How much % does one give up by never doubling 11 vs. 10 or Ace

Just curious what the long-run negative impact is if you only double 11's against 9's and lower, and never double 11's vs. 10 or A regardless of the count.

Thanks.

If you are playing shoes, it isn't basic strategy to double an 11 versus an Ace. According to the charts in Professional Blackjack, by Stanford Wong, the six-deck expectation for doubling 11 versus 10 ranges from 0.174% for a 9-2 to 0.178% for most of the other hands. Hitting gives you 0.117% to 0.118%. So, you lose around 0.57% each time you hit instead of stand. If you are in a single deck, h17 game, the percentage for doubling an 11 versus an Ace is 0.189% for the 9-2, to 0.240% for the 5-6 combination. Hitting ranges from 0.170% to 0.173%. In that single decker, you lose from 0.019% to 0.067% by hitting that hand. For an 11 v. 10 in a single decker, the expectation ranges from 0.130% to 0.141% for hitting, while doubling yields from 0.181 to 0.233, so you would lose from 0.051% to 0.092% for hitting.

If you look up the hand frequencies, you could compute the overall loss for not making those plays.

Perhaps the .57% should be .057%?

Never seen Bootlegger make a typo before, but he is human too. See what eventually happens when you are as helpful as Boot!

The numbers in Appendix E are in units of original bet, not %. The following is for 6 decks. For doubling 11 vs 10, the advantage is 17.4% to 17.8% of the original bet. And for hitting 11 vs 10, the advantage is 11.7% to 11.8% of the original bet. So you average winning 5.7% more of your original bet each time you double, instead of hit, your 11 vs. dealer's 10. If your original bet was $100, you would average making $5.70 more by doubling, than by hitting.

It is basic strategy to double 11 v. Ace in shoes with H17.

You are right. I was in error when I expressed it in terms of percentages. My bad, and I apologize for the mistake.

I was assuming an s17 game, since I never play h17 shoe games, but you are correct.