Three Card 16s

This is likely a stupid question, but indulge me. Most of basic strategy is intuitive to me. Even something as "odd" as hitting A7 vs an A or 10*. But what I can't seem to understand is why a BS player should hit a two card 16 vs. 10, but with three or more cards totaling 16, the player should stand. I assume it is a composition dependent thing (i.e. you have three smallish cards, so it is more likely to get a big card and bust), but it's never seemed obvious - and it if is composition dependent, does it make a large difference in a shoe game?

*-A23 help your hand, 9T still give a made hand, that's 8/13 possible cards. At least that's how it makes sense intuitively to me.

The Wizard has explained it before on his site, but 16 vs. 10 is one of the most borderline (and worst) hands in Blackjack. Most BS charts calculate the best play based on initial two-card hands. There are two ways to make a hard 16 vs. 10 with two player cards (excluding 8+8 which should be split): 9+7 or 10+6; in both cases all three cards that are assumed to be visible are ones that would bust the player, which means that the proportion of helpful cards is slightly higher than usual. However, a 3 (or more) card 16 usually has more smaller cards; in many cases this tips the odds toward standing (especially if there is a 4 or 5 in the hand, since those two cards would be the most helpful). I once read one of the "ask the Wizard" questions, and he did an experiment to see which option would be better for playing a 16 vs. 10 in a perfectly neutral deck/shoe, and standing came ahead by a very tiny margin (IIRC, it was so close that he'd have to take it out to more decimal places than shown). Yet when playing from an infinite deck (where the removal of one card has no effect on the distribution of the card values) hitting comes out ahead slightly.

Another borderline hand in BJ is one that you mentioned: Soft 18 vs. A. The reason you should hit a soft 18 vs. a 9 or 10 is because the dealer is likely to get a hand of 19 or 20 (following the "assume there is a ten in the hole" logic), and since you have an opportunity to improve your hand without the risk of an immediate bust (unlike with a hard 18) you should do so. When the dealer has an A up the "ten in the hole" logic does not apply (since if there was one it would be a Blackjack and an immediate loss). Instead, the Ace's strength for the dealer is the same one that makes soft hands better for the player than hard ones of the same total (the ability to "go around twice"). The soft 18 vs. A is a closer call than with a dealer 9 or 10, but when the dealer hits soft 17 is still a clear hitting situation. If the dealer stands on soft 17 he/she is more likely to finish at 17 than with H17, and in these cases it's a close call for the player. For an initial two-card hand (A7) standing comes out ahead only in single-deck S17 (which BTW is extremely rare to find, at least with what I consider a real Blackjack game). With two decks and S17 hitting is favored on the first two cards, but the Wizard mentions that when figuring all possible ways to make such a hand standing comes out ahead slightly (and awhile back when I was playing the Wizard's Blackjack game with two decks and S17 I got an ace and 7 for my first two cards vs. a dealer Ace; the game said that standing was the better option). In a shoe game hitting (except for maybe a few composition-dependent exceptions) would always come out ahead for the soft 18 vs. A scenario.

There are several other cases where there is a logical composition-dependent explanation for certain BS exceptions (comparing bust vs. helpful vs. neutral cards). Here's two more that I can recall:

12 vs. 4: The only card that can bust such a hand is a 10-value. If the player's 12 does not have one (7+5, 8+4, or 9+3; I don't include 6+6 since splitting would be better for that one) then standing is the better option. In the case of 10+2 though (at least under a certain number of decks) the 10 that the player has slightly reduces the chances of busting and tips the odds towards hitting.

Surrendering 15 vs. 10: There are three ways to make such a hard hand: 8+7, 9+6, or 10+5. In the first case both cards the player holds are ones that would cause a bust upon drawing one, and below a certain number of decks should not be surrendered. With the latter two the player's had has one bust card and one helpful one, and with fewer helpful cards in play tips the odds toward surrendering. This same logic applies when against an Ace if the dealer hits soft 17 (if the dealer stands on soft 17 the odds are in favor of hitting this hand).

Thanks, you really have to dig deep into the archives at WoO to find the answer, but here is the link.

Let me ask you these simple strategy question. The question is which side of even money bet should you take (for or against something happening) or does it not matter for the following three questions?

(1)If I flip a coin 2 times in a row, the bet is that there will be a streak of 2 or more in a row.
(2)If I flip a coin 5 times in a row, the bet is that there will be a streak of 3 or more in a row.
(3)If I flip a coin 11 times in a row, the bet is that there will be a streak of 4 or more in a row.

Question (1) is determined by looking at the space of all possible outcomes of the experiment. There are four possible outcomes HH, HT, TT, TH. Now 2 of the four possible outcomes have a streak of 2 or more (HH, TT) and two do not have a streak (HT, TH). So the answer is that it doesn't matter what side of the bet you take since the odds of winning or losing are 50/50.

This question is usually considered a "degenerate" case, because the outcome space is so very small that anyone can figure out the odds.

Question (2) has 32 possible outcomes. Once again if you are careful about writing all 32 possible outcomes on a piece of paper and counting the ones with streaks of at least 3 or more in a row you will get 16 with streaks and 16 without streaks.

Like the simpler question above the technique is the same, but it takes a little more time and effort. (HHHTH has a streak of 3 or more, as does HHHHH). You can try it on a piece of paper.

Question (3) has 2048 possible outcomes. The answer is no longer 50/50 but the problem space has grown so large that it is no longer possible to do it on paper. The mathematics is somewhat complex as there is no simple algebraic formula. The favored decision is better by 16 out of 2048 possible outcomes.

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There is nothing intuitive about the correct answer for question #3 above. In blackjack the potential outcome space is even bigger. The closer and closer the probabilities get to even the more likely it is that the answer will not be intuitive. But the principal is the same. The computer program simply lays out all the possible outcomes and determines which ones will win and which will lose. That is the decision you should make.

There are actually many compositionally dependent strategies that you can use that will increase your expected values by very small amount. For example you always hit a two card 14 against a dealer 2. However, in reality you should usually stand on a two card 14 where one card is a 10 against a dealer 2. The word "usually" is that it may depend on the number of decks. Very small increases in EV are rarely worth memorizing the dozens of extra rules.

Oh, trust me, I believe the math. It's just hard to explain to friends that are new to blackjack why you should hit a two card 16 but stand on three or more cards, whereas some of the other esoteric BS rules are somewhat easy to explain when you lay out all of the options.

Interestingly, one of the Wizard's columns that I found while researching the problem showed that even that rule is too vague, since it depends on the exact composition of your three card 16, which I didn't realize before.