[switters]

Dave - your math is flawless, as near as I can tell - but you are missing a key point...

You are right that "in a vacuum", my opponent will only have 2 [img]/images/graemlins/diamond.gif[/img]s with a predictable probability, and that the math will always wash out to be the same as if I assumed there were 9 [img]/images/graemlins/diamond.gif[/img]s left in the deck. But what if I have extra information about my opponent that indicates the cards in his hand better than strict probability would allow?

for example, let's say I open raise from middle position with 7 [img]/images/graemlins/diamond.gif[/img] 8 [img]/images/graemlins/diamond.gif[/img], I'm called by Total Unknown Guy and three-bet by Rocky Tight Guy. both me and TUG call the three-bet.

flop: 2 [img]/images/graemlins/diamond.gif[/img] J [img]/images/graemlins/diamond.gif[/img] Q [img]/images/graemlins/club.gif[/img]

I decide to bet out my flush draw, and Total Unknown Guy raises me. Now, Rocky Tight Guy *calls* the two bets cold. Let's say I know rocky tight guy well enough to know the following:

* a three-bet pre-flop means AA, KK, AKs, or QQ, and nothing else.
* on this flop, he would raise with a set of queens to protect his hand. He would three-bet with AA or KK one time to see if he's still ahead, and then shut down (fold or check-call, depending on the action) unless he improved on the turn.
* BUT, he does understand pot odds, and since we can't ever make it more than 2 bets to him cold... if he held a big draw to the nuts (12, as far as he knows), he would call any number of bets, two at a time, all the way to try to hit.

now, this is a *little* contrived, although I have certainly played against players this tight and this readable, and they're pretty common at the 10/20 - 20/40 level. The point is, I think in this situation I could say with about 90% certainty that the Rocky Tight Guy is sitting on A [img]/images/graemlins/diamond.gif[/img] K [img]/images/graemlins/diamond.gif[/img], and since Total Unknown Guy is likely to hold a Q or a J on that flop, I'm probably dead to running 8's and 7's - if I'm lucky.


The math you did is still really useful, though -- you showed how you can arrive at the "all the cards are unseen" numbers by figuring out the probability that your opponents are on a flush draw, using no additional knowledge. If I can use extra table knowledge, however, to say that player X is 50% likely, given his play, to have 2 of my [img]/images/graemlins/diamond.gif[/img] outs, rather than the 3% that pure probablity allows for, I can plug that number in instead, and recalculate my odds accordingly.