Blackjack basic strategy table question?

For anyone more intimately familiar than I with blackjack strategy or probability, I have a question about a basic strategy table linked from this site (at www.blackjackinfo.com).

Referring to the table for "6 decks, H17, DAS, Late Surrender, Peek", I'm confused by one recommendation. It says to surrender a pair of 8s against a dealer A, but to hit a single 8 against a dealer A. Why don't the recommendations match? What's different about splitting and then playing two of the 8 vs. A hands?

You stand to lose up to twice as much.

I suggest you go to www.bjmath.com in the reference section, and see the expectations for all the card combos in the different games.

According to the tables in Appendix E of Wong's Professional Blackjack, splitting a pair of 8s versus a dealer Ace results in a loss of .518 of your original bet total in the game you describe. Since surrender is a loss of .500, then surrender is the proper play. However, a two card hand total of eight is a different proposition. If you hit the eight against an Ace, your loss is either -.266 or -.264, depending upon the composition of the hand. In either case, such a loss is less than the loss of .500 for surrender.

Now you may ask; why is the expected loss on a two card hand total of eight more than half of the loss for splitting 8s? Primarily, it is because you may be able to double down if you draw a 3 on one or both of your eights, which you cannot normally do once you hit a two card hand which totals eight. In a six-deck, H17 game, doubling an 11 versus an Ace is a positive play, so you gain back some of the loss from your one hand total of eight.

I don't understand why that matters at all. If an 8-8 against an A is such a losing hand that you should surrender rather than split, why play out a single 8 against an A? Why not surrender that hand also?

Again, I think I must be missing something fundamental. It sounds like you're saying that a one-card 8 is better than a two-card 8 because of the possibility of a favorable double-down if you get 3s. So if it's better, why are you told to surrender it (the two one-card 8s if you split) while playing out the less favorable two-card 8?

Please take another look at my post. It gives the specific expected values for the hands and plays. Simply put, you lose less by making the proper play. Remember that when you split 8s, you are doubling your bet. This makes your total loss larger than what will happen to your two-card eight, even though the expected loss for a single hand total of eight is less.

Here's what happens with some numbers plugged in.

Let's assume you have a $10 bet down. If you split the eights, you will lose, on average, $5.18 for the total hand. If you surrender the eights, you will lose $5.00. So, you're 18 cents ahead by surrendering.

Now let's look at a two-card total of eight. If you hit the hand, you will lose, on average, either $2.66 or $2.64, depending on the hand composition. If you surrender, you will lose $5.00. In this scenario you are either $2.34 or $2.36 better off than you would have been by surrendering.

Do you understand now?

I think I understood this concrete explanation much better, thanks. I didn't quite grasp the asymmetry of the benefit of surrendering the pair of 8s. And the idea of split 8 still losing less, $2.59, per hand than two-card 8 also fits with my intuition of a reduced chance of getting stuck with 16s. Thanks for your persistence.