Is this an edge?

A friend of mine offered me a proposition bet.

He takes 6decks of cards takes out 13 cards which count to a +8

He deals the 1st card off the top of the remaining decks then shuffles the +8 cards back into the 6 decks and completes a hand of bj standard Vegas rules.

H17 is there an edge?

S17 is there an edge?


I say no but I maybe wrong.

If I read the scenario correctly, then I would generally answer no for H17 and S17 (at least for a shoe*** game).

***: If it was done with a CSM, similar to the link here, then there could be a small edge.

Quote: Salthouse

He takes 6decks of cards takes out 13 cards which count to a +8
link to original post

Does this mean the 13 card have a count of +8 or the remaining cards in the deck?

The 13 cards have a plus 8.

So the remaining decks therefore are minus 8 for the first card dealt.

I may also mention that I use wongs halves count as well the + - count so the count varies slightly depending on the 2, 5, 7 , 9 in the 13 card mix.

Quote: Salthouse

(snip)
So the remaining decks therefore are minus 8 for the first card dealt.
(snip)
link to original post

That would increase the house edge.

Not sure I understand.

I would like the remaining decks to be dealt out to be minus 50

Sounds poor to me if you are down 8 high cards off the top.

Ok for clarity the count would be +8 +9 or + 10 then you get to bet.

You get dealt 1 card and the discards get then shuffled back in.

The dealer then completes the hand.

Quote: Salthouse

Ok for clarity the count would be +8 +9 or + 10 then you get to bet.

You get dealt 1 card and the discards get then shuffled back in.

The dealer then completes the hand.
link to original post

If the count is +8 in a six deck game the true count would be about 1.25 for one card. I don't think that gives you enough of an edge to be positive.

Quote: Salthouse

Ok for clarity the count would be +8 +9 or + 10 then you get to bet.

You get dealt 1 card and the discards get then shuffled back in.

The dealer then completes the hand.
link to original post

I still don't quite understand.
But below is a link that shows "What is the change in advantage from TC to TC" (TC = True Count)

link

Note: If you are only getting one card from the "hi-count shoe" and one card from the "roughly neutral shoe", then just halve the figure in the graph titled - "What is the change in advantage from TC to TC" (see example below)

Example: Say the TC is +1, then the graph shows an advantage change of a bit less than +0.8%, but since you are only being dealt one card from that shoe, then the overall advantage change should be +0.4% (so take away this from the base house edge, to see if there is an overall advantage).

Note: This assumes that no matter how many decks are left, that all true counts are roughly equal.
Note 2: The count used for those graphs was a "hi-lo count"
Note 3: Someone else better than me at this, may be able to prove that what I said is true, close, or completely wrong.

You have summarized the question well:

If you are only getting one card from the "hi-count shoe" and one card from the "roughly neutral shoe" do you have an edge.

I don't know what "standard Vegas rules" are, so if you can tell me the rules of the game, I may be able to help a bit more.

Note: The minimum rules I need to know are in the link here.

Lets make the rules not so good:

decks 6

dealer stands soft 17

play can double after split

double on 9 10 11 ( no soft hands )

resplit to 4 hands ( not aces )

no hitting on split aces

players lose 1 bet to BJ

no surrender

BJ on $10 pays $15

Quote: ksdjdj

Quote: Salthouse

Ok for clarity the count would be +8 +9 or + 10 then you get to bet.

You get dealt 1 card and the discards get then shuffled back in.

The dealer then completes the hand.
link to original post

I still don't quite understand.
But below is a link that shows "What is the change in advantage from TC to TC" (TC = True Count)

link

Note: If you are only getting one card from the "hi-count shoe" and one card from the "roughly neutral shoe", then just halve the figure in the graph titled - "What is the change in advantage from TC to TC" (see example below)

Example: Say the TC is +1, then the graph shows an advantage change of a bit less than +0.8%, but since you are only being dealt one card from that shoe, then the overall advantage change should be +0.4% (so take away this from the base house edge, to see if there is an overall advantage).

Note: This assumes that no matter how many decks are left, that all true counts are roughly equal.
Note 2: The count used for those graphs was a "hi-lo count"
Note 3: Someone else better than me at this, may be able to prove that what I said is true, close, or completely wrong.
link to original post

The main issue here is you ignore the dealer cards.
When the count is +1, it gives you a 0.8% advantage because high cards are better for you than for the dealer. One assumes (not sure) that high cards are also good for the dealer but the 0.8% is just the difference of how much more advantageous they are to you than to the dealer.

The key thing here is you get 1 high count card one normal, but the dealer gets both medium card. The advantage might be much worse or much better I have no idea to be honest but I don't think that your estimate is accurate.
Gonna wait for one of the math gurus here to hopefully post sim results

Quote: Salthouse

Lets make the rules not so good:

decks 6

dealer stands soft 17

play can double after split

double on 9 10 11 ( no soft hands )

resplit to 4 hands ( not aces )

no hitting on split aces

players lose 1 bet to BJ

no surrender

BJ on $10 pays $15
link to original post

Under those rules, I think it is not enough to turn it into a small player edge (but again wait and see what the "math gurus" say).

Well lets hope the guru also gives us good rule numbers:

decks 6

dealer stands soft 17

play can double after split

double on any 2 cards

resplit to 4 hands ( not aces )

no hitting on split aces

players lose 1 bet to BJ

no surrender

BJ on $10 pays $15

Quote: tyler498

(snip)
The main issue here is you ignore the dealer cards.
When the count is +1, it gives you a 0.8% advantage because high cards are better for you than for the dealer. One assumes (not sure) that high cards are also good for the dealer but the 0.8% is just the difference of how much more advantageous they are to you than to the dealer.

The key thing here is you get 1 high count card one normal, but the dealer gets both medium card. The advantage might be much worse or much better I have no idea to be honest but I don't think that your estimate is accurate.
Gonna wait for one of the math gurus here to hopefully post sim results
link to original post

Thanks, you are likely correct.

I still think the advantage of one card to the player from a "~ +1 TC " shoe and one card to the player and dealer from a "neutral count" shoe, will not be enough to overcome the house edge for most 6 - deck games.

---
Update:
I stand corrected, see my next post below

Quote: Salthouse

Well lets hope the guru also gives us good rule numbers:

decks 6

dealer stands soft 17

play can double after split

double on any 2 cards

resplit to 4 hands ( not aces )

no hitting on split aces

players lose 1 bet to BJ

no surrender

BJ on $10 pays $15
link to original post

Quote: Salthouse

Lets make the rules not so good:

decks 6

dealer stands soft 17

play can double after split

double on 9 10 11 ( no soft hands )

resplit to 4 hands ( not aces )

no hitting on split aces

players lose 1 bet to BJ

no surrender

BJ on $10 pays $15
link to original post

After doing it a different way (using more accurate, but longer method) I now think you are more likely correct (that there is a small house edge).

For the "double on any 2 cards" rules you mention above, I get a ~0.17% player advantage.
For the "double on 9 10 11 ( no soft hands )" ,,, above, I get a ~0.07% player advantage.

"Incomplete" Proof (see below):

For the chances of the first player card (pc) I used these figures below:

22/299 for: 2 to 6 pc (each card value)
23/299 for: 7 and 9 pc (each ...)
24/299 for: 8 and Ace pc (each ...)
95/299 for: 10-value pc

Then I multiplied those chances by the "player EV" when receiving those cards as the first card and used "neutral deck EV" (see "first card EV" figures below)

2 = -13.15% , 3 = -15.27% , 4 = -17.62% , 5 = -19.75% , 6 = -20.98%
7 = -18.15% , 8 = -8.37% , 9 = -0.87% , 10-K = +14.34% , Ace = +50.21%

Note: I think this is one*** way to get a running count of +8, using Wong Halves (I have never used Wong Halves before)
***: Since the edge is so small, I don't know if it goes back to negative for any +8 running counts with 299 cards left in the shoe.
Note 2: These "proof figures" are just for the "double on 9 10 11 ( no soft hands ) " / 0.07% player edge example.
Note 3: I used the link here to get the "player EV" figures for each "1st player card".

^ Before reading through down to this reply, I'd have probably approached it the same way (looking at the range of starting cards and their EVs) and think the ratio of missing cards seems a very fair estimate.