Impact of Card Removal

In a strictly mathematical sense, given a finite number of decks, does removal of any card impact the results of future hands. I know their are situations where it obviously will, such as blackjack, but are their situations where removal of cards will have absolutely NO impact whatsoever. I know baccarat is a game where counting is irrelevant, but does removal of cards have a minutely finite impact? For example, does removing an ace from an 8 card shoe impact, in any microscopic amount, the likelihood of a tie or banker bet?

Do you have a reason in mind as to why it wouldn't?

Intuitively, I would suggest that if removing any given card didn't have an effect for some game then removing all cards wouldn't have an effect?

I don't know Baccarat, but it sounds like you might be asking why it isn't countable if removing cards does have an effect? The top link on a quick google showed an article by Eliot Jacobson ( http://www.worldgameprotection.com/the-catwalk/casino-ology/BOSVIEW/Card-Counting-in-Baccarat/ ) which says the game is hard to count because it's symmetrical and, unlike BJ, cards don't have the same tendancy to favor either the player or dealer.

To satisfy your mathematical curiosity, imagine a simplified version of war, where you and the dealer are each dealt one card. If there's a tie, you keep flipping until someone wins. Wins are charged a 5% commission.

I think there is no effect of removal in that game.

Quote: rKelly

In a strictly mathematical sense, given a finite number of decks, does removal of any card impact the results of future hands. I know their are situations where it obviously will, such as blackjack, but are their situations where removal of cards will have absolutely NO impact whatsoever. I know baccarat is a game where counting is irrelevant, but does removal of cards have a minutely finite impact? For example, does removing an ace from an 8 card shoe impact, in any microscopic amount, the likelihood of a tie or banker bet?

card removal in baccarat absolutely has an effect. you can see for yourself at the wizard's new baccarat game. it's just that the game is not very countable because the removal of cards don't ever overcome the house edge.

Quote: dwheatley

To satisfy your mathematical curiosity, imagine a simplified version of war, where you and the dealer are each dealt one card. If there's a tie, you keep flipping until someone wins. Wins are charged a 5% commission.

I think there is no effect of removal in that game.

there would be an effect of removal in this game. the same card coming out more than 1/13 of the time would lower the house edge. It would never lower it below 0, but it could get all the way to 0, if, for example, the first 48 cards of a 1-deck game contained no 9s, and they dealt all the way to the bottom of the deck.

Quote: sodawater

there would be an effect of removal in this game. the same card coming out more than 1/13 of the time would lower the house edge.

In this game, a tie is not a push, the game will just continue till there is a winner.

Since the game is completely symmetrical for any possible composition of the deck, house edge will always be 2.5%. There is absolutely no card removal effect.

This is a good point. War is symmetrical, and Baccarat is nearly symmetrical.

Because some games are asymmetrical (like Blackjack, where the player can hit or stand at will), the effect of card removal (EOF) depends on how important any rank of card is to the play of the current and next hand. You have a 16 against a dealer's 9, and the count is high (lots of tens), you stand, so as not to bust. The same 16 hand against the dealer's 9, when the count is low (lots of small cards), you hit for a shot at a pat hand. If the count is high going into the next hand, bet big to collect on a 20 or a dealer's bust.

Some games, like side bets for Blackjack and Baccarat, have a "scattered" group of winning hands, where differing hands within a wide range can be winners, getting rid of key-card weighting.

In Blackjack, only 19, 20, and 21 are the good hands, (and where the high cards can also bust the dealer), the key card effect is very concentrated to high-low card groups.

Somebody once said, "ANY game drawn from a shoe can be counted." Almost, but not quite. If the game is asymmetrical, OR the player has voluntary card-drawing options, then it may be indeed. Blackjack is a hallmark game here.

I've been trying to come up with an eloquent way to state this, and you guys helped a lot. This is the basis of the idea:

"Any symmetrical card game is uncountable"

and more interestingly stated:

"Any non-symmetrical card game is countable"

I would like to expand this concept to the point where I can prove as fact that any formulaic dealer system can have its edge shaved by opportune play. I wonder if their is more rigorous mathematical exploration of these concepts in formal logic. It seems like it's been explored very practically by countless players, but I rarely hear of people exploring it on a purely mathematical level and creating axioms leading to theorems, etc.

Quote: MangoJ

In this game, a tie is not a push, the game will just continue till there is a winner.

Since the game is completely symmetrical for any possible composition of the deck, house edge will always be 2.5%. There is absolutely no card removal effect.

Maybe we are imagining different rules for this game. In the extreme case of a deck having only the same card left, there can be no winner. Maybe if they keep the bet locked in the betting circle while they shuffle the next deck... but that would be ridiculous.

Quote: rKelly

"Any non-symmetrical card game is countable"

(A => B) <=> (not B => not A)

Thus the correct logical statement conclusion is: "Any countable game must be non-symmetrical". It doesn't say that any non-symmetrical game is countable, for this to prove you would need much more work to do.

Quote: rKelly

I've been trying to come up with an eloquent way to state this, and you guys helped a lot. This is the basis of the idea:

"Any symmetrical card game is uncountable"
and more interestingly stated:
"Any non-symmetrical card game is countable"....

No. The inverse is not always true.
I think:
A symmetrical game is not countable,
but a asymmetrical game may also be non-countable.
What this means instead, is that if a game is countable, it must be asymmetrical AND have key card groupings, even if some asymmetrical games are also uncountable.

An Asymmetrical game that can also form key card groupings is countable.

You can also cancel key card groupings in an asymmetrical game, such as blackjack, to make it uncountable, or less countable.

If you had in Blackjack:
1. Blackjacks pay 2:1 if the dealer has a 6 or less showing, but paid blackjacks even money if the dealer had a 7 or better showing, (so that the average payout on a Blackjack was still 3:2), you'd at least partially nullify the High-low count or advantage, as the blackjack payouts would inversely follow the high-low count of the unmodified base game.

2. If you had the dealer push when he busted on a 10-card bust (high card), but had him pay double to the players if he busted with a 6-card (low card), to compensate, you'd also counteract the high-low count effect, or the "key card" effect.

It is not enough for a game to be asymmetrical, the card ranks of the depleting shoe must also have "+" and "-" values for the player sides, in terms of "side worth" when depleted from the remaining shoe composition. By re-balancing key tag cards to more neutral tag values through new game rules, you can create a harder to count games that are still drawn from a shoe. The Blackjack side bets "21+3" and "Bust Bonus" distribute "winning key card values" more widely, to make them harder to count. The "Lucky Ladies" side bet, - which is fundamentally easy to count - just increases the base house edge to such a high level, that it is rare to find a positive player-advantaged count.