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#1
08-29-2005, 08:44 PM
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Re: An Update.
I think its very similar, if you have a board showing a pair and only 1 player playing, they will have quads about 1/1100 times. If they play ten hands, on this board, they will hit quads about ~1/110 times. If there are 10 players at the table, they will each have quads about 1/1100 times 1*10/1100 = 1/110. For every case the same thing can be done whether there are 3 of a kind on the board one pair or 2 pair.
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#2
08-30-2005, 02:16 AM
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Re: An Update.
[ QUOTE ]
I think its very similar, if you have a board showing a pair and only 1 player playing, they will have quads about 1/1100 times. If they play ten hands, on this board, they will hit quads about ~1/110 times. If there are 10 players at the table, they will each have quads about 1/1100 times 1*10/1100 = 1/110. For every case the same thing can be done whether there are 3 of a kind on the board one pair or 2 pair. [/ QUOTE ] I think this reasoning is flawed due to independence issues. Can someone else chime in? -RMJ
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#3
08-30-2005, 08:49 AM
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Re: An Update.
You definitely need to apply the inclusion/exclusion (or similar) principle here to get an exact answer. I was just reaching for a ballpark estimate that might be within the first standard deviation when propogated over a few thousand hands.
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