If it showed nothing new, I don't think I'd have posted it.

*Originally posted by DukeFreedom*

Then you'll see that EV[b/f] > EV[c/c] **> EV[c/f]** then.

As for your second question about the 1/(bets in pot)...

Lemme first just quote what Cornholio wrote

If you assume your opponent will __check behind all worse hands__ in case you check, but bet for value all better hands, it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot). If the opponent has a better hand, you would lose one bet, and you won't get a bet from a worse hand.

He worded this very poorly and actually it's even wrong when this is what he really ment, because as I said before: if our equity <50% and we assume your opponent always checks behind worse hands when checked to

**then c/f is always the best move**.

Instead, he should have written

If you assume your opponent will **play optimally** in case you check, it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot).

Now note that

**playing optimally is *NOT* the same as always checking behind worse hands**. To see this quickly, take your example and assume that your opponent will bluff 1/11 times when he bets (when you check to him). In that case you can't call (actually: you'll be indifferent between calling or folding), because 10/11 times you'll loose 1 BB and 1/11 times you'll win the pot of 10 BB, giving a net EV of 0, the same as c/f.

In fact, "playing optimally" is very precisely defined for your opponent according to game theory, but I really don't want to get into explaining the exact theory behind this... Basically it just says that your opponent will bluff with such a frequency that you'll be indifferent between c/c his bet or c/f.

Hence this "optimal playing strategy" for your opponent dictates that both c/c and c/f will have an EV of 0 for you, as with the example I gave above.

However, b/f will have an EV of > 0 if your opponent calls worse hands with a probability P > 1/B, where B = bets in pot. This follows, because:

with probability (1-P) we lose 1

with probability P we win B+1

Now the net EV of b/f will be P*(B+1) - (1-P)*1= P*B - 1.

Now P*B - 1 will always be > 0 if P > 1/B.

Hence b/f will be better then c/c and c/f if we assume our opponent calls worse hands with probability > 1/(bets in pot) and would play optimally if we check.