I don't understand what you're driving at then.

Sorry there is a possible confusion. I'm considering 'no hole card' games where the dealer doesn't get their 2nd card uintil after we play.

Ok, just give me the scenario. "I get x, dealer's up card is y, next card in the deck is z. My question is..."

We dont know what z is. We just know either we take z or dealer gets z

I gave you some examples
You have 15 vs a dealer T (no hole card dealt)

e.g if next card was a T then you bust but would have lost anyway when the dealer got it.
e.g if next card was a 5 then you improve a lot but the dealer hand would have been weakened significantly if they had got it.

Ok, so the EV calculation assumes z is uniformly distributed from the set (2-A), works out how many units we win (or lose) when we either hit or stand for each z (obviously if we stand, then dealer's first card is z), then averages out the results and tells us EV(hit) and EV(stand). What's the question?

What if dealerdoesn't share the same next card?

so now if I woudl have hit a 5 but stick then dealer might get a Ten

Since we're not considering card removal effects, just consider a toy version of the game.
Edit: sorry, example was wrong, I'll re-post when I've fixed it.

I'm considering the answer for any game with this 'shared card' property.

Despite the intuition of many - the common exclamationt that dealer woud have beat them anyway after they bust, it doesn't matter if we share an (infinite) deck or have multiple different (infinite) decks. Sharing the cards to be dealt cannot matter

I'm not offering a dull proof or ev calculation.

Consider a deck consisting only of cards saying "AUTO WIN" and auto "AUTO LOSE". If player chooses to draw, he either wins or loses depending on which card he gets. If player passes, the dealer draws and the player wins if the dealer draws AL and loses if the dealer draws AW.

If there is a 50/50 chance of either card, it's straightforward to see that the chances of winning if you draw first are 50% and the chances of winning if you pass are also 50%. It might also be easier to see in this example why the shared cards make no difference. Whether the dealer draws from the same deck or from a fresh deck, his chances are 50%. If you choose to hit, the times when you hit and took the dealer's AL card are exactly balanced by the times you hit and took the dealer's AW card.

A given scenario in blackjack is more analogous to when there is a skew in the distribution of these cards. For example, if there is a 60% chance of drawing AW, obviously you are better off hitting than you are passing, and the times you do hit and "take the dealer's AL card" will be balanced by the times you hit and "take the dealer's AW" card for an overall 60% chance of victory. Where the analogy breaks down is that often in blackjack, the composition of this deck is >50% AL, but the dealer doesn't have to draw. There is some probability calculation based on his upcard that defines whether he has to draw or he just wins anyway if you pass, forcing you to draw from this deck even in cases where you have a >50% chance of losing. The principle in question remains the same, though.

i love that this thread has devolved to the most stringent iq defenders competing to signal their high iqs

Says bro who's got "smart" in his user name. Anyway, are you even allowed to post here if you're not on the list?

both fair points

Yet another way of looking it this is to go back to the original game and consider a rule change where the dealer shuffles the deck before he hits. All the calculations for us remain the same, the EV of standing an the EV of hitting remain the same, but we now no longer "take" or "leave" the dealer's next card. This is pretty much the same thing as the dealer drawing from a different deck as you (chez) mentioned before.

Ok. On#ce you relaise shuffling makes no difference you're sorted although asking if shuffling makes a difference is another way of stating the problem

Another way of looking at it is that in practice we can never be sure that it is a shared card rather than a shared probability distribution. Therefore it can't change the probabilities even if it is a shared card (if that even means anything). Or think of it as shrodinginger's card - it's a probability wave until we force it to collapse by showing it. Or chezlaw's card - it's a an algorithmic pointer until we have to run the algorithm to determine it's value.

Another way of saying this is "an unseen card is an unseen card", which is what I said previously.

As in the unseen cat is the unseen cat.

Well yes but somehow lacking l

Or even the unseen rabbit

Unseen isn't sufficient for shared cards not to matter. The river in holdem is also unseen but the shared card matters.

The point about the shuffle is that it's between the player hitting and the dealer drawing. If that shuffle between possible events doesn 't matter then the shared card cannot matter.

I don't think you understand what the phrase means. It means that all unseen cards are fungible therefore changing their order cannot affect any probability calculation based on their distribution, i.e. shuffling a deck of unseen cards, or replacing it with a different deck with the same distribution, or drawing a card from the middle or bottom of it rather than the top, cannot make a difference.

ok