spoonitnow's 5000th Post: Game Theory and Poker

Originally Posted by MadMojoMonkey

The only 100% consistent result is that the lowest hand which can be bet for value is always the average of the limits of the caller's calling range. I.e. if the opponent will make the call with the range [0.2, 0.4], then the lowest number to V-bet into that range is (0.2 + 0.4)/2 = 0.3. Very intuitive.

This kinda translates into "I need at least 50% equity against villain calling range to make a v-bet", doesn't it?

Originally Posted by daviddem

This kinda translates into "I need at least 50% equity against villain calling range to make a v-bet", doesn't it?

Exactly.

Some interesting things about EV calc's:
1) All EV calc's start positive (nut hands are winners), and
2) always have a slope less than or equal to 0 (good hands win more than, or at least as much as, bad hands),
3) whether Hero bets, calls, checks or folds (no line will make a worse hand have more EV than a better hand when both are played correctly).

For the mathematically literate:
Given that EV is a function of hand x and betting line L (where L includes all prior actions and the current action), then:
All EV(x|L) are monotonically decreasing functions.

conclusions, taking into account that a fold is always 0EV:
If the EV of a call falls below 0 for any value, then a fold is less -EV than a call for all greater values. (If it's -EV to call with hand x, then it is -EV to call with any hand that can't beat x.)

If the EV of any bet is below 0 for any value, then Hero can not bluff for exploitation. (When Villain calls too much, Hero exploits by not bluffing.)