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#1
12-19-2005, 03:20 PM
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Re: Absence of ACES before the FLOP
Assume 10 handed. In the first example, you have no knowledge whatsoever, including your own cards. In the second example you have knowledge of your cards.
First: C(48,20)/C(52,20) Second: C(46,18)/C(50,18) 
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#2
12-19-2005, 06:58 PM
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Re: Absence of ACES before the FLOP
[ QUOTE ]
Assume 10 handed. In the first example, you have no knowledge whatsoever, including your own cards. In the second example you have knowledge of your cards. First: C(48,20)/C(52,20) Second: C(46,18)/C(50,18) [/ QUOTE ]
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#3
12-20-2005, 09:33 AM
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Re: Absence of ACES before the FLOP
OK, I'm slow but I'm almost there and I'm still missing the obvious.
First: C(48,20)/C(52,20) Second: C(46,18)/C(50,18) I see what the 20 and 18 are - unseen dealt cards. Where do the other numbers come from? This newbie thanks you for your patience and help.
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#4
12-20-2005, 10:09 AM
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Re: Absence of ACES before the FLOP
First: Number of combinations you can deal 48 non-aces when you deal 20 cards divided by the number of combinations you can deal 52 cards when you deal 20 cards.
Second: Number of combinations you can deal 46 non-aces when you deal 18 cards divided by the number of combinations you can deal 50 cards when you deal 18 cards.
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