Questions about SD EOR and a carnival rule

Betsson Group have initiated an interesting SD game in their online casinos, which offer the best blackjack conditions I have seen so far in the industry. The game is a machine-shuffled SD, H17, DAS with some unusual rules, and in regard to this I have som questions about EV and deviations.
The casino advertise a 99.95% RTP, in other words 0.05% house edge, in their information tab for this game. Im sure this number is based on x computed hands, however there seems to arise some situations that are unclear to me.
The conditions are:
- You only get to play 1 hand. After the hand, the deck will be shuffled
- Split as many times you want
- No RSA
- No peek at 10. If you double, or split, and lose to a BJ, you will lose both bets. However, if you catch a 21 you will push against a natural.
- If you catch 2 concecutive tens after splitting aces, dealer will open his hole card and not deal any more cards, whether  it’s a pat hand or not, and immideately pay you.
Im looking for the optimal play in the following scenarios:
1: 11 vs 10

2: You split 77 vs 3 and catch a 5 on the first one, and an 8 on the second. Do you hit or stand with your hard 12?

3: 16 vs 10, when your hand is built up by 10-3-3. Remember to account for that catching a 5 will potentially tie a natural. In general Im curious whether it’s optimal or not to hit all 16s against 10, since there is a 1/13 chance that the dealer has an ace underneath.

4: 16 vs 9 when your hand is built up by 4425A or 6325

5: You split 88 vs A and receive a 3 on the first one, and 10 on the second. Do you double the 11? 

There are plenty of «close call» situations like these that can arise after splitting, and I have failed to find appendices in the litterature. Im aware that these spots are super marginal, but I cant help myself thinking whether they are accounted for, or if the casino calculated the house edge based on a fixed basic strategy.

Any suggestions will be greatly appreciated. 


 

Morgan Kane,

You wrote, "2: You split 77 vs 3 and catch a 5 on the first one, and an 8 on the second. Do you hit or stand with your hard 12?"

Does this mean that after you split, the dealer gives a card to BOTH of your split cards BEFORE you make any playing decisions on your first split hand? The usual procedure is to give a second card to your first split hand, then wait for you to finish playing that hand before giving a card to your second split hand. Same question about Scenario #5.

Dog Hand

Dog Hand

Correct, you are dealt a card on each before taking action. Once you get a pair vs 7 and below things can get very interesting. You can end up with a pretty juicy count. I have had those rare double oppertunities like 8 vs 4, and even 9 vs 8 on the second hand.
Do you really think the casino implement all these deviations when they determine the house edge? 

 

" Do you really think the casino implements all these deviations when they determine the house edge?"

No, certainly not.

Don 

Don

Thank you for engaging in this topic. My next task will be to gather all possible deviations (based on exact hand composition, like standing with 7-5 vs 3 or hitting 10-2 vs 4 etc.) that can occur, and write them into an appendix myself. This will obviously be a formidable job, and I probably will not do it if your answer to my next question is negative.

Based on your conclusion, and given the already mentioned house edge of about 0.05%, would you intuitively suppose the total of all possible deviations is enough to offset, or possibly overcome, the house advantage? 
 

"Based on your conclusion, and given the already mentioned house edge of about 0.05%, would you intuitively suppose the total of all possible deviations is enough to offset, or possibly overcome, the house advantage?"

Tough one. 0.05% isn't a lot to overcome, but a lot of the plays have extremely low frequencies, which means that you simply won't get to make them often enough. When you make the play, it can be worth something, but when you factor in the rarity of the occurrence, I'd say even if you can overcome the 0.05%, it won't be by much.

Don

Don

I clearly overestimated their frequency rate (in particular the pairs), which were surprisingly low. In fact, wizardofodds.com states that a "perfect composition dependent strategy" will produce a microscopic 0.0075% house edge. The problem with that site is the lack of parameters in the settings. I needed the "no peek on 10 only" column. Without any software tools I decided to search for this rule in the literature, and subtract/add the loss/gain from/to the published results. And whereelse to look than in BJA?!

First I thought this rule was a bad thing, since you would have to refrain from splitting 8's. And well, now I guess I have the answer to my first scenario in my initial post, the 11 vs 10 play, which too has to be abandoned from the regular strategy (Schlesinger, (2000) BJA2; Chapter 3: Evaluating the new rules and bonuses; Rule #5: Dealer 10-up Natural Counts as Plain 21: p38-39). It was a pleasant surprise to discover that the gain from catching up with a dealer's natural superseeds the loss of abandoning the 2 plays mentioned above (a net gain of .176-.082=.094). Unfortunately, the casino's rules states that a 10-up natural will count if the player happens to get a blackjack himself, so this aspect of the chapter is ignored (hence my use of "carnival rule" in the original headline).

My head is spinning a little bit. Wizzardofodds.com states a 0.046 % house edge, using basic strategy and continuous shuffler, but have not accounted for the "no peek 10-up rule". By adding the extra gain from this somewhat twisted rule, we all of a sudden have a 0.05% player's advantage off the top! And this is without the use of any deviations!

What happened here, what am I missing? Could the wizard be wrong? My only explanation is that the casino have used the wrong strategy when calculating the edge (then again, how is this possible?), with a specific reference to the 88 and 11 vs 10, respectfully. I should probably not speculate any more than I already have, but regardless of the issue discussed, I have one more question about the strategy in this game: Even when holding a multi-card 16 vs 10, would hitting be optimal play, since catching a 5 could potentially tie a natural? If that's the case, then I guess we lose an important deviation.
 
Thank you again for sharing your knowledge with "The Community" Don!



https://wizardofodds.com/games/blackjack/calculator/

   

"Even when holding a multi-card 16 vs 10, would hitting be optimal play, since catching a 5 could potentially tie a natural? If that's the case, then I guess we lose an important deviation."

Yet another close play. I make the value of this play about 0.01%. First, you have to get the multi-card holding of 16 vs. T. Then you have to draw a 5. Then the dealer has to have an ace in the hole. In short, doesn't happen very often! Once very 10,000 hands. And when it does, you save one bet. Does that outweigh the value of standing on all multi-card 16s vs. the dealer's ten? I simply don't know. Difference can't be much either way.

Oh, and by the way, BJA2? Really? Shame on you!

Don

every, not very.  Finally gotcha!  This day will go down in history.

Not my first! Won't be the last. The way I type, it's a miracle there aren't more. But, that's no excuse for not proofreading more carefully. Good catch.

Don

Being a non-english native speaker I appreciate you guys going easy on me, since I probably have lots of inaccuracies 

As for the topic of discussion: I checked the discription of the rules more closely, and found this: "...Blackjack beats 21". Only that it does not. And so I will conclude that a software-error has to be blamed for all my confusion. We should indeed be looking at a player's edge of about 0.05% off the top, right Don? You also did admit your estimates were conservative, when looking for the frequency how often the player draws to a 21 vs a 10 up (You gave this value 7.3%), and I quote:
"Actually, the 7.3% is a conservative estimate based upon the dealer's probabilities. The player will achieve slightly more 21s by virtue of hitting soft 17 and soft 18, options not available to the dealer"
(Schlesinger (2000) BJA2: 38)
Could this minor detail explain the standard error of .02% from Wong's findings? It seems the edge could be pushed further to a total of 0.1%, using composition dependent strategy.

Everything considered, and though the numbers are in the microscope-range, I do find the idea of having an edge off the top against an online-machine, somehow a bit romantic. If something can be done from a practically point of view, remains to be seen. But I would like to point out that it is not difficult to play 3-4 times faster online than playing live. 

When sharing the reference with the author himself Im slightly embarassed, because obviously I should have the 3rd edition by now. I have ordered it and look forward to dig into new (at least to me) appendices, and use it in future discussions!

The chart (Table 4.1) at the bottom of page 50 in BJA3 gives the final player probability for drawing to 21 to be 7.16%. Obviously, somewhat rule dependent.

Don

This has been discussed here before. Another important number is the portion of the 7.16% player probability for drawing a non-blackjack 21 being used for doubling down.  We need both numbers to calculate the final EV.

1.64%.

Don

Do you remember from what year I can find that discussion in the archives?

I don't understand why it's important to figure out how often a non-blackjack 21 is being tied with a double down, in order to calculate the final EV. I mean, we are not gonna execute the play, so why calculate it?
 

I haven't followed this thread closely and haven't got into the exact logic in this post, but let me still add my input here.  Here is the link for the post I mentioned: Free Forum - I need someone who knows math (bj21.com)
Based on Dog Hand's CVDate simulation result on the a 6-Deck Hit-17 DAS game, I calculated the player's probability for an undoubled non-blackjacks 21 is 3.0%, and the player's probability for a doubled non-blackjacks 21 is 2.2%.  Adding these two numbers together, we get the total probability for a player's non-blackjack 21 is 5.2%. This is not even close to the number of 7.16% above. Am I right?

Yes and no. But you're closer to right than the numbers I posted, above, which were a misinterpretation of my own chart, which is somewhat embarrassing.

Table 4.1 on page 50 has the correct values, again, depending on rules. I double-counted the "% via doubling number." That 1.64% is rolled up in the total 5.52% value, so it can't be added to that value, as I did, to obtain the incorrect 7.16%. Sorry about that.

So, to summarize, the basic strategist achieves a non-BJ 21 5.52% of the time. 5.00% of those are untied, and 0.52% of those are ties with the dealer's 21. Of the 5.52%, 1.64% are achieved via doubling.

Don

The player’s probabilities depend on the game rules.  For a 6-Deck Hit-17 DAS game, I used Dog Hand's CVDate simulation result to calculate the player's probability numbers again. These are my latest results: the player's probability for a non-blackjack 21 via non-doubling is 2.9% and that via doubling is 2.1%, and the player’s probability for hand splitting is 2.7%.

I studied your Table 4.1 on page 50 of BJA3 and found that your probability numbers for doubling and splitting were rather conservative. This makes me think that you might use a Stand-17 game without doubling-down after splitting and this can well explain the discrepancy. The good thing is all these numbers are converging.

Thank you for your hard work!
 

I appreciate the numbers that's been given, but Im having a hard time converting them into the EV for this specific game. To me, it looked simple to use the number for the dealer's probabilities of getting a non-blackjack 21 (this number is 7.29% according to Don, BJA3: 50), and multiply with the dealer's probabilities of getting a 10 up natural (2.41% of the time). Regardless what the player has, we are gonna follow the same rules as the dealer when we are up against the 10 (not double down, nor split, with the exception of AA. Just hit all the way until we get at least 17). I do get that 7.29% might be a little bit off, depending on the rules. If anything, I thought this value should be slightly higher, since we are gonna get more chances reaching 21 by hitting soft18, which the dealer can not. Thus it translates into 7.29% x 2.41%=.176%. 

Im actually just repeating Don's work from BJA2, so my assumptions should be safe. 

If this is wrong, I hope someone will correct me. Thanks for all the help so far!