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#1
12-07-2004, 06:49 AM
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Stupid Question - Hold\'em: Do you get AKs less often than AA?
I had always just assumed that AA came less frequently than AKs, but after looking at my PT stats, and having AKs come in 1/2 the time AA did, I checked the math, and was surprised.
AA seems to be drawing to 4(aces) then to 3(aces): <font class="small">Code:</font><hr /><pre>4/52 * 3/51 = ~.0045</pre><hr /> and AKs seems to be: Drawing to 8(aces and kings) and then 1(matching A or K): <font class="small">Code:</font><hr /><pre>8/52 * 1/51 = ~.0030</pre><hr /> Is this right? 
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#2
12-07-2004, 08:59 AM
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Re: Stupid Question - Hold\'em: Do you get AKs less often than AA?
yes
there are 12 ways of dealing AA in two cards, 4 ways of dealing AKs, and 4 ways of dealing KAs thats 12 to 8, or 3 to 2 frequency your answer is .0045 to .0030, or 3 to 2
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#3
12-07-2004, 06:00 PM
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Re: Stupid Question - Hold\'em: Do you get AKs less often than AA?
[ QUOTE ]
yes there are 12 ways of dealing AA in two cards, 4 ways of dealing AKs, and 4 ways of dealing KAs thats 12 to 8, or 3 to 2 frequency your answer is .0045 to .0030, or 3 to 2 [/ QUOTE ] I get the same result, but I don't get your math. 6 combos of AA - AsAc, AsAh, AsAd, AhAd, AhAc, AcAd AKsuited = 4 combos suited in hearts, diamonds, spades, clubs the way you calculate it, they are doubled because you are counting AsKs different from KsAs....they are the same hand.
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#4
12-07-2004, 06:16 PM
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Re: Stupid Question - Hold\'em: Do you get AKs less often than AA?
the way you calculate it, they are doubled because you are counting AsKs different from KsAs....they are the same hand.
yes they are that is why i said they were the ways of dealing two cards you can deal AcKc, KcAc, AdKd, KdAd, etc - same hand, different ways of dealing them out in two cards same end result your result is 6 to 4, or 3 to 2 as long as you are consistent throughout, the proportions will be equal whichever way you choose to look at them
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#5
12-07-2004, 06:17 PM
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Re: Stupid Question - Hold\'em: Do you get AKs less often than AA?
Sossman,
He's doing the math based on ordered pairs of starting cards. That is, A[img]/images/graemlins/diamond.gif[/img]K[img]/images/graemlins/heart.gif[/img] is different from K[img]/images/graemlins/heart.gif[/img]A[img]/images/graemlins/diamond.gif[/img]. Personally, I find it more intuitive to work with unordered cards when calculating probabilities like these, but both methods will work. gm
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