Hi. Assuming you are in a 3handed poker game and BT opens, which of these methods do you think is more correct to determine how often neither of the blinds defend?:

1)
(1 - (SB call% + SB 3bet%)) * (1 - (BB call% + BB 3bet%) = combined fold.

2)
% 3bet = 1 - ((1- SB 3bet%) * (1 - BB 3bet%))
% call = (SB call% * %BB doesn't 3bet) + (BB call% * %SB folds to BT open)

Combined fold = (1 - % 3bet) * (1 - % call)

Your equations do not explicity account for the possible dependence existing between the SB and BB actions. Why not --

Pr(SB and BB fold) = (% SB folds) * (%BB folds given SB folds)

How would he input the figures for: (%BB folds given SB folds)?

The same way he gets data on any other percentage. Observe or use an appropriate filter if using an on-line tracker.

Ok, so to get %BB folds given SB folds, do I just need to write down the % BB would fold if it was folded to him after BT opened? Or do I need to make it like equation 2) where BB fold% * SB fold% is 'BB fold given SB folds'?

Not sure I understand your question.

Assume you have the following history of 100 hands when the button opened:

70 hands where SB folded and 30 hands where SB called

Of the 70 folded hands by SB, BB called 35 times, folded 35 times.

Of the 30 hands where SB called, BB called 10 times, folded 20 times.

Then,

Pr (SB folds) = 70/100 = 70%
Pr(BB folds | SB folds) = 35/70 = 50%

Also,
Pr(BB folds | SB calls) = 20/30 = 67%

Pr(SB and BB fold) = P(SB folds) * Pr(BB folds | SB folds) = 70%* 50%= 35%

Of course we get this result by just counting the 35 hands where both folded, but I wanted to write an equation consistent with the forms OP used, which was to have a fold probability for each player.

Note the possible dependence here, where BB is more likely to fold if SB calls (67%) compared to the folding chance when SB folds (50%).

Oh right, I understand now. Thanks a lot.