My EV calculation
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The EV of the KQ draw is:
91% of the time: -1 SB
9% of the time: 11 SB (4 initially in the pot, plus 11 put in by AK post flop)
EV KQ draw = .91(-1) + .09 (11) = .08 SB (note that there is no a priori reason this number had to be positive)
Hence, EV (AK) = pot - EV (draw) = 4 - .08 = 3.92 SB
I claim that, under the assumptions for this problem, this is the EXACT EV for AK, not an estimate, and that any system of calculating profit which wants to be taken seriously must produce exactly this number.
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Eric,
I might take issues with your assumptions here, but under those assumptions the math seems to be right. I calculated the same 3.92 EV under these assumptions looking at pot equity on a street-by-street basis:
AK's Pot Equity of Pot on Flop = 91% x 4 = 3.64
AK's PE of Flop Bets = (91% x 1) - (9% x 1) = .82
AK's PE of Post-Flop Bets = (91% x 0) - (9% x 6) = -0.54
Total = 3.92
Mind you, I think your method, or any other similar one (such as what I did), becomes much more difficult (if not impossible) in multiway pots, particularly if you factor in what happens when players make other non-nut hands with which they might choose to continue. I would point out that neither of our EV calculations factor in the possibility of a split pot when a Jack comes on a turn and a Quenn comes on the rvier, but for simplicity sake I think that is fine.
All the best.
Colgin
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