11 vs. 10 and 11 vs. 11 question

I have a question regarding basic strategy that struck me recently. In a 6 deck game, you are supposed to double down on 11 vs. 10, but hit on 11 vs. 11. It seems to me that these situations are similar if the dealer checks his hole card and does not have blackjack. In the 11 vs. 10 case, if the dealer does not have BJ, his hole card is (2-10). In the 11 vs. 11 case, if the dealer does not have a BJ, his hole card is (A,2-9). I realize these cases are not the same, but seem somewhat similar. Is the reason that 11 vs. 11 calls for a hit that the flexibility of an A is greater than that of a 10? In other words, both hole cards have the possibility of being a (2-9). The only difference is that the 11 vs. 10 may have a 10, and the 11 vs. 11 may have an A. I am guessing that the dealer is more likely to bust in the case of 11 vs. 10 than in the case of 11 vs. 11, which is why you double on the 11 vs. 10, but not the 11 vs. 11. Is this correct, non-mathematically speaking? I realize that it is not right to consider what the play should be intuitively, because everything is based on probability. However, if someone asked you to explain this non-mathematically, would my thought process be correct, in any way, shape, or form?

I hope this is not too jumbled. If it is I apologize. :)

-- Thanks, Homer J.

I'll try a non-mathematically explanation without saying "the math has been done" or "the percentages of what you'll win/lose has been figured" according to your decision. Your guess is right in that the dealer is more likely to bust with a Ten up than with an Ace up because of the "flexibility" of the Ace.

A real short non-mathematical explanation: When a dealer shows an Ace and the dealer has to draw (he flips up an A-5), he has 2 chances to make his hand, making him less likely to bust.

Longer explanation: When you double on 11 and get a bad card (A-5), you must hope the dealer busts in order to win. If a dealer has a 10 showing and flips a 2-6 causing him to draw, all it takes is one Ten to bust. However, if a dealer has an Ace showing and flips up and Ace or 2-5, drawing one Ten will not cause him to bust - he gets 2 chances to "make" his hand: soft 17-21, then hard 17-21. He needs two Tens to bust.

That's as non-mathematical as I can get.

But if you count cards and the TC is +1, make sure you double your 11 against the Ace. Also, basic strategy says to double 11 vs. Ace if the rules are H17, even for 6-deck games.

11 vs 10 add'l adv by double over hit is aprox 6%.

11 vs A add'l adv by hit over double is 1-2%.

While the full answer is obviously the expected value, the following thoughts might give you a clearer idea where that value comes from.

When you have a two card total of 11, you will reach 21, 4/13 = 30.77% of the time...When the dealer has a face card, he will make a three or more card total of 21 only 3.72% of the time...He will also bust 23.08% [approx.]of the time...If the dealer has an Ace, he will reach 21, 7.78% of the time and bust 16.74% [approx.]

It is those kind of percentages in combination with the other probabilities involved in this situation which cause the correct answer to be exactly what the computers tell us it is.

Every time I double 11 against a ten and draw an ace I wonder why we don't double a ten against a ten.

That is why the I18 contains double 10 vs 10 and double 10 vs Ace in it. You make both plays at HiLo +4, (+7 and +5 risk-adverse). Two very profitable plays.

I find the double 10 vs Ace particularly rewarding in high counts, you almost always get 20 or 21, and if you don't, the dealer's chances of busting are even better than normal. (Helps you win back that insurance bet, since he didn't have blackjack.)

10 vs 10 and 11 vs 10. The dealers most likely hand and best reasonable likely hand to get is twenty.
When you double 11, you most likely get 21, almost sure to beat the dealer. When you double 10, you most likely get 20 for a likely push making a big difference between the two cases.

Part of the problem is selective memory. Remember 10 is 4 times more likely than ace.