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Of course I factored them into the equation. If you have a 57% chance of drawing 2 3 4 5 6 7, then you have a 43% chance of catching all the other cards (A 8 9 T J K Q). I'm just interested in the cases where you catch good and he catches bad. There are 4 branches to this probability tree and we're only interested in how often 1 particular branch happens.

And yeah, technically if he catches a T and you catch a 9 your hand looks better. But I don't know how often he'll fold there and you want him to, because his hand IS better.

But this is kind of balanced out by the fact that there are a couple cards each of you can catch that look good but aren't. If you want really comprehensive results, you'd probably need to enumerate all the cases and consider how often you'd get cases you like. The math I did above is kind of shorthand... like you're not going to be sad at all if you get a 9 and he gets a K even though he's still ahead, because he might fold.

The cases are enumerable but I don't think you'd want to do it by hand. There are 160-ish distinct 3rd streets (that's kind of just a ballpark guess, actually probably a bit fewer)

And yeah of course I am not factoring in known up cards. These can considerably sway the decision one way or the other.

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Here's the problem the way I see it. Let's say I have T2/A v xx/A. For me to at least appear to outdraw my opponent, I must catch a 2, 3, 4, 5, 6, 7, or 8 while my opponent catches an A, 9, T, J, Q, or K. So that's 27 cards for me and 21 for my opponent. Assume it's an eight-handed game, so there are six other cards out, but they're all random, so I'll just consider them unknowns. So that's: (27/48)*(21/47) = .56*.45 = .25 = 3-1. If I assume my opponent will still call me if he catches a 9 and I catch an 8 or better, then the odds drop to 4-1. But is this a reasonable assumption?

I guess this kind of goes back to what I was trying to say before but might not have said very well. There are few instances where you catch "good" and your opponent catches "bad" but not "bad" enough to warrant a fold. Those instances might be: xx/A8 v xx/A9, xx/A8 v xx/AT, (maybe) xx/A7 v xx/A9 . But in all other instances, you'll have xx/A[insert 2,3,4,5,6,7 here] v xx/A[9,T]. So far more often that not, 9s and Ts are bad cards for your opponent. But again, I'm not sure how to put an exact figure on that.

In any event, I think I learned something here. Thanks for the responses.