both hole pairs the same

What are the odds of 2 players getting the same pair in the hole in Texas Hold Em?

Quote: hunterbao

What are the odds of 2 players getting the same pair in the hole in Texas Hold Em?

The odds of any two specific players getting a pair of the same hole cards is (13*COMBIN(4,2)/COMBIN(52,2))*(1/COMBIN(50,2)), which is 1 in 20,825.

Quote: hunterbao

What are the odds of 2 players getting the same pair in the hole in Texas Hold Em?

There was a great knockout in the WSOP last year - it's kind of famous and can be found on youtube. Worth a watch if you haven't seen it; a million-dollar buy-in eats the sickest bad beat ever.

Quote: hunterbao

What are the odds of 2 players getting the same pair in the hole in Texas Hold Em?

Depends how many players are at the table. Heads-up it's what the Wiz said, though since all configurations of the 4 cards are valid, you can arrive at the answer more simply:
13 / C(52,4) = 1 in 20825

Quote: Wizard

The odds of any two specific players getting a pair of the same hole cards is (13*COMBIN(4,2)/COMBIN(52,2))*(1/COMBIN(50,2)), which is 1 in 20,825.

If the player has XY (not a pair) then the probability of any other given player having those same two card ranks is (6/50)*(3/49) = 1-in-81.7. This does not distinguish suited from unsuited. If a player has a pair XX, then the probability of any other given player having that same pair is (2/50)*(1/49) = 1-in-1225.

If you ask the question about how often the situation XX vs. XX comes up between two given players, that's 13*(4/52)*(3/51)*(2/50)*(1/49) = 1-in-20825, as Mike said.

I think I am not understanding the question.

No, I never saw this. So, so sick.