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Saborion
07-25-2003, 12:09 AM
How do I calculate the chances of my pocket Q's holding up against either AA, KK or AK? And so on.

Seen some earlier post a while ago when people used something like C[x,x]. Anyone care to make a short explanation of that one? Seems like something one should know. /images/graemlins/smile.gif

Copernicus
07-25-2003, 12:13 PM
C(x,y) stands for the total number of combinations possible drawing y cards out of x total cards. It is an Excel function COMBIN, or can be calculated from Factorials or manually when y is small. The formula is x!/(y!*(x-y)!). For example for your pockets its C(52,2)=52!/(50!*2!). Note that since y is small, almost all of the 52! is cancelled out by the 50!, so all that is left is 52*51/2=1326, the number of possible pocket cards you could have.

Your problem requires several different steps to determine whether the Qs hold up when you consider the long shot hands such as 4 of one of your suits and no one else having them, making a straight that doesnt stretch to the AA, etc. The majority of times it holds up, though, it will be because no A or K appears on the board, and is a close enough approximation. That would be calculated vs AA for example, as C(46,5)/C(48,5). In English, the total number of combinations for the board is 5 cards out of the 48 unknown (after the QQAA). The winning combinations are 5 cards out of the 46 after removing the other two Aces.

You would get the same for the KK. For AK you have to remove 6 cards from the deck instead of just 2, so it becomes C(42,5)/C(48,5).

When it gets as complex as finding the complete answer, you are far better off using software that does it for you. Ive seen www.twodimes.net (http://www.twodimes.net) recommended here, but I dont like it. I use Hold'em Analyzer, but its not free.

Saborion
07-26-2003, 12:46 AM
<font color="blue">"That would be calculated vs AA for example, as C(46,5)/C(48,5). In English, the total number of combinations for the board is 5 cards out of the 48 unknown (after the QQAA). The winning combinations are 5 cards out of the 46 after removing the other two Aces."</font>

Is this really true? If I`m going all in with QQ vs AA and we neglect straights and flushes, then only 2 cards will help me, while the remaning 46 will destroy me. Or?
Shouldn`t it therefore be C(46,2)/C(48,5)?

Someone has AK vs my QQ (and we neglect straights, flushes etc). Then that`s C(42,5)/C(48,5) right? Calculating that manually I got the answer ~0.49679 Should be 49.77 % that they hold up right? Not a lot. =)
Going all vs AA, KK or AK pre-flop is clearly not a wise move.

And that percentage should be the same if someone holds say 66 and goes all in against A7+ right?

And since all other cards except A or K won`t help the other player, this calculation (vs against AA or KK the way you stated it), should be correct.

If so, then if no A or K comes up at the flop, QQ will win against AK C(39,2)/C(45,2) = ~74.8 % of the times?

Am I understanding this now?

Copernicus
07-27-2003, 08:11 PM
You apparently understand it better than I do. /images/graemlins/blush.gif

I was so hung up on explaining the notation I spaced out on the problem itself. Thats why I use software to do the easy work (computation) that is also easy to screw up. Formulating meaningful questions and applying the results is the challenge.

Eg. knowing that AK is only a tiny underdog to lower pairs doesnt mean that you should go all in any time you get heads up with them and the blinds give you the pot odds you need.