Re: Count Cards, Not Card Ranks
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Yes, I said in the OP that on the flop you are "about 50/50". Your numbers are more correct though, as I knew you would still be ahead most of the time. 6:5 seems more accurate.
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Checking with Pokerstove, it turns out that on the flop {ATs, Ato} is almost exactly a 6:5 favorite over {AA, AKs-A2s, AKo-A2o} on a flop of Ac 9d 6h -- 54.8:45.2, or 6.06:5.
If you think that a 6:5 favorite is "about 50:50", then I invite you to lay 6:5 odds on tossing a coin, 100 times over, and see how close to even you are afterwards.
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But when the turn and river also bring uncoordinated undercards, you become something like a 2:1 dog by showdown. Right?
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IF we start by specifying that the five cards on the board are A 9 8 6 4, and then say that one player holds AT and the other player has one ace and a random second card, THEN the player holding AT is a 3:2 dog to hold the winning hand.
(I see what the problem here might be: if you think that 3:2 is "something like 2:1", then it is understandable that you might see 6:5 as "about 50:50." [img]/images/graemlins/wink.gif[/img])
But remember this: AT is a 3:2 dog to an ace with a random kicker when you start out by specifying the board as A-9-8-6-4. But when you start with a flop of A-9-6, the odds are very close to 3:1 against winding up with a board with an ace and four other non-pairing cards lower than your kicker.
In setting up this situation you are literally stacking the deck against the player holding AT.
Instead of posing this situation as some sort of "proof" that medium kickers are bad news, you might ask the question of what's better for a player holding a bare ace: for the flop to come with an ace and big cards or with an ace and small ones?
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