Deucekies
Deucekies
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July 17th, 2014 at 2:42:29 AM permalink
An off-hand remark from a player today got me wondering something.

Theoretically, what is the highest advantage a player can have going into a hand of blackjack? Assume the best rules you can find, and the most astonishing count possible. I assume it would come when the shoe was nothing but face cards and aces remaining.
Casinos are not your friends, they want your money. But so does Disneyland. And there is no chance in hell that you will go to Disneyland and come back with more money than you went with. - AxelWolf and Mickeycrimm
RS
RS
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July 17th, 2014 at 3:07:00 AM permalink
I'd venture to say if all remaining cards are aces or faces.......or if every remaining card was an 8. However, neither are realistic.

Counting doesn't give you a big edge, just a tiny sliver of an edge.

You'd have a much greater game if you had hole-card and/or next/first-card information. That would be astonishing.
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odiousgambit
odiousgambit
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July 17th, 2014 at 3:11:30 AM permalink
Quote: Deucekies

nothing but face cards and aces remaining



From my own experience with knowing a lot of Aces were left, I can tell you the thing that kills this from being unbelievably +EV is that the dealer keeps getting BJ too [g]

I think I've seen indication that the best game you can find [without something like a triple-down-allowed promotion] is about 0.20 % HE. I'd have to guess at how +EV the deck could get from there.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
ksdjdj
ksdjdj
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July 17th, 2014 at 6:06:24 AM permalink
Since this is a hypothetical question, i am going to propose this 'impossible' scenario (it would be interesting what the odds of this happening would be)

parameters: remove all 7's and 9's from the deck, only start with 24 x 10's , and with 6 full decks for all other cards, dealer hits soft 17,bj pays 1.5/1, no doubles, no splits, no surrender,

this game would have a starting player edge of about 1.13...%

in the above scenario the count would be easy, just keep an eye on the 10's and 8's left in the deck, because in this scenario the LESS 10's and 8's the better it is for the player, (i am guessing because the dealer has no choice but to stop hitting on soft 18 and hard 17, but the player can keep hitting those?!?)

with no ten's left, the edge would be about: +8.96...%

with no 8's left, the edge would be about: +6.42...%

with no 8's or 10's left, the edge would be about: +20.586%

to show how strong no 7's, 8's, 9's and 10's is, i even calculated it with 'dealer wins ties', which is still about: +6.64% player edge

------

the main basic strategy would be to hit all 17's and also hit all soft 18's, see exceptions below

1. with no 8's left (exceptions)

(a) hit on soft 19 against a dealer 10

(b) hit a hard 18 against a dealer Ace

2. with no 10's left (exceptions)

(a) hit on soft 19 against a dealer Ace

(b) hit on hard 18 against a dealer Ace

-------
ps, you could go a step further and remove all the 6's too, but if you remove that from the 'theoretical' game, there is no chance of a dealer bust, and it just gets closer to a game of 'war' with more player input
------

Edit:
for dealer stand's on all 17's the edge would be a lot better,
for example, with no 7's-10's, the edge would be about 27.92%
mcallister3200
mcallister3200
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July 17th, 2014 at 6:46:12 AM permalink
Quote: RS

I'd venture to say if all remaining cards are aces or faces.......or if every remaining card was an 8. However, neither are realistic.

Counting doesn't give you a big edge, just a tiny sliver of an edge.

You'd have a much greater game if you had hole-card and/or next/first-card information. That would be astonishing.

I don't know the point at which it happens, but at a certain point when the deck is very saturated with 10 value cards the advantage actually begins to decline, the reason being 20-20 pushes being the most likely outcome. I want to say the advantage counting on a specific count tops out somewhere in the 6% range.
dwheatley
dwheatley
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July 17th, 2014 at 7:02:38 AM permalink
In the ridiculous situation of having a deck that was 50-50 aces and tens, and standard liberal rules (split aces receive 1 cards, split 10s up to 4 hands), I estimate a player edge of over 56%. Turns out most of the return comes from splitting 10s against a dealer hand of A,A or T,T.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
dwheatley
dwheatley
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July 17th, 2014 at 7:06:57 AM permalink
But this edge is dwarfed by the equally unlikely situation of having all 8s as mentioned by RS, which you could always split to 4 hands, and the dealer would always bust. Guaranteed +400%. That's probably the theoretical upper limit, I can't think of a way to double split hands into winners.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
Ibeatyouraces
Ibeatyouraces
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July 17th, 2014 at 7:11:32 AM permalink
deleted
DUHHIIIIIIIII HEARD THAT!
AxiomOfChoice
AxiomOfChoice
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July 17th, 2014 at 10:55:45 AM permalink
Quote: RS

I'd venture to say if all remaining cards are aces or faces.......or if every remaining card was an 8. However, neither are realistic.

Counting doesn't give you a big edge, just a tiny sliver of an edge.

You'd have a much greater game if you had hole-card and/or next/first-card information. That would be astonishing.



At extremely high counts, the edge actually starts to drop a bit because of the large number of pushes (20 v 20).

It is possible for a player to have a 400% edge in certain deck compositions if they know the composition and play accordingly. Your example of all 8's is a perfect one -- the player will split to 4 hands and have 4 16s, and the dealer will bust, for a net win of 4 bets. However, the player has to know that this is the distribution, and they have to deviate from basic strategy (once they have maxed out splits, stand on the 16 v 8)
AceTwo
AceTwo
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July 17th, 2014 at 11:51:38 AM permalink
The highest EV on a BJ Hand that I have encountered was 150%.
That was a Special Promotion game that lasted only a few hours with a joker which counted as Automatic BJ if you got it.
I was playing first base and I knew that the very next card was a joker.

But under Normal BJ rules he biggest EV attainable in a Hand of BJ is around 50%.
That is when you know your first card (or second) will be an Ace.
This could be because of exposed first card or bottome steering or other AP method. Although in most AP methods you will not be 100% where exactly the Ace is so the EV will be considerably less than 50%.

Mispayments also could produce EVs of upto 200%. ie 200% on been paid on a losing hand, 100% on pushing a losing hand.

Exposed 3rd cards also would have very High EVs. For example the previous player signalled Stand and a Fast Dealer proceeded to Hit.
In some places I played, the exposed card would stay in play for the next player.
I once Doubled a 13 with an exposed 8.

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