Quote Originally Posted by Pyroxene
Quote Originally Posted by SonOfAkira
The question is...
The odds of a specific table being dealt the exact same two card hand that is in play at another specific table is:

1 in 52!/((50!)*(2!)) or 1 in 1326 (.0754%)

the odds of the same hand happening and then matching the 3 card flop are:

1 in 1326 * 50!/((47!)*(3!)) or 1 in 25,989,600.

the general odds of the same hand happening and then numerically matching the 3 card flop with 2 matching suits and 1 unmatched (and no pairs) are:

1 in 1326 * 50!/((47!)*(54)) or 1 in 2,887,733.3 (the key here is that there are 54 combinations of the same ranks with 2 matching suits and 1 unmatched).

the more specific odds of all that happening and flopping two pair matching your hole cards

1 in 2,887,733.3 * 50!/((47!)*3*3*44*6) or 1 in 142,928,213

Quote Originally Posted by pokerroomace
The odds of being dealt the exact same hand on both tables in any order is 1 in 2,300,000. The same as being dealt a royal flush of spades (or a particular suit) with 5 cards.
I want to point out that this is not quite the questsion that was asked. While I agree that the odds of being dealt the same hand on both tables are 1 in 2,598,960 (52!/(47!*5!)), that would match situations where the hole cards on table one are :As: :Ks: and the flop is :Qs: :Js: :Ts: and the hole cards on table 2 are :Ts: :Js: and the flop is :Qs: :Ks: :As:. His question was a bit more specific that the hole cards had to match and then the flop had to match; which, of course, is less likely by a fair amount.
Ye. I didn't think of that. I meant to write 1 in 2.6mil. But it's higher than that as you said, because the two cards in your hands on both tables have to match as well.