How are the pots computed when in a tournament deal when the big blind posts his full ante but then doesn't have enough to post his big blind?
To take a concrete example, here's a hand history from a tournament deal where the big blind posts the ante but then doesn't have enough to post the big blind, so he's all in (he has 45 chips in total before starting the deal). Ante/SB/BB are 40/200/400.
The deal reads like this:
villain1 posts ante 40 villain2 posts ante 40 villain3 posts ante 40 hero posts ante 40 villain4 posts ante 40 villain5 posts ante 40 villain6 posts ante 40 villain7 posts ante 40 villain1 posts big blind 5 and is all-in Dealt to hero [3c Js] *** PRE-FLOP *** villain2 folds villain3 folds hero folds villain4 folds villain5 folds villain6 raises 1606 to 1611 and is all-in villain7 calls 1611 *** FLOP *** [Qh Qd 2s] *** TURN *** [Qh Qd 2s][8h] *** RIVER *** [Qh Qd 2s 8h][Jh] *** SHOW DOWN *** villain1 shows [Tc 4h] (One pair : Queens) villain7 shows [As 6s] (One pair : Queens) villain6 shows [Ad Kh] (One pair : Queens) villain6 collected 3547 from pot
In this deal I've got,
villain1 isn't winning anything, so I'm not sure what he could have won should he have had the best hand.
Here villain6 wins 3547 (1611*2 + 5 from the big blind + 8*40).
But how would the pot(s) have been computed if villain1 (the big blind) had had the best hand? Would he be winning 15 chips (5*3)? Or 15 + 8*40, that is: 335 chips?