Why not double down 11 against an Ace?
BobbyShaw
double down all 11's except against the dealer Ace? You already know the
dealer doesn't have 21, so why not get more money out and hope for your own 21?
Robert Shaw
CAM
Basic Strategy for single deck indeed does call for doubling 11 VS Ace.
It's multiple deck Basic Strategy that calls for hitting 11 VS Ace.
We know the dealer's down card is not a ten and the dealer's up card
is an Ace. Those two cards are not available to be received on a
double down. The player's hand of 11 contains two non-tens and are
not available to be received on a double down. In a single deck game,
those four non-tens not being available tips the balance toward
doubling. However, in a multiple deck game, there are other cards
identical to those four undesirables still left in the deck and that
tips the scales toward hitting instead of doubling. Computer simulations
have proved it.
DERMADMAN
>Why do you
>double down all 11's except against the >dealer Ace?
I do double down on all 11's versa a dealers ace. It's part of a one
strategy fits all strategy. However, you make a little more in multi deck or
european no hole to just hit it; or maybe you lose less. So, if you play in
all sorts of games like me, then you might just want one comprehensive
strategy. If you always play in multi deck or european no hole, then you might
want to hit.
*** . . . --the mad man
Mark (ATL)
if the game is Hit Soft 17. Hit 11 vs Ace if the game is Stand on all
17.
On Fri, 4 Aug 2000 04:29:49 PST, bobb...@aol.com (BobbyShaw) wrote:
>This is one of the only basic strategy situations that I question. Why do you
>double down all 11's except against the dealer Ace? You already know the
>dealer doesn't have 21, so why not get more money out and hope for your own 21?
>
>Robert Shaw
>
Rob M Rosengard
So, if you are given the chance to DOUBLE on someone else's
11 Vs. Ace you should do so.
I believe that in a H17 game, even multiple deck, you should
always double 11 Vs. Ace.
R
Mark (ATL)
than Hitting 11 vs Ace. There are different potential outcomes and
therefore different expectations on returns.
VBSlinger
20, or 21.
Of course that's optimal, but you are also betting on the dealer busting if
that fails. If the dealer is showing a 10 and you have 11 - I don't have
the numbers memorized, but the dealer is more likely to bust than you are to
draw a 21(I think).
The Ace is dangerous because the dealer's chances of busting are reduced
dramatically. Your chances of winning the double now hinge on having a pat
total more so than the dealer busting.
Just a toss it,
The Slinger
"BobbyShaw" <bobb...@aol.com> wrote in message
news:pgpmoose.2000...@shell9.ba.best.com...
Al Krigman
<vbsl...@hotmail.com> writes:
>When one doubles on a 10 or 11 they automatically assume it is to get a 19,
>20, or 21.
>Of course that's optimal, but you are also betting on the dealer busting if
>that fails. If the dealer is showing a 10 and you have 11 - I don't have
>the numbers memorized, but the dealer is more likely to bust than you are to
>draw a 21(I think).
>
This type of reasoning is convenient for picturing the situation in your mind
and getting comfortable with the extra bet. But Basic Strategy (and card
counting spreads and variations) aren't predicated on *qualitative* thinking
but on *quantitative* analyses. When all possible combinations of player and
(as you note) dealer results are considered, along with their probabilities, it
turns out that a bettor's "expectation" -- the "expected value" of the money at
risk -- is greater when "the book" says to double and you do than when you play
a particular hand any other way, and so forth for all other situations. (And
conversely when 'the book' says to hit rather than double, etc.) In an 8-deck
game, for example, doubling on 5-6 vs A has an expected profit of 12.4 cents
per dollar of initial bet, whereas hitting the same starting hand has an
expectation of 14.6 cents on the dollar (this is for 'playable' aces -- dealer
does not have blackjack). For 9-2 vs A, the figures are 11.8 cents/dollar for
doubling and 14.6 cents/dollar hitting.
I show the two for comparison because it heightens the emphasis on the origin
of strategies with "maximum expectation" as the optimization criteria. With
5-6, the two cards (out of 413 unknowns) that are unavailable will both be
relatively undesirable (you don't want to finish with a 16 when you double, and
a 17 isn't all that much better). With 9-2, one of the unavailable cards is
useless (the 2 -- finishing with 13 is no different than finishing with 16),
but the 9 is relatively strong, and its being missing hurts you. So doubling on
9-2 is weaker than doubling on 5-6, although neither is as good as hitting.
Note, though, that expectation is positive either way. And by doubling, you'd
increase the variance (ergo the bankroll swings) in a favorably-skewed manner.
So, if your performance criterion isn't maximum expectation but is something
based on short-term bankroll swings, you can make a case for doubling 11 vs ace
-- and that case gets stronger as your starting hand goes from 9-2 down
(through 8-3 and 7-4) to 6-5. I'm not advocating this, or setting it up as a
straw issue to be shot down, but only pointing it out as an option with pros
and cons.
Alan Krigman
Syndicated gaming columnist;
Publisher of WINNING WAYS newsletter (for a free no-SPAM no-strings-attached
subscription to the e-mail edition, send request to punte...@aol.com or
check http://hometown.aol.com/punterpress/signup.html)
Al Krigman
newsletter, I did the analysis based on the good old (i.e., easy) infinite shoe
approximation and found the following as the probabilities of various end-point
totals for the player hitting and doubling, as well as for the dealer, in this
situation. These totals gave the next table -- the probabilities of win, loss,
and push for either option. Expectation (in percent) when hitting is then
probability of win minus probability of loss; expectation when doubling (based
on the initial bet) is twice the difference. For the infinite shoe, the
expectations are +0.143 hitting and +0.108 doubling
Note, the expectations get closer together as the number of decks in play gets
smaller (hitting goes down slightly, doubling goes up). According to Griffin (I
didn't check this against the detailed expectation tables -- something has to
be left to the gentle reader -- there should be a crossover (doubling is better
than hitting, expectationwise) for all 11s in a one-deck game and for 5-6 and
4-7 but not 3-8 or 9-2 in a two-deck game.
(If you don't subscribe to the free e-mail edition of Winning Ways, see below
about how to do so ... mention in your "opt-in" request that you'd like the
11-vs-ace issue and I'll send it to you when I acknowledge your subscription
request.
(to have the following align properly, block the text and select a
"letterspaced" font such as Courier)
========================================
Probabilities of various final
point totals
for player doubling or hitting 11
against a playable dealer ace-up
final hand double hit dealer
12 to 16 38.4% 0.0% 0.0%
17 7.7% 11.1% 18.9%
18 7.7% 11.1% 18.9%
19 7.7% 11.1% 18.9%
20 7.7% 11.1% 18.9%
21 30.8% 34.3% 7.8%
bust 0.0% 21.3% 16.6%
========================================
Probabilities of various outcomes
for player doubling or hitting 11
against a playable ace
outcome double hit
player win 48.6% 51.6%
player lose 43.2% 37.3%
push 8.2% 11.1%
========================================
Emilywrite696777
a 42% chance (5 in 12 - the Ace doesn't count since it would be a blackjack for
the dealer) that the dealer would have to draw.
The likelihood of a dealer busting with an Ace showing drops dramatically,
since there are a number of ways the dealer can make his hand.
My best reason for doubling down 11 is to save myself from a bust (since you're
only drawing one card). How often have we seen a bad draw on the double down -
only to be seen when the dealer busts with a 10 up? Doubling down 11 hoping to
catch a 10 is difficult statistically - around 31% (4 out of 13).
MIDIGOD
<<A bunch of well-thought-out tables and specs, for which I am grateful.>>
Hi Al:
I don't post much to this group, since I prefer to sit back and learn
something, but I have to push you further in this analysis.
I'll respect your numbers as a "given" for this discussion; but the part I
don't see is how the win amount factors in. For example, if both plays were a
win of one unit, it would be obvious that the hit would be the proper play.
But, since the double still has the win advantage, and wins twice as much, how
does this affect the *final* outcome?
I understand that you can't simply double the winning expectation, since you've
also wagered more, but since the expectation is ultimately positive, it seems
to me that the double would give you a better percentage, as far as final
outcome goes, and not simply cancel itself out.
Can you tell me how that would factor in? It looks as though your tables (the
second in particular) deal only in wins and losses, and ignore bet size-related
outcomes.
Thanks!
-Craig
Al Krigman
(MIDIGOD) writes:
>I'll respect your numbers as a "given" for this discussion; but the part I
>don't see is how the win amount factors in. For example, if both plays were
>a
>win of one unit, it would be obvious that the hit would be the proper play.
>But, since the double still has the win advantage, and wins twice as much,
>how
>does this affect the *final* outcome?
>
Good question. A lot of people are confused about this... and most won't admit
it! Here's a repeat of the second "table" from my post:
outcome double hit
player win 48.6% 51.6%
player lose 43.2% 37.3%
push 8.2% 11.1%
It's telling you the following:
a) if you double -- you've got two units bet. Your net expected win will be
2*(.486 - .432) which equals 2*(.053)=.108. So, if your bet at the beginning of
the round was $1, your "expected" win -- factoring in the amounts and the
probabilities -- is 10.8 cents.
b) if you hit -- you've got one unit bet. Your net expected win will be 1*(.516
- .373) which equals .143. So if the initial bet was $1, your expected win --
again factoring in the probabilities -- is 14.3 cents.
Think of the probabilities as follows. Say you are in this situation 1,000
times. If you experience the "statistically correct" proportions of results,
doubling you'll win 486 times, lose 432 times, and push the other 82 times. So
your net profit will be wins minus losses -- times two because of the double
(i.e., 486 $2 wins and $432 $2 losses -- so you end up with $972 minus $864 or
$108). Just hitting, you'll win 516 times, lose 373 times, and push the other
111 times. So your net will be wins minus losses and you'll end up with $516
minus $373 or $143.)
See how it works? You've got to factor in the probabilities of winning and
losing as well as the money. This gets you the "expectation." And Basic
Strategy is the set of decisions that gives the player the greatest expectation
for every situation.
Of course, there are other factors you might choose to consider. Your bankroll
swings will be much greater if you always double than if you always hit --
because you've got twice as much at stake. This means that on an individual
session basis, when you have a statistically small number of decisions, the
inherent volatility of the game (measured by statisticians as "variance" or
"standard deviation") swamps the expectation, so the doubles could make you a
lot of money or could kill you. You can play this way -- using volatility as
your performance criterion. This would give you more doubles and splits than
you get under Basic Strategy. Long term, it wouldn't be profitable. But short
term, doubling whenever expectation was positive to do so, even though it was
less positive than hitting or standing, could make you a lot of money -- or, as
stated, could wipe you out before you knew what hit you. I don't recommend this
-- I'm only pointing out that your criterion for optimizing your play does not
necessarily have to be based on expectation alone.