chances of two players with two pair by the river, 9 handed holdem without the board pairing. if it makes a difference, with the players sharing 1 card? and with both players having unique cards?

I don't think this can be answered in a meaningful way without restricting the problem. First, no player can have a pocket pair and meet the criteria. With 9 players, there is about a 42% chance at least 1 has a pair. Ok, you can actually figure out a distribution and work with that. The more critical issue is how many of the non-pocket pair players go to showdown.

So, to provide an answer, I'll consider heads-up and assume neither player has a pocket pair and both go to showdown.

No shared rank: P(exactly 2 pairs for each player) = (C(3,1)^4)*36/C(48,5) = 0.17%

One shared rank: P(exactly 2 pairs for each player) = C(2,1)*(C(3,1)^2)*C(40,2)/C(48.5) = 0.82%

Note: Above does not exclude flushes and straights and for the shared rank case, the board can have a pocket pair but the players do not need it for a second pair.

With 9 players, the results will be higher but I don't think that they will be much higher. Most games don't get to showdown or have more than 2 or 3 players seeing the river, so I doubt that if you tracked actual games the results would be over 1% or 2%.

Setup:
- 1 million simulated games in which the board is not paired
- 9 players
- all players go to the river
- only two pair as a hand are counted (better hands are not counted!)

0 players having exactly two pair as best hand:44,088 % of games
1 players having exactly two pair as best hand:40,6066 % of games
2 players having exactly two pair as best hand:13,2911 % of games
3 players having exactly two pair as best hand:1,8958 % of games
4 players having exactly two pair as best hand:0,1155 % of games
5 players having exactly two pair as best hand:0,003 % of games
6 players having exactly two pair as best hand:0 % of games
7 players having exactly two pair as best hand:0 % of games
8 players having exactly two pair as best hand:0 % of games
9 players having exactly two pair as best hand:0 % of games


Please take this with a grain of salt. This is my own code (i.e. not reviewed)
The thing about shared/not shared cards might take a bit of effort, tho.

Here are the results of a simulation I just performed over 1,000,000 deals of a 9-player table at which everybody goes to showdown.

Each deal: the board has five different ranks (not paired)

Hole cards of the 9 players are randomly dealt.

Tally of deals with at least two players making two pair (flushes and straights are ignored):

0 ranks overlap = 63,276

1 rank overlap = 84,695

2 ranks overlap = 9,274

TOTAL = 157,245 (which is, of course, 15.7245%)


Note that if more than two players have two pair, I tallied the first two of them only.

I let the simulation run out 10,000,000 deals just to give more credibility to the results.

Tally of deals with at least two players making two pair (flushes and straights are ignored):

0 ranks overlap = 630,555

1 rank overlap = 847,709

2 ranks overlap = 94,161

TOTAL = 1,572,425 (which is, of course, 15.72425%)