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#1
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I know this is probally an easy question, but how many unique hold'em starting hands are there? Taking into account the different suits.
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#2
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169
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#3
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1,326
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#4
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Wait a minute. Are you looking for useful information or a statistics question?
Are you counting AT and TA as the same hand? |
#5
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Formula:
n! ----------- x!(n-x)! 52 choose 2, so: 52! ----------- = 1,326 2!*(52-2)! sorry the spacing sucks |
#6
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[ QUOTE ]
Are you counting AT and TA as the same hand? [/ QUOTE ] AcTc and TcAc are the same hand, and 8h9h and 8d9d are different hands |
#7
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![]() EDIT: I just re-read that he does want different suits. Please excuse my ignorance! For the sake of answering this person's question... the answer should be 169. This is just considering the hands suited or unsuited. I doubt he cares if he gets AhTh vs AcTc. List of hands expected EV Although I'm sure you are technically correct Vandal. |
#8
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The correct answer is 1326, as previously stated.
The formula is simpler than it was described. Since there are only 2 hole cards in holdem, the formula is 51+50+49...+1. The logic is simple. There are 51 possible combinations that contain a given card (example Ace [img]/images/graemlins/heart.gif[/img]); the ace and the 51 other cards. Next take the K [img]/images/graemlins/heart.gif[/img] there are 50 combos (not including the A [img]/images/graemlins/heart.gif[/img] K [img]/images/graemlins/heart.gif[/img] combo that has already been accounted for). Now take the Q [img]/images/graemlins/heart.gif[/img] ... you get the idea. I hope that helps. PG |
#9
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Or just 52 x 51 / 2.
GoT |
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