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
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
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.
----- You want the truth! You can't handle the truth!
cardshark
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
Administrator
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.
It's not whether you win or lose; it's whether or not you had a good bet.
cardshark
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
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?
----- You want the truth! You can't handle the truth!
cardshark
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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