I tried searching through 2p2 as well as google but couldn't really find anything definite.

So I ask... Can anyone compute, or tell me how to compute the odds of quads over quads happening at a 9man table with the 2 players each being dealt a pocket pair?

It really depends on how you want to define the situation.

Are you asking, "Given that two people have a different pocket pair, what is the probability that they both make quads by the river?"

That is different from this question, "Given a 9 handed table of poker players, what is the probability that (at least?) two people will be dealt a pocket pair, and that both of these people will see the flop, turn, and river, and that when their hands are turned up they will both have quads."

The second question is different for a few reasons. First it is different mathematically because it requires that we compute the probability that at least two people get dealt a pocket pair at a 9 handed table. Second, it is different practically because it also requires that they see a flop, turn, and river. That is, neither player can fold his hand. In practice, people sometimes fold a naked pocket pair when stacks are deep, either on the flop or preflop.

So if you want a question answered, your really need to be very specific about the scenario of interest and exactly what probability you want to know. Different questions have different answers.

I suppose that's why I can't figure it out myself. I wasn't thinking in that much detail.

I guess I'm just curious about the odds of 2 pocket pairs making quads by the river on the same hand. So assume 2 players start with 2 pocket pairs, and they're going to see all 5 community cards. How often would it be expected to see both players hit 4 of a kind?

44/C(48,5) = 1 in 38,916

There is just 1 way to choose 4 of the board cards, and 44 ways to choose the 5th board card, out of C(48,5) total possible boards.

those odds are bigger than I expected, you sure you've calculated correctly?

BruceZ is pretty much ALWAYS correct.

If you would like a semi-intuitive way of verifying these odds, how about...

There are 48 cards left, and 4 cards need to hit the board, so their are 44 non-trip/quad cards.

Lets say the first card can be any non-trip card: 44/48
...and the second card has to be one of 4 cards out of 47: 4/47
...and the third card has to be one of 3 cards out of 46: 3/46
...and the turn has to be one of 2 cards out of 45: 2/45
...and the river has to be the last card: 1/44

So multiply these up we get 1/194580

But the first card doesn't have to be the blank... it can be in five different spots. So divide the odds by 5...

And you get 1/38916.

If you don't understand either BruceZ's answer or mine... then I can't help you.

How come you guys make it sound so simple and when I try to figure this stuff out on my own I end up with a headache and 100 wrong answers?

Because it is simple.

No idea. Experience, maybe?

To put it in a nicer way, it is because we sit here and do this all day. You can do it too.

When you're first starting out, it's easy to justify an incorrect solution. Then over time, you learn which techniques work and which do not, and you stick with ones that work. That's one level. Another level is to be able to do any problem a variety of correct ways, and to recognize specific errors in other's calculations. For example, alot of people could do the problem jomony asked, but it would be harder for them to pinpoint the exact error he made. Once you can do that, you can fix his error, and give him a working solution that is as close as possible to the one he attempted so he can learn from his mistake. That is what I do, and I get alot of satisfaction from it.