...except a player BJ.Originally Posted by DSchles
Don,
The original post seems to exclude a BJ for either side.
Dog Hand
Yeah, I admit I tend to be lazy after so many decimals such as the frequency of a hand being 0.00063359 and multiply that by some odds fraction. Actually, Don it's because I tend to lose confidence in my math at that level. I have to work on that. I presume that the slight edge remaining is due to the soft doubling.
You can always count on me to ask "strange" questions!!! One of the reasons of my OP is that I find that with a surplus of aces at neutral or slightly negative counts, combined with a lack of 789s,is quite bad for the player.
Let me ask a more of a "street" question. When the above situation happens, splitting aces may not be the best play strategy. Is it worth "not splitting aces" the rare times when it's not the better play? Does that bring too much attention?
Not really, in slightly to negative counts, even with a surplus of Aces remaining. However in an example of a TC of -5 or 6(L2) with additional Aces left in the deck and No re-splitting Aces is allowed you might want to hit vs. a dealers Ace.
http://bjstrat.net/cgi-bin/cdca.cgi
Secretariat,
Ok, I plugged all of the numbers from the WoO's Appendix 9 into Excel and made two copies: one for Player A-up; the other for Dealer A-up.
On each I removed the Player BJ's (the WoO's numbers already assume no Dealer BJ). On the Dealer A page, I removed all hands where the Dealer's upcard is not an A; on the Player A page, I removed all hands where the Player did not have at least one A.
On each page I then adjusted the probabilities so they would sum to 1 on each page.
On each page I used the MAX command in Excel to find the EV-Maximizing move for each matchup, then calculated the SUMPRODUCT of the MAX times the Adjusted PROBs.
Here are the results:
Given that neither the Player nor the Dealer has a BJ for a 6D, S17, DA2, DAS, NoRSA, NoS game:
A hand Overall Probability EV Player 0.09603826 +0.134798 Dealer 0.05073824 -0.143477 Combined 0.14677650 +0.038602
So we see that if the Player has the A (a roughly 9.6% chance), he has a 13.5% edge, but if the Dealer has the A (a roughly 5.1% chance), the Player's EV is -14.3%. However, the Player is nearly twice as likely as the Dealer to have the A, since the Player has 2 cards to the Dealer's 1 upcard, so overall the Player has a nearly 3.9% edge on these hands.
By the way, in an earlier post I gave the EV for Dealer A-up as -9.5%, but I forgot that the WoO's listed probabilities already assume no dealer BJ.
Hope this helps!
Dog Hand
Now that's getting very interesting Dog Hand. So if we consider that the dealer's upcard is not an ace but that his down card is an ace, we would get the same overall probability which means that the player is practically at a 1% disadvantage, right?
Not saying you're wrong -- it's Secretariat's question -- but I don't understand why the player's probability to have an ace as his first card is double the dealer's. Where did it say in the original question that the player had at least one ace in his hand? That is, where did it say that the player started with two cards? My understanding was that the player started with an ace, but that he was not permitted to turn that ace into a blackjack. That's not the same thing that you calculated.
Don
Don,
That's a fair point: I just re-read the thread and I see that Secretariat never actually states whether he meant that the player's first card is an Ace, or that the player simply has an Ace among his first two cards. Let's see if he'll clarify the matter.
Either way it will not effect the calculated EV of +0.134798 for the player Ace, but instead it will change the probability. If the 1st card has to be the Ace, then the probability will match that for the dealer Ace.
Dog Hand
I meant that the player had an ace, either as a first card or as a second card against any dealer hand.
I also meant the dealer had an ace either as a first card or as a second card against any player's two cards.
All dealer and players blackjacks are excluded.
The basic question is: who has the OVERALL edge with non-blackjack aces?
My basic assumption was that it is a close call for the first two cards and that maybe the player has a slight edge because of the soft double downs but it may not be the case.
Now when you add an ace as a 3rd, 4th, 5th, 6th card, I presume that overall it helps the dealer more
than the player since the dealer has to draw more often because we stand on some of our stiffs while he must hit until he gets to 17. Even if we draw an ace on any 2-3-4-5-card 16, we are in a bad position at 17. On the other hand a dealer who hits 16 and gets an ace is in very good shape at 17 when we stand on our stiff against a low dealer upcard. Plus with a bunch of baby cards out, the dealer will come up with some crazy 19s, 20s or 21s more often than we do since he hits more often than we do. However, I realize how difficult it is to come up with 3-4-5-6- card hand probabilities.
In other words, I presumed that our beloved aces act somewhat as traitors when we don't get blackjacks. The ace is a non-busting card. It may only occasionnally lead to a dealer bust, most often when he gets an ace on his eleven total and that is followed by a Ten for 22.
I thus presume that we are slightly at a disadvantage overall considering how the non-blackjack aces are played out and and from what I understand from Dog Hand is that we are close to -1%.
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