View Full Version : Stupid Question - Hold'em: Do you get AKs less often than AA?
JoshuaD
12-07-2004, 06:49 AM
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?
Mike Haven
12-07-2004, 08:59 AM
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
SossMan
12-07-2004, 06:00 PM
[ 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.
Mike Haven
12-07-2004, 06:16 PM
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
gaming_mouse
12-07-2004, 06:17 PM
Sossman,
He's doing the math based on ordered pairs of starting cards. That is, A/images/graemlins/diamond.gifK/images/graemlins/heart.gif is different from K/images/graemlins/heart.gifA/images/graemlins/diamond.gif.
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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