Interesting Stud 8/B Theory Question

[arcticfox]

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But to answer seriously, if you started with 6A54 then bricked twice, you are looking to catch a 2, 3, 7 or 8. You have 12 outs, but seat 1 has one of your outs. Of 52 cards in the deck, 14 are known, so your 11 outs are in the remaining 38. So it's 27:11 = 2.5:1. Since you are only playing for half the pot you need to be getting about 5:1 I think.

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Thats some bad maths in your head! You have 15 outs (4x4 less the one card already out). Secondly you need to factor in the fact that you are betting once to win 1.5 bets on top of what the pot is already laying you, assuming both highs will call a river. So you basically need around 3:1 pot odds to make this call correct assuming you only get one call on the river, and less assuming you get both highs to call the river when you hit.

So if the pot is 7BB with you closing the action on 6th, your EV of folding is obviously 0.

Your EV of calling is -1BB when you call and don't hit 23/38 times = -0.61 BBs

Your EV of calling and hitting with 7 BBs in the pot (8 BBs including your bet and assuming you only get one call on the river) is 15/38*3.5=1.38 BBs.

Meaning making this call has an EV of 0.77 BBs in your favour, plus the chance you get 2 high calls on the river where you make your low which increases your EV significantly. (these calculations obviously ignore rake)

You basically break even on this where there are 2 BBs in the pot prior to your bet, suggesting that you should be usually chasing lows where you are guaranteed a lock on the low if you hit unless the pot is tiny. In reality it is rare that you have a lock on the low you are drawing to on 6th which makes folding on 6th correct in most other circumstances unless the pot is big.