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Part 15: Break Even Fold Percentage for a Pure Bluff
Suppose we're heads-up against a single opponent. We have a pot size of P and we're considering making a bet size of B as a pure bluff, meaning we have no equity when called. Our opponent will fold F % of the time. There are two possible outcomes. In the first outcome, if he folds F% of the time, we win the pot P. In the second outcome, if he doesn't fold (1-F)% of the time, we lose our bet size B. Therefore, our EV equation is
EV = PF + (1-F)(-B)
Now let's suppose that we wanted to look when this pure bluff is break even. A break even play by definition has an EV of 0, so we let EV be 0 in our equation.
0 = PF + (1-F)(-B)
Now we can solve for F to find what fold % we need to have to make our pure bluff break even. From there, we'll know that if we can make our opponent fold more often than whatever that break even value is that our pure bluff will be +EV. So we have
0 = PF + (1-F)(-B)
0 = PF - B + BF
0 = PF + BF - B
0 = F(P + B) - B
B = F(P + B)
B/(P+B) = F
This shows that our break even fold percentage for a pure bluff is bet/(bet+pot), something a lot of you probably already knew.
Part 16: Break Even Fold Percentage for a Semi-Bluff
In part 13 we found that our equation for the EV of a semi-bluff shove is
EV = PF + E(1-F)(P+B) + (1-F)(1-E)(-B)
where P is the pot size, B is our bet size, F is how often our opponent folds, and E is how much equity we have against our opponent's calling range. If we let EV be 0 like in part 15, we can solve for our break even fold percentage in the same way, though it's a little more difficult, and a little more difficult to interpret the results. So here we go:
0 = PF + E(1-F)(P+B) + (1-F)(1-E)(-B)
0 = FP + E(P + B - FP - BF) + (1 - E - F + EF)(-B)
0 = FP + EP + BE - EFP - BEF - B + BE + BF - BEF
0 = FP + EP + 2BE - EFP - 2BEF - B + BF
0 = BF - EFP - 2BEF + FP - B + EP + 2BE
0 = BF - EFP - 2BEF + FP - B + E(P + 2B)
B - E(P + 2B) = BF - EFP - 2BEF + FP
B - E(P + 2B) = BF - EF(P + 2B) + FP
B - E(P + 2B) = F(B - E(P + 2B) + P)
(B - E(P + 2B))/(B - E(P + 2B) + P) = F
When our opponent’s fold % is greater than that big glob of stuff at the end (which we’ll simplify immensely in a moment), our semi-bluff is +EV. So now we have the task of figuring out how to do something with this. If we let B – E(P+2B) be an entity of its own (call it x), then we’d have F > x/(x+P). This is exactly like the equation we solved for earlier when looking at a pure bluff! Now let’s figure out what B – E(P+2B) means since it seems like it might be important. If you notice, (P+2B) is the total size of the pot after we bet and Villain calls. Then, E(P+2B) is our equity % of that total pot size. Finally, B – E(P+2B) is our bet size minus our equity % of the total pot size. So let’s see this in an example.
Say it’s 25nl and the pot is $30 and we have $20 behind heads-up. If we shove our $20 and we think we’ll have 20% equity when we’re called, how often does Villain have to fold for it to be +EV? First, we’ll find the total pot size after Villain calls, and that’s $70. Second, we’ll find our equity’s percent of that pot, and 20% of $70 is $14. Now we subtract $14 from our bet size of $20 to get $6, which is our x value. To finish, our Villain’s fold % has to be greater than x/(x+P) for our semi-bluff to be +EV, which is 6/(6+30) here, or 1/6 = 16.7%.
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