Odds of a specific dice distribution occuring when playing Catan (2d6) So this is messing with me, because I have no idea how I would even go about starting calculating this.
I have played a game of Catan. For those that don't know, it's a board game that involves rolling two D6 per turn. So there are 12 possible outcomes for each roll, but each number has a different probability of occuring.
During a game, we had the following dice rolls occur over a 69 turn game:

*

*2: 5

*3: 6

*4: 5

*5: 11

*6: 7

*7: 9

*8: 8

*9: 9

*10: 0

*11: 8

*12: 3

This lead to the conversation about what are the odds of no 10s occuring (easy to calculate, it's (33/36)^69 = 1 in ~400). This further then lead to a conversation about how rare any given combination of rolls is. I just wanted to calculate some lower and upper bounds, but I honestly had no idea how to: A: Deal with the fact that the odds of each number are different
B: Deal with the fact that [A, A, B] = [B, A, A]
So my question is how do I go about calculating the probability of any given combination of numbers? Just getting me started looking in the right place would help a lot.

An upper bound is easy. The worse case is rolling all 2s (or 12s), for which the odds are 1/36^69 = 4e-108, but beyond that I'm stuck.
A "simpler" solution that I would also be happy with, would be the odds of the "perfect" distribution occuring during a 72 move game. So two 2s, four 3s, six 4s, etc. Every number perfectly matching its frequency. This also represents a lower bound for the probability of any given dice combination (in a 72 move game)
No idea how hard or complicated this is. But I thought I would ask. If there's somewhere better to ask, just redirect me and I'll ask there!
A: First of all it is required to compute the probabilities which are:
$$
\begin{array}{c|ccccccccccc}
i&2&3&4&5&6&7&8&9&10&11&12\\
p_i\times\frac1{36}&1&2&3&4&5&6&5&4&3&2&1
\end{array}
$$
The probability of a specific outcome $n_2$ of '2', $n_3$ of '3' and so on after $n$ rounds ($\sum_i n_i=n$) can be computed as:
$$
n!\prod_{i=2}^{12}\frac{p_i^{n_i}}{n_i!}.
$$
