Donate SIGN UP

Hole'em poker 1on1 same pocket pair

Avatar Image
SimonVigor | 02:35 Wed 28th Sep 2005 | Science
7 Answers
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?
Gravatar

Answers

1 to 7 of 7rss feed

Best Answer

No best answer has yet been selected by SimonVigor. Once a best answer has been selected, it will be shown here.

For more on marking an answer as the "Best Answer", please visit our FAQ.
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.
Question Author
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 .
Question Author
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.

1 to 7 of 7rss feed

Do you know the answer?

Hole'em poker 1on1 same pocket pair

Answer Question >>

Related Questions

Sorry, we can't find any related questions. Try using the search bar at the top of the page to search for some keywords, or choose a topic and submit your own question.