odds any given player is dealt a PP:
First player: First card can be any card in the deck, but second card must be the same number card, of which there are 3.
52/52 x 3/51 = 3/51 = 0.0588 = 5.88%
Which is what you said, except that you didn't quite remember how decimal to percentage changes work. If you're only talking about a few players you could multiply it by the number of players, but this is not quite correct. Assume the first player was NOT dealt a pocket pair. Now for two numeral sets there are only 3 cards of each of those number remaining in the deck, so things get tricky.
Second player: Again, the first card doesn't matter (50 cards remaining). For the second card, 11/13 times they have 3 ways to draw it, but 2/13 times they have only 2 ways to draw it.
50/50 x ((11/13 x 3/49) + (2/13 x 2/49)) = .0508 = 5.08%
Notice that it's a bit smaller. This calculation gets more complex the more players there are, and depending on what assumptions we make. I'd be curious if anyone has a simpler, more elegant way of tackling the question.
