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Poker is a huge function of math. However, you are only looking at one part of the math. Let's say I have 20% chance of winning the pot, and villain bets $1 into the already $1 pot, thus making the pot $2, and me being faced with a $1 bet. The pot odds I'm being given is 2:1, which means I need 33% equity, yes? Unfortunately, I only have 20% equity, so a call would be incorrect. So it seems as if a fold is correct, as a fold has an Expected Value of 0 [0EV], and a call has a negative Expected Value [-EV]. However, that doesn't tell the entire story. We could consider implied odds, meaning the amount we stand to win if we hit, which could push this to a call. Better yet we can look at bluff/semi-bluff raising, and it could very easily be +EV if villain is folding with a high enough frequency. This doesn't mean we are going against the math, it just simply means we are considering more of the "equation".
Here's some math to look at the previous example, in which we are faced with a potsized bet, and we assume we only have 20% equity against his betting range. To make it easier, let's assume we have a draw, and therefore, 20% equity against the majority of his range.
EV[call] - The EV of us calling his bet
EV[call] = (Equity we have)(total pot) - (amount we called)
EV[call] = (0.20)(3) - 1
EV[call] = 0.6 - 1
EV[call] = -$0.40c
This shows that us calling his bet is -EV given the immediate pot odds we have, and our pot equity. And we know a fold is 0EV, so folding is a better option than calling here, assuming no implied odds (which likely isn't the case, but for examples sake).
Now let's see how often he need to fold to a raise to make raising profitable. We are assuming we raise his bet of $1 to $3 in this example. Different bet sizing from him or us would change the EV of a raise some.
EV[raise] = (% he call)*(Ev[call]) + (% he folds)*(EV[fold]), where X = % he folds
EV[raise] = (1-X)*(-1.6) + (X)*(2)
0 = -1.6 + 1.6X + 2X
1.6 = 3.6X
X = 1.6 / 3.6 = 0.44 = 44%
EV[call] - This is different than the before EV[call] function. This is now dealing with our EV when he calls our bet.
Ev[call] = (0.20)(1 + 3 + 3) - 3
Ev[call] = -$1.60
Ev[fold] - The EV we gain everytime he folds to our raise. Which everytime he folds to our raise we win the pot of $2.
EV[fold] = $2
Unless my math is incorrect, this shows that if villain if folding 44% of the time to your raise, then your raise is breakeven, and is therefore just as good of an option as folding. If he folds more often, then it's +EV to raise here. It needs to be noted, that this depends on other factors. The ability for villain to 3bet us, and not allow us to realize our equity, or re-bluff us, or our outs being tainted will play a part, and make the EV less and less. Those are assumptions you would need to take into account when you have enough information about the villain.
So as you see, then though one part of the math says that a fold might be correct doesn't make it so. You have to consider all possible options, and consider all of the math that comes with it. Here we considered the Equity we would have when called, the frequency with which he folds to a raise, etc. So, your not going against the math, you are simply using more math.
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