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Semi-Bluff Stack Math
50/100
UTG opens large to 400 and 8500 chips.
I call in the cutoff with QJo and 2300 chips.
The button, sb, and bb all have 3000 chips but they fold.
Heads-up
KTT rainbow
UTG bets 350 into 950. I raise all in for 1850 effective.
Calculate the expected value of semi-bluffing
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Fold EV = (% opponent folds) * (amount hero wins if fold)
Fold EV = (%F) * (950 + 350)
Fold EV = (%F) * (1300)
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Call EV = (% opponent calls) * [ (% hero win) * (showdown win) – (% hero lose) * (amount raised)]
Call EV = (%C) * [(%W) * (950 + 1850) – (%L) * (1500)]
Call EV = (%C) * [(%W) * (2800) – (%L) * (1500)]
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Total EV = Fold EV + Call EV
Total EV = (%F) * (1300) + (%C) * [(%W) * (2800) – (%L) * (1500)]
Total EV = 1300F + C * (2800W – 1500L)
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*** %W + %L = 1 ***
*** %F + %C = 1 ***
Use these equations to replace C with (1 – F) and replace L with (1 – W).
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Simplify
Total EV = 1300F + (1 – F) * (2800W – 1500 * (1 – W))
Total EV = 1300F + (1 – F) * (2800W – 1500 + 1500W)
Total EV = 1300F + (1 – F) * (4300W – 1500)
Total EV = 1300F + 4300W – 1500 – 4300FW + 1500F
Total EV = F(1300 – 4300W + 1500) + 4300W – 1500
This was simplified to two variables (%fold and %win) or (F and W). If we make an estimation of one, we can calculate the break-even value of the other.
Break-Even occurs when Total EV = 0.
In the example, we have 8 outs to hit a straight. We can maybe assume that we win about 25% of the time with two cards to come. Thus, let’s estimate that %W = 0.25 and calculate how often we need the opponent to fold to break-even.
0 = F(1300 – 4300W + 1500) + 4300W – 1500
0 = F(2800 – 4300W) + 4300W – 1500
F = (1500 – 4300W) / (2800 – 4300W)
F = (1500 – 4300 * 0.25) / (2800 – 4300 * 0.25)
F = (1500 – 1075) / (2800 – 1075)
F = (425) / (1725)
F = 0.246 or 24.6%
In other words, we show a profit when
— the opponent calls less than 75% of the time
— the opponent folds at least 25% of the time.
I need my opponent to fold 1/4 of the hands in his range. For example, I expect to get called by all K’s and T’s but a lot of pocket pairs and some Ace highs that missed should be folded. The paired board makes it difficult for my opponent to nail the flop with their pfr range and very few combos are for value.
If my opponent c-bets no matter what, then re-raising all-in should generate plenty of folds to make this move overall +EV.
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*** Note ***
If you don’t plug in chip stacks, it would look something like this:
Total EV = Fold EV + Call EV
Total EV = (%F) * (P + B) + (%C) * [(%W) * (P + B + R) – (%L) * (R)]
0 = FP + FB + CWP + CWB + CWR – CLR
Substitute C for (1 – F)
0 = F * (P + B) + (1 – F) * (WP + WB + WR – LR)
0 = FP + FB + WP + WB + WR – LR – FWP – FWB – FWR + FLR
0 = F(P + B – WP – WB – WR + LR) + W(P + B + R) – LR
Substitute L for (1 – W)
0 = F(P + B – WP – WB – WR + (1 – W)R) + W(P + B + R) – (1 – W)R
0 = F(P + B – WP – WB – WR + R – WR) + WP + WB + WR – R + WR
F = (WP + WB + 2WR – R) / (P + B + R – WP – WB – 2WR)
F = (W(P + B + 2R) – R ) / (P + B + R – W(P + B + 2R))
Lastly, you can introduce a new variable Z where Z = P + B + 2R to simplify
F = (WZ – R) / (P + B + R – WZ)
Example,
If the pot is 100, my opponent bets 50, and I raise to 300…let’s find the % fold in-terms of W.
Z = P + B + 2R = 50 + 100 + 2 * 300 = 750
F = (750W – 300) / (50 + 100 + 300 – 750W)
F = (750W – 300) / (450 – 750W)
Solve for W if you assume that F > 0
0 < (750W – 300) / (450 – 750W)
(450 – 750W) < (750W – 300)
150 < 1500W
W < 0.10
Not sure if this is right, I rushed it.
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