Why 10+2 Doesn't Equal 9+3 at the Blackjack Tables

Every experienced blackjack buff knows "the book" says to hit 12 versus a dealer's 2-up. Few solid citizens are sure why, because they seem to lose when they do. But that's another story. Suffice it to say for now that your soft underbelly is exposed in this situation no matter how you handle it, but expectation isn't quite as bad hitting as standing. The hand therefore illustrates a common blackjack predicament: choosing the lesser of two evils.

The dilemma also exemplifies another idiosyncrasy of blackjack. The chance of winning, losing, or pushing a hand against a particular dealer upcard depends not only on the point total, but on its composition. That is, ignoring the obvious difference between a 6-6 pair you can split and a 12 on which you must hit or stand against 2-up, 10-2 isn't the same as 9-3, 8-4, 7-5, or a multi-card set such as 3-5-4, 4-2-5-1, or 2-2-3-4-1.

I'll illustrate for a game played with eight decks. This is 8 x 52 or 416 total cards. The results would be different -- and more pronounced -- when fewer decks were agglomerated.

Say you start with 10-2 versus 2. You know these three cards, so 416 - 3 or 413 remain concealed.

What's the chance of drawing a 10 and busting? The original eight decks contained 16 x 8 or 128 10s; you already have one of them so the chance is 127 out of 413. If you get out your calculator or dust off your long division you'll find this is 30.75 percent.

Similarly, what's the chance of drawing anything from ace through four and not improving your position? The original eight decks contained four of each rank or 4 x 4 x 8 or 128 such cards. Of these, two -- your and the dealer's deuces -- are already removed so the chance is 126 out of 413. This is 30.51 percent.

As for cards that improve your hand, there are 4 x 8 or 32 of each rank. You therefore have 32 chances out of 413 of drawing a five and finishing with 17. Ditto for six and 18, seven and 19, eight and 20, or nine and 21. This is 7.75 percent for each case.

Here's the skinny for another two-card 12. Pretend you have 7-5.

What's the chance of drawing a 10 to 7-5 and busting? All 128 10s are still fair pickings so the chance is 128 out of 413 or 30.99 percent. Compare this to 30.75 percent for a 10-2 start. You're more apt to bust with 7-5 than 10-2. True, only by a fraction of a percent, but that and luck are what casinos are all about.

Likewise, what are your chances of drawing to 7-5 but not improving? Of the original 128 low cards, only one -- the dealer's deuce -- is gone so your chance is 127 out of 413 or 30.75 percent. This is higher than the 30.51 percent for the 10-2 so your prospects are greater for the draw not helping.

The chances differ with respect to enhancing your position when you hit as well. One five is missing from the shoe so your chances of getting to 17 are 31 out of 413 or 7.51 percent rather than the previous 7.75 percent. The same for drawing a seven and reaching 19. Probabilities of drawing a six, eight, or nine don't change. They're 32 out of 413 or 7.75 percent.

A complete analysis would follow analogous reasoning for the other combinations, considering effects not only of busting and not improving but also of ending with higher and correspondingly stronger totals. The conclusion would be that 9-3 was the worst way you could have a two-card 12, followed by 8-4 and 7-5, with 10-2 as the least sorry of an admittedly wretched lot.

Your homework assignment (er, you do do your homework before rushing to the tables and tossing money at the casino bosses, don't you?) is to try a few multi-card combinations and see whether they leave you weaker or stronger than you'd be with two-card 12s. While you're grinding out the numbers, remember the recommendations of the renowned rhymer, Sumner A Ingmark:

Ignoring the parts, viewing only the whole,
Can sometimes cause gamblers to bypass the goal.