AZDuffman
AZDuffman
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March 17th, 2010 at 8:02:35 AM permalink
Is this a standard side-bet with standard rules up and down the strip, or does it vary?

And is it beatable. I have heard if you get a true-count > +5 or so it is beatable. But if the bet needs more than a surplus of 10s maybe a team could do it? One counts the Qs and another does a true count. When either is positive that person asks the other a question, "Do you want to get a buffet later?" Then the other answers yes or no or something else to signal back.

Is this viable or is the bet so much a house edge to not even bother? Am I just dreaming I will be one of those "Breaking Vegas" specials??
All animals are equal, but some are more equal than others
Croupier
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March 17th, 2010 at 8:31:15 AM permalink
The Wizards info on Lucky Ladies implies the rules are the same but the paytables can be different.

With a house edge in the high 20+% my guess is that it would need a whole lot of work to get near break even, never mind beat it, even with counting.

Although it is a nice idea.
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AZDuffman
AZDuffman
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March 17th, 2010 at 8:55:48 AM permalink
Quote: Croupier

The Wizards info on Lucky Ladies implies the rules are the same but the paytables can be different.

With a house edge in the high 20+% my guess is that it would need a whole lot of work to get near break even, never mind beat it, even with counting.

Although it is a nice idea.



Lets just assume Table "C" for now.

Lucky Ladies - Pay Table C

Pair of queens with dealer BJ 250 to 1
Pair of queens 25 to 1
Ranked 20 9 to 1
Suited 20 6 to 1
Any 20 3 to 1

So here a A is your enemy in a high count since you lose the LL bet even if you get BJ. Three of us sit down. We will use aliases of Tom, Dick, and Harry and assume a six-deck shoe.

Tom: His job is to do a normal Hi-Lo count.
Dick: His job is to count how many queens are left.
Harry: His job is to count the decks remaining and the A's remaining.

Now, when we get to a count of say +8 or more, Tom says an arranged signal phrase. Maybe he has been flat betting and says, "Not sure if I feel lucky or not?"

Dick has been counting the Qs. Dick who has been betting 1 unit stays at 1 if it is bad or goes to 2 or 3 if it is good based on the math.

Harry returns a signal phrase or action based on the decks and As left.


Clearly you could go all day and getting caught is a possibility. But could the deck get favorable enough that you might get a chance or two an hour at this?
All animals are equal, but some are more equal than others
boymimbo
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March 17th, 2010 at 9:49:26 AM permalink
The 20 can be achieved in 2 ways: 10-10 and A-9. I am going to look at Table "A". 4-9-19-125-1000

In a six deck shoe,
A 20 occurs with 10-10 on (96/312 * 95/311) occurences = 9.39896%,
A 20 occurs with A-9 on (24/312 * 24/311)*2 occurences = 1.18724%.
This agrees with the Wizard's Table A - six deck, 20 nonoccurences of 89.41380% = (100 - 9.39896 - 1.18724)

The HA is 24.709% for the Table A outcome.

Since we count only 10s vs low cards, we would disregard the A-9 occurrence.

The break even point on Table A occurs when the number on non-winners is <.8594 (I evenly distributed the number and frequency of winners into Table A based on the non-winners to get the HA = 0).

At this point the ratio of A-9s remains the same at .018724. That means that the 10-10s must occur 1-.8594-.018724 = .12873.

Now we solve for the frequency of 10-10 to be at .12873. Simply take the square root of that number to get .35878. That means that there needs to be 18.656 10s per deck. (.35878 * 52)

At a true count of +1, there are supposed to be 16.5 10s per deck.
At a true count of +2, there is supposed to be 17 10s per deck... and so on.

So when the true count is +5, there are 18.5 10s per deck (and 13.5 3s to 6s).
And at +6, 19 10s per deck.

When there are 19 10s per deck, the odds of getting a 10-10 combination is approximately 19/52*18/51 = 12.8959% > 12.873%.

So my advice is that when the count is +6 or higher and you are playing pay table A, hammer the lucky ladies.
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cardshark
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March 17th, 2010 at 10:04:51 AM permalink
You can absolutely KILL the LL side bet using a simple hi-lo count. TC (true count) of +7 and up gives +EV in the 6/8 deck game. Don't bother with the pair of queen of hearts 1000-to-1 payoff, its worth very little in terms of EV when compared to the other "20"s payouts. Its not even worth it to keep a side count for the queen of hearts (despite the fact that this is very easy to do.)

I won about $10,000 off LL alone over my card counting "career". When the TC reached +10 I placed a max bet on the LL (usually $25, though I did see a few $100 maxes back in the day). I didn't even care about my main bet because I knew at TCs of +10 and more, my advantage was over 10%.

Unfortunately, all the casinos near me removed the side bet. A lot of the tables replaced LL with Perfect Pairs and Match the Dealer (all useless to the counter.) I guessed the casinos wised up.

I did notice when I was in Vegas a few years back that some of the single deck 6-to-5 tables had a version of the LL bet. It is different from the one I used to play - I'm not sure what the count would need to be to take advantage of it.
Wizard
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March 17th, 2010 at 10:05:40 AM permalink
There is a lot of literature already about Lucky Ladies:

Beyond Counting second edition has 9 pages devoted to it (pages 62-70)

The Big Book of Blackjack has a short chapter on it. Snyder says to make the bet if the Red Sevens count is +10 or greater, and only in the last two out of six decks.

The real Current Blackjack Newsletter indicates what casinos offer Lucky Ladies, and whether the liberal or stingy pay table.
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cardshark
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March 17th, 2010 at 10:09:52 AM permalink
Quote: boymimbo

The 20 can be achieved in 2 ways: 10-10 and A-9. I am going to look at Table "A". 4-9-19-125-1000

In a six deck shoe,
A 20 occurs with 10-10 on (96/312 * 95/311) occurences = 9.39896%,
A 20 occurs with A-9 on (24/312 * 24/311)*2 occurences = 1.18724%.
This agrees with the Wizard's Table A - six deck, 20 nonoccurences of 89.41380% = (100 - 9.39896 - 1.18724)

The HA is 24.709% for the Table A outcome.

Since we count only 10s vs low cards, we would disregard the A-9 occurrence.

The break even point on Table A occurs when the number on non-winners is <.8594 (I evenly distributed the number and frequency of winners into Table A based on the non-winners to get the HA = 0).

At this point the ratio of A-9s remains the same at .018724. That means that the 10-10s must occur 1-.8594-.018724 = .12873.

Now we solve for the frequency of 10-10 to be at .12873. Simply take the square root of that number to get .35878. That means that there needs to be 18.656 10s per deck. (.35878 * 52)

At a true count of +1, there are supposed to be 16.5 10s per deck.
At a true count of +2, there is supposed to be 17 10s per deck... and so on.

So when the true count is +5, there are 18.5 10s per deck (and 13.5 3s to 6s).
And at +6, 19 10s per deck.

When there are 19 10s per deck, the odds of getting a 10-10 combination is approximately 19/52*18/51 = 12.8959% > 12.873%.

So my advice is that when the count is +6 or higher and you are playing pay table A, hammer the lucky ladies.



Hey boymimbo,

Not to argue with you or anything, but at a TC of +6, the EV is slightly negative. Check here for more info:

http://www.blackjackincolor.com/blackjacksidebet1.htm

Like I said in my post above, I only played at TC +10 and more - it helps minimize variance.
boymimbo
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March 17th, 2010 at 11:09:57 AM permalink
Quote: cardshark

Hey boymimbo,

Not to argue with you or anything, but at a TC of +6, the EV is slightly negative. Check here for more info:

http://www.blackjackincolor.com/blackjacksidebet1.htm

Like I said in my post above, I only played at TC +10 and more - it helps minimize variance.



I got close enough. I approximated the square root. The frequency of 10-10 at .12873 still applies. But the frequency of that would I guess actually be n x (n-1) = .12873 x 52 x 51. At this point n still equals 19.

I still like my calculation that the HA is about even at +6. Really it's an approximation anyway because you don't take into account how many decks are left.

Still, I'll add lucky ladies to my repertoire for a shoe dealt game when the TC is +7 or higher.

How often is +7 or higher achieved in a 6 deck shoe anyways?
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cardshark
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March 17th, 2010 at 11:29:18 AM permalink
From same website: http://www.blackjackincolor.com/truecount1.htm

It's not often and it doesn't stay there for very long. If you are going to be playing solely for the LL (therefore only playing TC+7 and up) you will be playing the waiting game. However, if you use it in addition to your regular counting strategy, its excellent. I don't know why any counter would not add this to his/her game - you don't need to learn a new counting system and its EV is HUGE when advantage playing. If you don't like the volatility, like me, just play it at TC+10 and up. Very, very worthwhile. Hell, I'd still be playing it if any casino in a 500 mile radius of me still offered it.
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