joelshitshow said
At a 4-handed table, the average V holding will be worse than at a 9-handed table in which 5 people fold, because in a 9-handed game in which 5 people fold you’re playing with what’s essentially a 42-card deck that is more likely to have better hands among the remaining players. Am I thinking about this right, or is this effect less than the reduced antes’ effect?

Without any evidence to support it I would say no. I can’t find any information that the chance of being dealt a specific hand is increased with fewer players. If you get dealt AA in the CO on a 4-handed table, the chances of the remaining players to have picked up say 99+ is the same as if you had CO at a 9-handed table with 5 players folding to you. And as Andrew said, even it was true the effect is really really small, not enough to change how we approach short-handed play. Other aspects should have a bigger effect to our play.

Even if 5 players fold when cards are dealt, every hand is starting with a 52 card deck. So just because 5 players folded you are not playing a 42 card deck, you all started playing with a 52 card deck. You only remove your cards from the deck and then flop, turn and river since these are known cards. The rest is unknown cards. So in both cases you and your opponents are playing with a 50 card deck of unknown cards.

It’s because of this that we can’t discount outs either. Lets say we get dealt Ah3h, at a 9-handed table 18 cards being dealt preflop (two cards to each player). On average, since a deck only contains 13 cards of each suit, 4,5 cards being hearts is dealt out (18/4 = 4,5. 18 being cards dealt out and 4 being the different suits available). But we can’t discount outs we haven’t seen, we could in fact be drawing dead on a 5h8sJh board when our opponent has a set if all the remaining hearts was dealt preflop. Extremely unlikely though. And because of probability we could discount 2 outs when flopping a flush draw since on average 4 hearts are dealt into everyones hole cards each round, but we don’t count outs that way.

So when we count our outs with Ah3h on 5h8sJh, we count the outs the same way no matter how many players was dealt in to the hand. 47 unknown cards in total (despite only 31 cards are remaining in dealers deck) and 9 hearts that can make us a flush which is roughly 1 in 5 or around 19% chance.

We do this despite knowing that probability says that on average 4.5 cards of each suit is dealt out in hole cards each round. 

This translates well into 7-card stud. You only discount outs that has been visible cards on your opponents third and open card. You can’t discount any cards because of probability of being dealt to a starting hand. 

Say that we in a 7-handed 7-card stud game are dealt Ah4hJh(Jh open card) and the remaining players does not have an heart as open card. There is 21 cards in play, of which 9 are known cards (our 3 cards and opponents each open card, 6 in total) There is 43 unknown cards (despite the dealer only holding 31 cards) and we have 10 hearts we can hit on the fourth card. Our odds to get a fourth heart is 1 in 4,3 or around 23%.
But probability says this: If 21 cards is dealt: (21/4=5,25) 21 cards being dealt total meaning on average 5 hearts will be dealt each round in a 7-handed stud game, hidden or open. But we don’t discard these two probable cards unless they are on an opponents open card.

I hope this makes some sense. While this translated into how count outs I think it translates the same about probability of being dealt a hand. No matter if you are playing 9-handed or heads-up, 50 cards are unknown preflop and the chances of being dealt two specific cards doesn’t change since the unknown cards doesn’t change.