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6 deck, S17, das, early surrender, peek

Is all the surrenders something new? Just haven’t seen this before, surrender on hard 5, 6, 7 vs dealer A etc.

Thanks

What’s the difference between early and late surrender?

Early surrender is vs a dealer T or A *before* they check for blackjack — extremely valuable.

Late surrender is *after* they check.

This trainer is a joke.

Sorry you don’t enjoy it. I like it and beat it a lot. I like how it mimics a deck without replacement. I wish it allowed multiple players and allowed the possibility of playing multiple spots. I love it.

Sincerely,

The Milk Man

I have been asking this question. How many blackjacks can be made in an 8 deck shoe?

Anon is correct at 32.

Does anyone have an answer. I have gotten from 32 to 512. I feel the answer is 128. Does anyone have the right answer. This has been going back and fort with a few people for several days and can’t believe the answers I’m getting. Please respond. Thanks

Not complicated. There are 4 Aces per deck, times 8 decks. That’s 32 possible blackjacks.

Wrong.

Are you asking how many can be dealt from one shoe, or how many distinct ways to make blackjack (which would not depend on the number of decks)?

This seems pointless and stupid regardless. It’s not rocket science.

It was not my question but one given to me. Yes from one shoe how many ways.

Yet again you pose your question in a way that could mean multiple things. This will be my final reply regardless if you like my answers.

A) I don’t think this version is your question, but just in case: If you want to know how many blackjacks can be dealt from one 8 deck shoe, I already gave you that answer. It’s 32, because every blackjack requires an Ace, and there are only 32 Aces in an 8 deck shoe.

B) How many different forms of blackjack can be dealt? The answer is the same regardless of how many decks are in play. There can be 4 suits of Aces, 4 kinds of tens (T,J,Q,K) with 4 suits of each kind of ten. So, 4 X 4 X 4 = 64 different “kinds” of blackjack. For example, Ace of clubs and ten of clubs is one kind. Ace of clubs and ten of diamonds is another. If you write them all down, you’ll find that you have 64 versions. This is true in 1 deck, 8 decks, or a million decks.

C) There is one additional way to interpret your question. Let’s say you wrote a number on the front of every card in an 8 deck shoe. The first card gets number 1, the second 2, and the last card will be card number 416 (which is 8 X 52). Now there are more “ways” to get blackjack because you can tell the difference between otherwise identical hands like Ace clubs/Ten clubs because their numbers will be different. How many of these ways to make blackjack are there? 32 different Aces times 128 different ten-value cards = 4096 “ways”. That answer seems completely useless to me, but maybe that’s what you wanted to know. In real life, we can’t tell the difference between two identical hands.

Further questions will be ignored, but thanks would be merited.

Don’t think that’s correct. Just one deck. 4 Aces and 4 Tens (picture cards + 10) =16 different ways to get 21.

4 aces x 4 (10)=16 per deck. So 8 decks x 16= 128. So again is that correct or not. Some people have come as high as 512.

If you don’t see patterns in how the cards come out of the deck, then you aren’t paying enough attention. And if you don’t know the precise mathematical reason for the valuations assigned by the authors of this game to the J,Q,K, then you can’t really speak intelligently about this particular game.

If you don’t see patterns in the cards coming out of this deck, then you’re just not paying enough attention. It goes both ways, if you are pre-disposed (conditioned?) to dismiss any possibility beyond random. then I guess recognition skills need not apply – grin. And if you can’t rightly say why the authors of this game designated the values of the J,Q,K as 10’s, mathematically speaking, it’s hard to see how one can speak intelligently to this game. But that’s just me.

“can you believe I’ll have you zipping through a deck in less than 20 seconds?”

This is hard to believe. I timed myself going through a whole deck without counting and the fastest I did was 25 seconds. Hard to believe someone is going to go through the whole deck while counting in less than 20 seconds.

I encourage you to do the math. A simulation is actually what is required. You will find that none of the spots you are betting on will ever have a positive edge, even if you were to apply full indices while playing. So the optimal bet for each is zero. Having a less negative edge on each successive spot is not a reason to increase your bet.

If the principle of your strategy was sound, then you would be better off finding a full table and simply flat betting the last spot. Your would have the maximum amount of information available to you and would never have to bet on a spot with a negative edge. This is, of course, not a valid approach. If it were, we would all be literally fighting one another for that last spot.

With respect to the potential disaster of doubling your bet on each spot, going from $10 to potentially $640 on the final spot, the harm is that will, over time, lose the house edge multiplied by the total of your bets. Betting more on the later spots will only increase the rate at which this will happen. It’s not clear what you mean by “brought to ruin”. For most, it means having lost enough that you quit playing for good. Perhaps you have a large bankroll, or perhaps you replenish it from another source, such as a job. If you have been playing for as long as you imply however, then it is extremely unlikely that you’re in the black.

Perhaps I’m still missing the point of your blog post. I’m hoping that the next post will say that this was one of your biggest early mistakes. But it feels like you’re leading up to a claim that this is a winning, long term strategy.

It is not.

Well Gronbog, it turns out that any benefit in betting more with each seat is very slight. Hardly worth doing the maths. Thanks for your comments!