5 pocket pairs in one hand

[EigorNosam]

Saw something last evening I thought must be rare and I've been trying to find the odds of it happening. Saw 5 players in a 9-ring game table pocket pairs in the same hand: 8-8, 9-9, J-J, Q-Q and A-A. Thinking the odds must be extremely low but was wondering how to calculate.

 

[Vlaad]

My approach would be (3/51)*(3/49)*(3/47)*(3/45)*(3/43) = 0.0001069 %.

My thought process: The first card of a hand is irrelevant. The second one however has to be the same as the first for it to be a pocket pair. There are 3 cards left in the deck (which now only consists of 51 cards, because the first card is already dealt) which give you a pair, so the probability of you getting a pocket pair is 3/51, or 5,88 %. You do this for every hand (with accounting for less cards in the deck) and multiply each probability.

I might be wrong though.