Could you still count with this procedure?

Let's say you are playing a DD card game where every card is worth the same except for the King and the Queen. If you get a King you get an extra point and if you get a Queen you get a point subtracted from your score. Obviously it would be easy to devise a counting system for this game. But what if before dealing the dealer splits the decks into 3 piles and asked you to pick one. He puts the piles you don't pick into the discard rack and deals the pile you pick down to the last card.

Could you still use a counting system on this game to track the money cards as if the discarded cards were still behind the cut card and would this procedure render your system a waste of time? What if a blackjack dealer did this on a blackjack game? Could you still count as if the discarded cards were still behind the cut card?

Quote: Brazen1

Could you still use a counting system on this game to track the money cards as if the discarded cards were still behind the cut card and would this procedure render your system a waste of time? What if a blackjack dealer did this on a blackjack game? Could you still count as if the discarded cards were still behind the cut card?

There's obviously the question of effective deck penetration, but the answer is generally yes. As long as all the cards were initially together before being separated into piles, and you have no information on the contents of the piles themselves, the fact that the cards are separated has no bearing on the probabilities as relates the remaining cards as more cards become known.

Let's simplify the problem:

I have one deck of 52 cards. I shuffle them and count 26, then separate them into two piles of 26 apiece.

If we fan out 25 cards of one pile and see all four kings, the probability the other pile contains a king is zero. The probability the 26th card in the pile from which we took 25 is a king is zero.

If we fan out 25 cards and see zero kings, the probability the 26th card is a king is 4/27. This is true even though at least three kings must be in the other pile.

Nothing matters in your scenario the penetration
is horrible rendering the game useless.

I see. It just seemed to me 1/3 deck might have been effective penetration in such a game. Let's say we were playing a dice-style game with a DD with a couple extra cards thrown in. So the rules state if you draw a 7 all your bets lose but if you get a joker you can use that as a safety card to cancel a future 7 or maybe the joker causes all your bets to win, whatever. So the decks contain 9 Sevens and 3 Jokers and you're tracking these cards and you have the same scenario.

And the first 2 cards to come out of that 1/3 deck are 2 Jokers or 2 Sevens. It seems to me that edges in such a game could shift quickly. Are you saying 1/3 deck would still be useless in pretty much any scenario for this type of game?

Quote: Brazen1

And the first 2 cards to come out of that 1/3 deck are 2 Jokers or 2 Sevens. It seems to me that edges in such a game could shift quickly. Are you saying 1/3 deck would still be useless in pretty much any scenario for this type of game?

What I am going to suggest for the game you mentioned in the OP is that you provide the full set of rules for how the game works.

So far, I have explained that cards are removed when cards are removed and all of the other cards are unknown no matter where they sit. I then provided an example of same.

Anyway, in the new scenario that you describe, yes that kind of shift could theoretically be enough to swing the odds either in your favor or dramatically against you. I guess it depends on what the other rules are. I don't know why you are playing with only 1.5 decks as relates Jokers and 2 and 1/4 decks as relates sevens, but oh well.

If I am understanding this right, this is the same as if, instead of putting the other two piles in the discard pile, they are put into the deck, but behind the cut card (i.e. there's only 1/3 penetration). It shouldn't affect counting, regardless of the system, but, as others have pointed out, you only get to work with the top third of the deck.

Thanks. I'll suggest a new a bet because I don't have a game fleshed out for the Kings example. What I am trying to determine is if an operator would ever deliver a game more vulnerable to blackjack out of anything but a CSM.

So let's use the Joker/7s example and just bet on a color. So we have 2 53 card decks: 48 red cards and 48 black cards as well as 10 naturals, eight 7s and two Jokers. The rules are if you get a 7 your black/red bet loses but if you get a Joker the bets pays 3 to 1. Winning B/R bets of course pay even money. I'm thinking the HE on this game should be fairly low. If you could show me how to properly write out the equation to calculate this I would greatly appreciate it.

How much penetration would you need to gain an edge on this bet and would a casino only deal it out of a CSM or 8 decks with minuscule penetration?

Quote: Brazen1

So let's use the Joker/7s example and just bet on a color. So we have 2 53 card decks: 48 red cards and 48 black cards as well as 10 naturals, eight 7s and two Jokers. The rules are if you get a 7 your black/red bet loses but if you get a Joker the bets pays 3 to 1. Winning B/R bets of course pay even money. I'm thinking the HE on this game should be fairly low. If you could show me how to properly write out the equation to calculate this I would greatly appreciate it.

How much penetration would you need to gain an edge on this bet and would a casino only deal it out of a CSM or 8 decks with minuscule penetration?

Okay, this is actually pretty easy. You're going to make a bet of one unit with an initial probability of losing that is 56/106, the probability of winning a single unit is 48/106 and the probability of winning three units is 2/106, ergo:

(48/106) + (2/106 * 3) - (56/106) = -0.01886792452

The expected loss is .01886792452 units, which means that the House Edge is 1.886792452%.

I'm not going to get into every possible scenario, because you don't need me to in order for you to figure out possible scenarios. You only need me to show you how to figure out the scenarios for yourself.

In this game, the house edge is so low that I wouldn't think it would take much to swing it in the favor of the player. For example, suppose that the first round (because each player would get his own card, I assume) has four players and no sevens or Jokers come out, but two cards of each color come out, here is the expected value of the next round:

(46/102) + (2/102 * 3) - (54/102) = -0.01960784313

That's obviously worse, because sevens comprise a greater percentage of the deck such as not to be offset by the enhanced value of Jokers. However, what if four red cards came out and no black cards? Obviously, black would be a better bet than red:

(48/102) + (2/102 * 3) - (52/102) = 0.01960784313

That represents a player advantage of 1.960784313%, and we didn't even touch the Jokers or the Sevens!

How likely is that scenario, though? Let's find out.

In this case, it doesn't really matter whether it is a card of a specific color or a seven, either way, a losing card is gone. If the card is a seven, that just improves the proposition (equally and by the same amount) for both red and black. Therefore, we start with our 56/102 probability and try to determine the probability of four, "Bad," cards coming out right away:

(56/106 * 55/105 * 54/104 * 53/103) = 0.07393577296

Therefore, the probability is 7.393577296% that the player would enjoy an advantage in excess of the house edge after only four cards have been removed of 106. That's about every 1 in 13.5 trials (rounded), thus, the game would be destroyed.

Let's say that only three, "Bad," cards come out, we'll make it a black and any combination you want of reds/sevens:

(47/102) + (2/102 * 3) - (53/102) = 0

Even Steven. No advantage either way after that, but that opens the house up to a very high probability of tossing out a hand with a player advantage on the next hand.

Anyway, like Blackjack card counting, you would do it on a, "True Count," basis for any number of decks. You couldn't just count Sevens and Jokers, generally, because you also need to know whether the deck favors red or black. You could count it based on, "Winning Cards," (whichever color there is the most of less three for every joker gone) and, "Losing Cards," (whichever color there is the least of, plus sevens).

So, let's pretend that there you can only bet red and that there is only one deck to simplify things:

(24/53) + (1/53 * 3) - (28/53) = -0.01886792452

The true count is as zero in this scenario. This is actually really simple, if the true count is +1, then it is an even money proposition, if the true count is greater than +1 (which can include decimals when dealing with multiple decks) the player is at an advantage. If the true count is negative, the player is at a greater disadvantage than usual from the top of a fresh shoe.

Anyway, that's a really nice and clean little game that you came up with which would make for a great kitchen table game for quarters with friends. Maybe it could be a casino game...but only if you ran 8 decks and either dealt it out of a CSM or cut several (I'm thinking at least six) decks off.

Thanks. I haven't digested all your information yet but I actually came up with a similar percentage but I was not sure if I did it right. I figured after the extra payout the House basically had an edge of 2 cards out of the 106. Then I divided by 2, then divided 53 into 1 but was not sure if I was doing it right. Math isn't my forte but I am laboring to improve in this arena.

Here is another question. Let's say this was a casino game. Do you think the gaming public in general--I'm not referencing the small percentage who might be APs--would actually prefer to have a game like this dealt out of say a 6-deck CSM so the extra-payout cards would always be in circulation?

Quote: Brazen1

Thanks. I haven't digested all your information yet but I actually came up with a similar percentage but I was not sure if I did it right. I figured after the extra payout the House basically had an edge of 2 cards out of the 106. Then I divided by 2, then divided 53 into 1 but was not sure if I was doing it right. Math isn't my forte but I am laboring to improve in this arena.

That's correct. Reds and Blacks start out equal, so that leaves:

Jokers: (2 Cards * 3 Units) = 6

Sevens: (8 Cards)

Total: 6-8 = -2

It's not hard to get better at math, and much of this can be done with simple math. You'll get the hang of my mechanism in the prior post, and it's always the same, you just have to adjust for the number of decks and the deck composition.

Quote:

Here is another question. Let's say this was a casino game. Do you think the gaming public in general--I'm not referencing the small percentage who might be APs--would actually prefer to have a game like this dealt out of say a 6-deck CSM so the extra-payout cards would always be in circulation?

I think they might prefer it because the game would be fast-paced with no interruptions. I don't know that your, "Average gambler," (for lack of a better term) gambles intelligently enough to concern himself or herself with the implications of missing jokers.

Quote: Mission146

That's correct. Reds and Blacks start out equal, so that leaves:

Jokers: (2 Cards * 3 Units) = 6

Sevens: (8 Cards)

Total: 6-8 = -2

It's not hard to get better at math, and much of this can be done with simple math. You'll get the hang of my mechanism in the prior post, and it's always the same, you just have to adjust for the number of decks and the deck composition.



I think they might prefer it because the game would be fast-paced with no interruptions. I don't know that your, "Average gambler," (for lack of a better term) gambles intelligently enough to concern himself or herself with the implications of missing jokers.

Thanks for your help, Mission. If you have some spare time, I was wondering if you could quickly show me how to calculate the HE on a bet where you guess the rank of your next card?

For instance, we have 8 full decks, 432 cards. So we are betting on the odds of getting 1 of 32 cards, so it looks like we have 400 losers. What would be the HE if you paid a winning wager 11-1 as opposed to 12-1?

Also, this question might sound stupid but bear with me. Lets say we were playing with 52 card decks, no jokers and paid this bet 12-1. Would this bet have no HE?

Quote: Brazen1

Thanks for your help, Mission. If you have some spare time, I was wondering if you could quickly show me how to calculate the HE on a bet where you guess the rank of your next card?

For instance, we have 8 full decks, 432 cards. So we are betting on the odds of getting 1 of 32 cards, so it looks like we have 400 losers. What would be the HE if you paid a winning wager 11-1 as opposed to 12-1?

Also, this question might sound stupid but bear with me. Lets say we were playing with 52 card decks, no jokers and paid this bet 12-1. Would this bet have no HE?

12/13 of the time you lose the bet for -1
1/13 of the time you win the bet for +11

EV = -1*12/13 + 11*1/13 = -1/13 = -0.0769

You should expect to lost 7.69 cents for every dollar bet. That’s a HE of 7.69%.

Quote: unJon

12/13 of the time you lose the bet for -1
1/13 of the time you win the bet for +11

EV = -1*12/13 + 11*1/13 = -1/13 = -0.0769

You should expect to lost 7.69 cents for every dollar bet. That’s a HE of 7.69%.

Thanks. But what would the HE be on a 54 card deck? Could you divide 1 into 13.5, since there are 2 extra jokers to get at it? Actually, this sounds wrong. I added what I thought the HE to be with 2 extra jokers to 7.69 and came out with 11.3 percent, but I might have made a wrong move somewhere.

Quote: Brazen1

Thanks. But what would the HE be on a 54 card deck? Could you divide 1 into 13.5, since there are 2 extra jokers to get at it? Actually, this sounds wrong. I added what I thought the HE to be with 2 extra jokers to 7.69 and came out with 11.3 percent, but I might have made a wrong move somewhere.

Close.

4/54 times win 11
50/54 times lose 1

11*4/54 - 1*50/54 = -6/54 = -11.1%

Quote: unJon

Close.

4/54 times win 11
50/54 times lose 1

11*4/54 - 1*50/54 = -6/54 = -11.1%

Off the top of my head, that sounds like the exact HE for a Hard 10/4.

Quote: Brazen1

Off the top of my head, that sounds like the exact HE for a Hard 10/4.

It is. Not so coincidental because 11.1% equals 1/9. Hard 4/10 should pay 8 to 1 but pays 7 to 1. That’s a house edge of 1/9 also.