man. United, Valencia, Bayer Leverkusen, Manchester City
Pot 3: Shakhtar Donetsk, Sevilla, Lyon, Dynamo Kyiv, Olympiacos, CSKA Moskva, Galatasaray, Roma
Pot 4: BATE Borisov, Borussia Mönchengladbach, Wolfsburg, Dinamo Zagreb, Maccabi Tel-Aviv, Gent, Malmö, Astana
the draw entailed teams from pot 1 all being drawn first, one after the other into 8 groups A, B, C, D, E, F, G, H. 
the process is then repeated for team in pot 2, then pot 3, pot 4 etc.
what are the odds of a team from the first pot meeting a team from the second pot in the(any) group ? (excluding special cases, where teams from the same association/country cannot meet each other)
e.g. what are the odds of BenFica and ManUtd being drawn into the same group ?
 A: You can't exclude the special cases because they influence all cases. Without the special cases, the probability would be $1$ in $8$ (BenFica is drawn into some group, and then ManUtd is drawn into that same group with probability $1/8$). However, since some cases are excluded, all other cases become more probable, so the probability for the cases that aren't excluded is somewhat above $1/8$.
A: I did a small numerical simulation taking the CL draw rules into account (I reject groups with two teams from the same country). This gives for example
$$P(\text{ManUtd in group with Barcelona}) = \frac{113}{549}\approx 20.5\%$$
higher than the naive $12.5\%$ found by neglecting them. For your particular question of Benfica we get
$$P(\text{ManUtd in group with Benfica}) = \frac{79}{549}\approx 14.4\%$$ 
just slightly higher than what we naively would expect.
Below is the simple Matematica code I used. Teams are given a number where the $100$-digit signifies the country.
pot1 = {100, 200, 300, 400, 500, 600, 700, 800};
pot2 = {101, 102, 103, 201, 202, 203, 301, 401};
nok = ntot = 0;

(* Check all possible groups *)
perm = Permutations[pot2];
Do[
 draw = perm[[i]];

 (* Check if current group is valid *)
 ok = True;
 Do[
  If[Abs[pot1[[j]] - draw[[j]]] < 10, ok = False; Break[] ]; 
  , {j, 1, 8}];

 (* Check if ManUtd (= 201) is in Barcelonas group *)
 If[ok, ntot += 1; If[draw[[1]] == 201, nok += 1]];
 , {i, 1, Length[perm]}]

 Print["P = ", nok / ntot];

