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Hole'em poker 1on1 same pocket pair

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SimonVigor | 01:35 Wed 28th Sep 2005 | Science
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In Texas holdem poker, a heads up (1on1) game, what are the odds that both players get the same pocket pair? For example in a game i played recently, we both had pocket nines, i say its about 280,000 to 1 for this to happen, but the guy i was playing claims it "WAY over 10 million to 1" Help us out... what are the actual odds?
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I don't know the format of the poker game you describe. How many cards are the two pairs selected from (i.e. are they among your hand of five, or are they somewhere else, or whatever?). Please describe the game and I'll try to help.
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The format is each player is dealt 2 down cards... so 4 cards are dealt between the 2 players total... 2 players-4 cards, all 4 cards of the same rank... ie, 4 nines .
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SO basically, what are the odds of shuffling a full deck and haveing four of a kind on the top?

I reckon that if you don't care what the cards are as long as all 4 are the same, then it's 3/51 x 2/50 x 1/49 which is 1/20825

Yep, hammer is correct.
I've just calculated the probability of you both getting a specified pocket pair, and that is around 1/280,000 so I assume that what you worked out SimonVigor. The other bloke was just talking rubbish!
Spot on. One chance in 20,825 for both players being dealt the same pair of any denomination. One in 270,725 for them being dealt the same pair of a nominated denomination.

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