16 vs 10 multicard in no peek

Basic strategy is to hit 16 vs 10, but with three or more cards it's correct to stand, the composition-dependent strategy coming into play with such tiny EV difference.

However, is this also correct in the European no-peek game? The generic EV numbers for the US peek game, stand -0.53888 and hit -0.53504, are much worse in no-peek (the dealer has a one in 13 chance of making blackjack by dropping an ace on the 10), but I don't know what they are. And therefore, I also don't know if that currently unknown difference between the EVs would be sufficiently great to make it correct to always hit any 16 vs. 10, irrespective of the number of cards.

I may not have phrased the question as well as I might, so feel free to ask for clarification. I appreciate any feedback.

The fact that a dealer-BJ is still possible has no influence on the strategy. It just moves the EV numbers down to -57% but the difference between hitting and standing remains roughly the same.
I'd like to point out that this multicard-16-rule is actually irrelevant for a card counter. Here the only rule is: with a positive running count you stand otherwise you hit.

Francis Salmon

Thanks very much, Francis - I remember your posts from years ago on BJ21. I had figured around -58%, wasn't sure about the EV differences between the two plays. Thanks for clearing that up.

I play occasional low-stakes BS blackjack and am always looking to eke out a bit extra, hence fussing around with 16 multicard vs 10 and some brief disquiet when it occurred to me I might be standing on multicard totals incorrectly. The games are all CSM, so no counting is possible.

This old saying from Napoleon (Impossible is not French) applies perfectly to CSMs. Since there are always some cards not readily available to be dealt, it is very well possible to count cards. It's just less efficient than in handshuffled games.
A last word to this 16multicard rule. For me this makes only sense if you are alone at a table playing one hand.
What's the point of standing on a multicard-16 when the table is otherwise covered with tens?

Francis Salmon

"A last word to this 16-multicard rule. For me this makes only sense if you are alone at a table playing one hand.
What's the point of standing on a multicard-16 when the table is otherwise covered with tens?"

He's a BS player; he doesn't consider cards on the table, belonging to other players, and he doesn't count.

Don

My point concerns BS-players or more presisely this special breed of BS-players who take the composition of their hand into account.Namgnos is a very illustrous representative of this species since he even investigates into EVs. I cannot imagine that he would want to ignore a bunch of tens hitting the felt.He wants to do the right thing with all the information available to him.
In most cases the situation on the table will not be that obvious and in this respect I remember a recepy of the late MathProf who for the situation 16v10 recommanded to stand if the fives on the table outnumbered the sixes otherwise hit.To me this makes more sense than the composition-dependant approach.

Francis Salmon

Haha, I'll take the reference to "species" on the chin as one of the risks of posting about BS on an advantage play board (self-deprecating pun entirely intended) and I do appreciate the response, Francis. Right, taking more cards into account is obviously better than fewer, but it seems a rather inexact science and a bit of a pain, which is why I've preferred to stick with my hand only, for which I have the three-card exceptions off pat. With four or more I stand on anything, the accuracy of which a table scan would improve. But I wouldn't be sure of exactly what I was looking for - "more 6s than 10s" seems misleadingly straightforward. I've seen some CSM games actually deal as far as 3 decks, which almost could be counted. But I'm not of the practising counter mindset, although I occasionally backcount with KO if I'm not playing, just for the mental exercise and to remind myself how badly I count backwards.

If you Google "hundred percent gambling" and take a look at my site, it might be a bit of an eye-opener about how some members of the BS blackjack player species think about blackjack and gambling generally - seriously. Not all BJ recreational players are retards, although I know that the vast majority refuse any and all attempts at education. I should add that the stuff of Stanford I've purloined is with his permission.

I didn't know MathProf was dead. Sorry to hear that.

There seems to be a universal assumption among card counters that csm's simply recycle every card perfectly. This is an easily demonstrated fallacy-certain cards are not immediately available for play, something you can prove with data tallying and statistical analysis. The card counting fraternity generally just dismisses this because it fucks up their calculations.

The problem is that certain decisions change the card order and the speed of redistribution, and this can result in causing certain cards not to be available for play on subsequent rounds. In that scenario a zero-memory strategy doesn't make sense since it would be sub-optimal overall on certain marginal hands. Does it make a major difference: probably not, we are talking about composition-dependent effects, but the point is that you just don't know.

In short: CSM BS isn't optimal in any meaningful sense and it wouldn't be trivial to work out what a truly optimal strategy is. It would be dependent on things like how fast you make your decision.

Probably better to avoid the things altogether unless you have some kind of meaningful edge from another technique.

The tens are too numerous to systematically include them in the appreciation. This would really "degenerate" into counting.MathProf's
idea was to only take the cards with the highest EOR (effect of removal)for the situation 16v10 in either direction.
The six is not only a bust card for you but it would also bring the dealer into difficulty if he got it as his second card.
The five is your most favorable card because it brings you to 21.
Not an exact science but very clever!

Francis Salmon

"The card counting fraternity generally just dismisses this because it fucks up their calculations."

This insularity of view reminds me of similar insularity in a few notable voices' response to floating advantage, as recounted in BJA. I may have misread what seems to me a pretty obvious logic behind FA, and that being the case I hope to not get shot down in internet flames, but this did occur to me as I was re-reading about FA a few months ago, so anyway:

The proposition, as basically noted by Revere and Uston years ago, was that casino edge reduces at a given count level with fewer decks remaining, and BJA gives the example of a zero count yielding c.-0.5% at the start of a four deck, "floating" to 0% with one deck remaining. It seems Snyder and even Griffin initially rejected the idea. But why is it even controversial? In the first deck of a six-deck a zero count yields -0.5% because the shoe is balanced (what "0" means, small cards and large cards in equal abundance) and the basic strategy edge at that point is -0.5%. Extend that to the last deck and with a zero count you have a moreorless normal single deck remaining. There's no guarantee there aren't more 6s and fewer 2s etc, but small-to-large is that of a fresh single deck, and any discrepancies obviously even out over the long run. Logically, the casino edge at count zero is now 0%, this being what basic strategy yields. So, the fact that count zero at six deck level is -0.5%, becoming -0% at one deck, seems to be stating the obvious.

Was it the case that the luminaries in question were so focussed on counting, where "every point of the count represents 0.5% to the player" is a sort of universally imutable law, that they failed to see the obvious wood for the trees? Or have I missed something? Uston simply lists the numbers in Million Dollar BJ, so it doesn't seem he thought there was anything remarkable or comment-worthy about it.

You asked if you missed something, and the answer is yes. The whole story is very clearly described in BJA3, in the preface to the FA study.

BS is NOT synonymous with playing at a count of zero. They are not the same thing. And, once five of six decks of a shoe have been dealt, the BS for playing that last deck is identical to the BS for playing the first deck. I'm sure you understand that.

It is specifically the edge for a counter that is the basis for the discussion of FA. BS has NOTHING to do with it. And no, it was not apparent to the luminaries at the time (at least not all of them!) that there was such a phenomenon.

Don

"BS is NOT synonymous with playing at a count of zero. They are not the same thing. And, once five of six decks of a shoe have been dealt, the BS for playing that last deck is identical to the BS for playing the first deck. I'm sure you understand that.

It is specifically the edge for a counter that is the basis for the discussion of FA. BS has NOTHING to do with it. And no, it was not apparent to the luminaries at the time (at least not all of them!) that there was such a phenomenon."

Maybe I wasn't clear enough.

Example: you've got six sealed decks in a shoe - we won't open them for the purposes of this. We're about to start playing (right, we can't because their sealed, but that doesn't matter), and the count is obviously zero. The house edge is 0.5%.

Now we ditch the first five sealed decks, and leave the last one. The (true) count is obviously still zero as we haven't actually touched any cards. The house edge is now 0%. It's "floated" up 0.5% - specifically because we're now playing a single deck.

To quasi-normalise things, let's unseal the decks, shuffle each in turn and stick them, back to back, in the shoe. All we're not doing at this point is mixing them together. At the top (six decks to go) the house edge is 0.5%. Once back to the last 52 cards of the remaining deck, the house edge is 0%. THIS much is incontrovertible.

Does this not explain floating advantage? If we now mix all the decks together and kick of again (ie. normal play), at the start we're again looking at the 0.5% disadvantage of the six-deck game (and a zero count obviously). If we play through the first five decks and, arriving at the last deck, find we're back at a zero count, we're basically looking at a single deck. Of course, that remaining deck will not contain the exact number of each suit, there may be a few more 6s and 3s but fewer 4s and 5s, maybe a few more 10s but an equal number fewer aces, but the big cards and small cards are balanced in number - the zero count tells us that. At this point, Uston tells us that the edge at zero has "floated" up to zero, which to me seems intuitively obvious because in essence we're facing a single deck, whose house edge in c. 0%.

I cannot understand what I'm missing. It also seems remarkably coincidental that the house edge "floats" up the exact same amount as dictated by the number of decks, (we start at -0.5%, and with one deck left and still TC zero we're at 0%), but not specifically because of the fact that six deck = -0.5% and one deck = 0%, again assuming TC zero for the last 52 cards. It also works for the last two decks, ie. Uston gives the edge floating down to -0.25% at the two deck level (TC zero), which matches the two deck house edge. To summarise all that: TC zero, 6 decks, -0.5%. 2 decks, -0.25%. 1 deck, 0%. Those are the "floating numbers" and also the house edge numbers for the given numbers of decks.

Basically, my reasoning brings me to the right numbers but for the apparent wrong reasons, which is bizarre.

You also said above, which I'll highlight here:

"And, once five of six decks of a shoe have been dealt, the BS for playing that last deck is identical to the BS for playing the first deck. I'm sure you understand that."

Yes I did. But now you mention it, I'm not sure.

TC zero and 52 cards left of the six-deck shoe. With 5/3 vs 5 we hit. But why not double as per one deck BS? Why wait for TC +3 (Wong)? I believe the reason for the more aggressive doubling in SD is the greater chance of a big card relative to the remaining cards, ie. we are relatively more likely to pull a 10 or ace with two 5s and a 3 already on the felt in this case. It seems logical that, with what is in essence a single deck remaining (TC 0), we should follow SD BS. What is specifically different in those remaining 52 cards, which we know contain balanced bigs to smalls with the TC 0, that says we shouldn't double that hand, whereas in an actual single deck, which appears to me identical to all intents and purposes (caveat different suits and one extra 6 here, one less 5 there etc etc) we should?

"Maybe I wasn't clear enough."

Crystal clear. Just wrong. :-)

"Example: you've got six sealed decks in a shoe - we won't open them for the purposes of this. We're about to start playing (right, we can't because their sealed, but that doesn't matter), and the count is obviously zero. The house edge is 0.5%."

No it isn't. As the decks aren't mixed together, the edge is 0% six times in a row, 52-card segments at a time, six times. Why would it be -0.5%, if the decks aren't ALL shuffled together?

"Now we ditch the first five sealed decks, and leave the last one. The (true) count is obviously still zero as we haven't actually touched any cards. The house edge is now 0%. It's "floated" up 0.5% - specifically because we're now playing a single deck."

Wrong. It was 0% all the time, six times in a row. It's not -0.5% until you MIX all six decks together. How could you imagine otherwise? Can you get six aces on the first two hands, doing it your way? Of course not.

"To quasi-normalise things, let's unseal the decks, shuffle each in turn and stick them, back to back, in the shoe. All we're not doing at this point is mixing them together. At the top (six decks to go) the house edge is 0.5%."

Again, no it isn't. Same problem. Shuffling them individually isn't the same as shuffling them ALL TOGETHER.

"Once back to the last 52 cards of the remaining deck, the house edge is 0%. THIS much is incontrovertible."

It's not BACK to 0%; it was 0% for all six decks the whole time.

"Does this not explain floating advantage?"

Not in the slightest. Not even remotely. Did you reread the chapter or not?

"If we now mix all the decks together and kick off again (ie. normal play), at the start we're again looking at the 0.5% disadvantage of the six-deck game (and a zero count obviously).

You are now looking at -0.5% edge for the FIRST time in this discussion!

"If we play through the first five decks and, arriving at the last deck, find we're back at a zero count, we're basically looking at a single deck."

You're looking at the edge for single-deck play, with a count of zero.

"Of course, that remaining deck will not contain the exact number of each suit, there may be a few more 6s and 3s but fewer 4s and 5s, maybe a few more 10s but an equal number fewer aces, but the big cards and small cards are balanced in number - the zero count tells us that. At this point, Uston tells us that the edge at zero has "floated" up to zero, which to me seems intuitively obvious because in essence we're facing a single deck, whose house edge in c. 0%."

You're being condescending. If you read BJA3, and realize that it wasn't "intuitively obvious" to Peter Griffin, then don't say that it's intuitively obvious to you. The light bulb is intuitively obvious to everyone today, but it wasn't when Edison was alive!

"I cannot understand what I'm missing."

Well, for starters, your entire starting premise was wrong.

"It also seems remarkably coincidental that the house edge "floats" up the exact same amount as dictated by the number of decks, (we start at -0.5%, and with one deck left and still TC zero we're at 0%), but not specifically because of the fact that six deck = -0.5% and one deck = 0%, again assuming TC zero for the last 52 cards. It also works for the last two decks, ie. Uston gives the edge floating down to -0.25% at the two deck level (TC zero), which matches the two deck house edge."

I'm not at all sure why you're quoting Uston. He had but the vaguest notion of all this. Have you or have you not read the FA chapter of BJA3?

"To summarise all that: TC zero, 6 decks, -0.5%. 2 decks, -0.25%. 1 deck, 0%. Those are the "floating numbers" and also the house edge numbers for the given numbers of decks."

Not, not a a law. The effect is not perfectly linear.

"Basically, my reasoning brings me to the right numbers but for the apparent wrong reasons, which is bizarre."

Happens all the time. Just make two mistakes that cancel out, and you have a right answer for the wrong reason. 2 + 3 = 6. and 6 + 4 = 9. So 2 + 3 + 4 = 9. Easy! :-)

"You also said above, which I'll highlight here:

"And, once five of six decks of a shoe have been dealt, the BS for playing that last deck is identical to the BS for playing the first deck. I'm sure you understand that."

"Yes I did. But now that you mention it, I'm not sure."

Well, you have a God-given right not to be sure, but I am.

"TC zero and 52 cards left of the six-deck shoe."

NO!!! NO true count at all. NO COUNT!! BASIC STRATEGY. Basic strategy for the last deck is the same as basic strategy for the first deck. NO COUNT INVOLVED! Either you play according to the count, or you play according to BS; you can't do both at the same time.

"With 5/3 vs 5 we hit. But why not double as per one deck BS? Why wait for TC +3 (Wong)? I believe the reason for the more aggressive doubling in SD is the greater chance of a big card relative to the remaining cards, ie. we are relatively more likely to pull a 10 or ace with two 5s and a 3 already on the felt in this case. It seems logical that, with what is in essence a single deck remaining (TC 0), we should follow SD BS."

Absolutely not true. BS and a count of zero have NOTHING to do with each other. You are making a fundamental error in thinking.

"What is specifically different in those remaining 52 cards, which we know contain balanced bigs to smalls with the TC 0, that says we shouldn't double that hand, whereas in an actual single deck, which appears to me identical to all intents and purposes (caveat different suits and one extra 6 here, one less 5 there etc etc) we should?"

The fact that there are trillions of ways you could have 52 cards remaining with a TC of zero, while, for a single deck, there is only one -- the original deck.

Don

You must feel down right now since Don was pretty harsh on you so I'll try to lift your spirits again.
Although your example with the sealed decks and the individually mixed decks was unfortunate I agree with your central message that FA shows the same characteristics and behaves the same way as the house edge according to the number of decks used.The evolution of the EV is not linear but obeys an empiric rule found by Wong (I think):
The improvement of EV over the infinite deck is 0.64% divided by the number of decks used. This gives 0.08 for 8 decks,0.11 for 6 decks,0.16 for 4 decks,0.32 for 2 decks and 0.64 for 1 deck.
I share your view that this applies not only to the number of decks in play but also to the remaining decks and therefore to FA. This can be verified in numerous simulation including the ones published by Don himself.
You asked:"Why don't we double when dealt 5,3 v 5 from 1 neutral remaining deck?" The answer is "we do!" since the TC has now risen to +3.
I wonder why a person with your devotion,knowledge and analytical talent is still not a card counter!

Francis Salmon

1. The example of edges that you gave is due to Griffin, not Wong.

2. To clarify, yet again: If you're at the last deck of six, and you know that the TC = 0, and you're dealt 5,3 v. 5, yes you would double. But, if you are a BS player, and you don't know the count, you would NOT double, because a) you can't just assume a neutral deck, and b) that play is not BS for SIX decks, and that is what you are playing -- no matter where in the shoe you are.

Clear to everyone?

Don

"You must feel down right now since Don was pretty harsh on you so I'll try to lift your spirits again."

Haha, I actually wrote a lengthy reply, previewed it instead of posting it...and lost it. I decided to give up for a few days because I simply could not be bothered to try and redo everything. Back now. Believe me, I've dealt and been dealt an awful lot worse.

Thanks for all that, I'll need to take a good look at what you said.

This forum software could use a radical overhall; Vbulletin comes to mind.

As noted to Francis, I spent an hour replying, hit the wrong button and lost everything. Needed to take a break before trying to retype that lot.

My example of six sealed decks was not one of my better moments. Why I missed that each was a single deck with a house edge of 0% (assuming normal rules) I don't know. I could have saved us both a lot of pointless posting, but hey. Brain drain.

"You're being condescending. If you read BJA3, and realize that it wasn't "intuitively obvious" to Peter Griffin, then don't say that it's intuitively obvious to you. The light bulb is intuitively obvious to everyone today, but it wasn't when Edison was alive!"

While I absolutely do not want this to descend into a slugfest, and I appreciate your responses a lot and I know what a luminary you are in the BJ business and at BJ21, it's a sight rich to accuse me of condescension when your replies - which again I highly appreciate - are shot through with condescension, arrogance and grandiose pomposity. Having pointed that out, and if you feel inclined to reply after those choice adjectives, I will continue to not respond in kind at any point. If you choose to do likewise, thanks. If not, whatever.

I am not being condescending. The light bulb is not analogous with what I said about it appearing "intuitively obvious" etc because the light bulb simply didn't exist and my intuitions are based on my BJ knowledge which most certainly does. To me it seems "intuitively obvious" - even if wrongly so - and I'm absolutely sure that the self-deprecating and mild-mannered Mr. Griffin would not think I was remotely pissing on his grave in so stating, even if you do. I will continue to so state, and if you take exception to that fact on the part of Griffin's reputation, OK.

"I'm not at all sure why you're quoting Uston. He had but the vaguest notion of all this. Have you or have you not read the FA chapter of BJA3?"

I have the previous edition. Sorry 'bout that.

However, a clarification on Uston: I incorrectly recalled that his "improvement" at the two-decks remaining level at TC zero was 0.25%. Having checked MDBJ I note it's actually 4%. House edge for off the top 2D is nearer 0.3%. However, Uston's figure for 52 cards remaining and TC zero, 0%, is exactly the house edge for 1D. I'm not sure if Uston's figures here are disputed? I repeat once more: the way I saw it, the house edge for 4, 2 and 1 deck is pretty much these figures (4D -0.5%, 2D -0.3%, 1D 0% at TC zero), which to me appeared to explain FA, as it seemed reasonable to conclude that with TC zero at 104 and 52 cards remaining we were essentially looking at double deck and single deck.

Also, although Uston may have only had "the vaguest notion", he and Revere would appear to have discovered the phenomenon, and if the figures are correct I'm not sure why I wouldn't quote them.

"Happens all the time. Just make two mistakes that cancel out, and you have a right answer for the wrong reason. 2 + 3 = 6. and 6 + 4 = 9. So 2 + 3 + 4 = 9. Easy! :-)"

That's just plain childish rudeness and you know it. You've certainly concluded at this point that I'm capable of a lot more than adding 2 to 3 and getting the right answer. Are you? Whaaaaaaaaatever, LOL.

Me: "TC zero and 52 cards left of the six-deck shoe."

You: "NO!!! NO true count at all. NO COUNT!! BASIC STRATEGY. Basic strategy for the last deck is the same as basic strategy for the first deck. NO COUNT INVOLVED! Either you play according to the count, or you play according to BS; you can't do both at the same time."

We are clearly at cross purposes. I'm am positing counting until (say) the last deck of a (say) six deck shoe, and being at TC 0. There is obviously nothing wrong with this. At this point, if we're using hilo, we know two things:

1) 2s to 6s and 10s / aces are balanced (TC zero).

2) We know nothing about 7s, 8s and 9s.

What also do not know is how many lows and highs there are; there may be none at all, those last 52 cards might contain only 7s, 8s and 9s. But equally, there may be none, with those remaining 52 cards containing only 2s to 6s and 10s and aces.

All we do know is that lows and highs are of equal number, whatever that number is. Is that why it is apparently not true that those last 52 cards have a single deck house edge of c.0%?

Supposing we keep a sidecount of the 7s, 8s, 9s and still have TC zero with 52 left? This must surely have a positive effect, as we would then know that we also had a full compliment of small and big cards, the only missing information being that we wouldn't know exactly how many of each there were. Heck, if we kept a sidecount of every card we would know we had an exact single deck at TC zero with 52 cards left.

Assuming that is true, then the FA, although the numbers do reflect the house edge (caveat Uston's 2D number does not reflect 2D house edge as closely as I'd thought) is down to "Factor X".

Me: ""With 5/3 vs 5 we hit. But why not double as per one deck BS? Why wait for TC +3 (Wong)? I believe the reason for the more aggressive doubling in SD is the greater chance of a big card relative to the remaining cards, ie. we are relatively more likely to pull a 10 or ace with two 5s and a 3 already on the felt in this case. It seems logical that, with what is in essence a single deck remaining (TC 0), we should follow SD BS."

You: "Absolutely not true. BS and a count of zero have NOTHING to do with each other. You are making a fundamental error in thinking."

Just to reinforce all that: I'm not saying they do. I was speculating that, with 52 cards left and TC zero, it might be correct to, for example, double 8 vs 5, ie. play 1D basic strategy.

The essential two points here are:

1) It is apparently not the case that 52 cards left and TC zero represents a single deck house edge (of around 0%); nor that, at this point, single deck BS applies. I'm struggling with that (OK, feel free to go to town on that last bit).

2) But that being the case, then the coincidental similarities are down to "Factor X". I do not know what that is (feel free...etc).

As noted at the top, I'm not indulging condescension or pomposity with condescension and pomposity, so dealing out more of that might be a good way to end the discussion. Admittedly, rather unsatisfactorily. But genuinely, thanks again.

"Having checked MDBJ I note it's actually 4%."

Meant 0.4%.