Blackjack : Playing 2 hands against Dealer

To specify : playing 2 hands against the dealer...with nobody else at the table.

they say it doesn't matter if your playing with perfect strategy...whether you're the only one at the table, or if you're playing 2 hands against.

Since it's generally 50/50 shot (not precise....but u know) when 1 on 1 against dealer.

But the odds should go in the player's favor when playing 2.

33% / 33% / 33%

since the player is covering 2 :

66% / 33% (generally speaking)

no...?

you know it can't be that easy, so this is a rhetorical question, right?

it's not 33% / 33% / 33%

That's not quite true. Let's simplify it to a coin-flip. 1 hand verse dealer would be 50%/50%.

2 hands verse dealer 1 hand you'd get something like:

8 possible outcomes: D-P1-P2 for dealer-player1-player2.

HHH : 0
HHT : -1
HTH : -1
HTT : -2
THH : +2
THT : +1
TTH : +1
TTT : 0

H is win, T is lose.

1/4 of the time you push both
1/4 of the time you win 1 push other
1/4 of the time you lose 1 push other
1/8 of the time you win 2
1/8 of the time you lose 2

Its not 66% / 33%.

but it's at-least slightly better than 1 vs 1 right ?

if u were to put a percentage on it ..?

Or consider a stupid game where you receive 19, 20 or 21.

Dealer	Player 1	Player 2	Win	Lose	Result	-2	-1	0	1	2
						5	6	5	6	5
19	19	19	0	0	0	0	0	1	0	0
19	19	20	1	0	1	0	0	0	1	0
19	19	21	1	0	1	0	0	0	1	0
19	20	19	1	0	1	0	0	0	1	0
19	20	20	2	0	2	0	0	0	0	1
19	20	21	2	0	2	0	0	0	0	1
19	21	19	1	0	1	0	0	0	1	0
19	21	20	2	0	2	0	0	0	0	1
19	21	21	2	0	2	0	0	0	0	1
20	19	19	0	2	-2	1	0	0	0	0
20	19	20	0	1	-1	0	1	0	0	0
20	19	21	1	1	0	0	0	1	0	0
20	20	19	0	1	-1	0	1	0	0	0
20	20	20	0	0	0	0	0	1	0	0
20	20	21	1	0	1	0	0	0	1	0
20	21	19	1	1	0	0	0	1	0	0
20	21	20	1	0	1	0	0	0	1	0
20	21	21	2	0	2	0	0	0	0	1
21	19	19	0	2	-2	1	0	0	0	0
21	19	20	0	2	-2	1	0	0	0	0
21	19	21	0	1	-1	0	1	0	0	0
21	20	19	0	2	-2	1	0	0	0	0
21	20	20	0	2	-2	1	0	0	0	0
21	20	21	0	1	-1	0	1	0	0	0
21	21	19	0	1	-1	0	1	0	0	0
21	21	20	0	1	-1	0	1	0	0	0
21	21	21	0	0	0	0	0	1	0	0

Quote: john2010

but it's at-least slightly better than 1 vs 1 right ?

if u were to put a percentage on it ..?

I would have to say, it depends. (I'm not a pro, just recreational). I think you have to look at what the counters do; they're not playing 2 hands when the deck is neutral or negative; they only spread to 2 or more when the game has a lot of 10's available. So that tells me that, overall, there's no real advantage to it by itself. Long term, anyway.

I think the big problem is, all your hands have to act before the dealer does; you have to bust first (if you're going to bust), and you have to decide on your full hand (if you don't bust) before you see the dealer's whole hand. Since an unseen card is an unseen card, what you do with one hand really doesn't affect any others, whether yours or someone else's. So I would have to say it's probably slightly negative most of the time.

Quote: john2010

but it's at-least slightly better than 1 vs 1 right ?

if u were to put a percentage on it ..?

No. Of course it's worse. Each hand the player gets is exposed to the house edge. And you're doubling the hands the player has...

i'm more inclined to believe beachbum on this one.

the reason..? I used this technique twice in vegas last week...and came back from being $2000 down both times.

2 hands against the dealer.

played perfect strategy for 1 exemption....a lot of surrenders from 12's and up. (13, 14, 15, 16)

for instance on a hand...

I have a 13 on one hand...12 on the other hand.

dealer has a 10.

I surrender on both. If I would of hit on both, I would of busted on both. (-2)

you keep the count high by surrendering even when you shouldn't. yea, u could say something like this isn't going to work over the long haul ..but for me it has.

If things start getting hairy, I either go back to 1 hand...or slowly up my bets.

Quote: john2010

i'm more inclined to believe beachbum on this one.

the reason..? I used this technique twice in vegas last week...and came back from being $2000 down both times.

2 hands against the dealer.

played perfect strategy for 1 exemption....a lot of surrenders from 12's and up. (13, 14, 15, 16)

for instance on a hand...

I have a 13 on one hand...12 on the other hand.

dealer has a 10.

I surrender on both. If I would of hit on both, I would of busted on both. (-2)

you keep the count high by surrendering even when you shouldn't. yea, u could say something like this isn't going to work over the long haul ..but for me it has.

If things start getting hairy, I either go back to 1 hand...or slowly up my bets.

Sounds like an awesome plan! I wouldn't go near it, but if it works... >_>

Quote: john2010

To specify : playing 2 hands against the dealer...with nobody else at the table.

they say it doesn't matter if your playing with perfect strategy...whether you're the only one at the table, or if you're playing 2 hands against.

Since it's generally 50/50 shot (not precise....but u know) when 1 on 1 against dealer.

But the odds should go in the player's favor when playing 2.

33% / 33% / 33%

since the player is covering 2 :

66% / 33% (generally speaking)

no...?

Welcome fellow blackjack player! Are you counting cards, John?

For a BS player I would view this as a means of reducing variance for the same amount of total bet.

If you are initially playing $10 / one hand, and then switch $10 * two hands, your long run -EV is doubled because you have doubled your total amount wagered.

But suppose you are going to bet $20 total no matter what. Either a single $20 bet, or two $10 bets.
You are exposing $20 to the -EV% no matter what.
But there will be more often hands with a net result of zero or +/- $10 (which is 0.5 betting units). So your variance is less, compared to a single hand of the same total $.


I would ask the counters out there, if you spread to multiple hands at a positive count, is there a *mathematical* advantage to the multiple hands versus just increasing your bet on a single hand?

Not better. Each of the two hands has the same edge as any one hand. Why would it matter if you were playing two hands versus you playing one and someone else playing one? How could this be different for the casino?

cyrus is right that is reduces variance. So it's actually less likely for you to win back your money this way. You just got lucky.

What's the difference if you or another person is playing the 2nd hand? Nothing.

There's a correlation on winning If the dealer busting a lot. There's a correlation on losing if the dealer keeps making hands.

For a normal player it won make a difference.

TBH if you don't see this, you might want to rethink playing BJ.

When the count is good, spreading gives you more chances at a natural 21.

If it is just one spot and the dealer, you both get the same number of hands.

If you spread to 2 spots, you get twice as many hands as the dealer.

More 10's = more likely to get 21. More hands on the table owned by you = greater chance of you getting the 21.

More chances for you to make good doubles, too.

Those things are supposed to be able to overcome the more frequent dealer 21s.

Quote: john2010

i'm more inclined to believe beachbum on this one.

the reason..? I used this technique twice in vegas last week...and came back from being $2000 down both times.

2 hands against the dealer.

played perfect strategy for 1 exemption....a lot of surrenders from 12's and up. (13, 14, 15, 16)

for instance on a hand...

I have a 13 on one hand...12 on the other hand.

dealer has a 10.

I surrender on both. If I would of hit on both, I would of busted on both. (-2)

you keep the count high by surrendering even when you shouldn't. yea, u could say something like this isn't going to work over the long haul ..but for me it has.

If things start getting hairy, I either go back to 1 hand...or slowly up my bets.

Hi John, and welcome to the forums. I'm going to give you some brutally honest advice/opinions! Surrendering randomly from 12-16 is not a good idea. I believe the Wizard did some analysis where the player always surrenders 12-16's against a dealers 10 and it jacked the house edge up a couple percent (if I recall correctly). There's a reason basic strategy is basic strategy and you should only surrender 15's and 16's at their appropriate times. That reason is math. It's not hunches, guesses, feelings, etc. It's proven mathematics. So while I'm am seriously glad it worked out for you 'this time' I hope you understand that surrendering your 12's and 13's is a very very poor decision (mathematically) and if you did that for the rest of your playing days it would absolutely cost you a lot of money.

Next, on to the question at hand... With your logic why wouldn't you play 6 spots against the dealer? Then you'd be 6/7 favorite, right??? Anyone whom tries this once or twice will learn very quickly you're not a 6/7 favorite when you play all the spots. As Babs said, you must play your hand before the dealer. Also, as she mentioned, what one hand does does not affect the other hands. They are all independent hands with the same house edge against them. Every bet you place in the separate circle is subject to the house edge of the game. Thus, playing more hands will do nothing more than make you lose more money per hour.

Example:

Let's say the house edge is .5%. If you place a $100 bet off the top of the shoe, then your expected value (loss) is EV = $100(-.005) = -.50. So if you play one hand you can expect to lose 50 cents on average in the long run. If you play two hands, they're independent from one another, but both subject to the house edge. Thus, the math is the same and you're simply adding another 50 cent loss, on average. So by playing 2 hands, you're doubling your expected loss.

The only time this changes is when one is counting and accurately notices the deck is in the players favor. Most true counts are worth about .5% to the player. So at True Count +2, the PLAYER now has a .5% advantage. Thus, your EV for one hand now would be EV = $100(.005) = .50. So now you're looking to MAKE 50 cents from every hand you play at this point in time (true count +2). Now adding a hand is a good idea because it becomes more profitable and helps you make more money.

Lastly, I just wanted to re-iterate what others have already said... Your 33%/33%/33% numbers are just absolutely incorrect. This correlates every hand as though there was only 1 winner per round. Not only that but if that were the case you'd still never be 67% because you'd always lose the other hand. Again, like mentioned above, if this was at all correct, why wouldn't you play six hands against the dealer to be an 86% favorite??? Hopefully this explanation helps, and I hope you keep having good luck.

Quote: Romes

Hi John, and welcome to the forums. I'm going to give you some brutally honest advice/opinions! Surrendering randomly from 12-16 is not a good idea. I believe the Wizard did some analysis where the player always surrenders 12-16's against a dealers 10 and it jacked the house edge up a couple percent (if I recall correctly). There's a reason basic strategy is basic strategy and you should only surrender 15's and 16's at their appropriate times. That reason is math. It's not hunches, guesses, feelings, etc. It's proven mathematics. So while I'm am seriously glad it worked out for you 'this time' I hope you understand that surrendering your 12's and 13's is a very very poor decision (mathematically) and if you did that for the rest of your playing days it would absolutely cost you a lot of money.

that's why I only do it when (lets say) I haven't seen a lot of 10's yet. (jack/queen/kings included)

Quote:

Next, on to the question at hand... With your logic why wouldn't you play 6 spots against the dealer?

Most casinos only allow you to play 2 hands anyway.

Quote:

if you spread to multiple hands at a positive count, is there a *mathematical* advantage to the multiple hands versus just increasing your bet on a single hand?

yes, when done at the right time. But i'm also not endorsing it. (proceed at ur own risk)

Quote:

Those things are supposed to be able to overcome the more frequent dealer 21s.

exactly why I surrender on *alot* of 12-16's. (not all)

Quote: john2010

that's why I only do it when (lets say) I haven't seen a lot of 10's yet. (jack/queen/kings included)

This still is a very poor playing strategy (in my opinion and mathematically). Let me give you a reason why... 14 surrenders to a dealer 10 when the True Count is +3. This means for every deck remaining there are 3 extra "big cards." If you have 4 decks remaining from a 6 deck shoe, then there needs to be an extra 12 face cards in the shoe. Simply not seeing some here or is very inaccurate. Not to mention I can't even FIND the index for when to surrender a 12 or 13 (it would probably be something like TC +10, which is astronomically high). You would basically never see even a professional card counter surrender 12's or 13's. Think about that... They're a professional at the game you're playing and you would still never see them do it. That's like folding aces pre-flop in poker because you have a 'hunch' you're going to lose. A real professional at the game would simply never do it (99.9999% of the time pending some tournament strategy).

Quote: john2010

Most casinos only allow you to play 2 hands anyway.

Most I've played at let you play more than 2, but make you wager more as well.

Quote: john2010

yes, when done at the right time. But i'm also not endorsing it. (proceed at ur own risk)

The whole reason for pointing out spreading to multiple hands in a positive count was to show that this is the only time going to multiple hands where it won't hurt you by doubling your expected value loss.

Quote: john2010

exactly why I surrender on *alot* of 12-16's. (not all)

I'm not sure who's quote this was, but you should REALLY NOT be surrendering 12-16's. Again, if you've had any kind of success with this then I'm glad for you, I genuinely am, but you must understand this is nothing more than pure luck that will eventually run out and converge to the mathematics of the game. You will absolutely lose a lot more money surrendering when you're not supposed to (15 to a 10, and 16 to A-10-9).

^ decently said.

I can't argue.

According to page 25-26 of Blackjack Attack 3, the optimal number of hands when playing heads up against the dealer is one hand, based on 100 maximum opportunities.

With one and two other players, two hands.
With three and four other players, three hands.

Quote: Romes

As Babs said, you must play your hand before the dealer. Also, as she mentioned, what one hand does does not affect the other hands. They are all independent hands with the same house edge against them. Every bet you place in the separate circle is subject to the house edge of the game. Thus, playing more hands will do nothing more than make you lose more money per hour.

This is "almost true" and is good advice for the OP, who seems to be in the early stages of learning and understanding game theory. But it is not strictly true in my opinion. particularly this:

Quote: Romes

They are all independent hands with the same house edge against them.

1. First, when you play two hands the outcome of the hands are not independent -their outcomes are coupled to the dealer's hand, which both player hands have in common. A player will lose each hand about 48.3% of the time - but will lose both hands more than (48.3 x 48.3) = 23.3% of the time. Similarly, each hand might be expected to win about 43.3% of the time but the player will win both hands more than 18.7% of the time - because the outcome of both hands depends strongly on the dealer's hand. When the dealer makes a 21 or 20, the player will lose both hands quite frequently. The opposite is true -when the dealer busts or makes a 17 the player will win both hands quite frequently. (This does not alter the average win rate but does affect variance.) I haven't calculated this effect. Can someone point us to site that has this calculation? or otherwise tell us how often both hands win and both hands lose?

2. Secondly, the shoe is not an "infinite deck." In a six-deck shoe, each card in the first hand represents 1/312 cards and its absence from the deck has the possibility of shifting the odds by as much as 0.32%. If your first hand is K-Q or A-A, then your second hand has a reduced chance of winning (no matter what the dealers card is.) Some posters express this as a issue of the "true count" but that does not fully catch the intricacies (particularly when your first hand has one or more Aces.) Imagine that the dealer has a 10, and both of the player's hands are 10-2. When hitting the 12's the odds of winning each hand are both elevated by the presence of the 10 and 2 in the player's other hand -despite the fact that the high-low count is unfavorable. So, again the outcome of the two hands are coupled by the presence of cards in the other hand, and hence the absence of those cards from the shoe.

[ Edit: Deleted a statement, upon additional thought]

3. The ability to use the info from hand #1 to alter your Basic Strategy play of hand #2 can, in theory, create a different winning percentage. This does not necessarily entail "card counting," per se:
- In a single deck game, if the dealer has a 10, and your first hand is 5-5 and you hit it twice and get the other two 5s (making a 20) and your 2nd hand is 10-6, then you should NOT hit the 16 on your second hand. This is just "basic strategy for two hands" because you look at nothing other than the cards you are dealt in your own hands.
- Similarly, in a double deck game, if the dealer has a 10, and your first and second hands are both 7-7, then you should stand on the 2nd pair of 7s (rather than basic strategy which says to hit 7-7 in a 2-deck game) because you are aware that 2 sevens are present in your first hand. Again, card counting systems do not track 7s, so this is not card counting.

These points are technical nits - again, the advice being given to OP was sound and appropriate.

Quote: gordonm888

...These points are technical nits - again, the advice being given to OP was sound and appropriate.

I'm still glad you pointed them out to anyone else who was curious. I'm aware of the co-variance/etc that comes with playing more than one hand. I thought it would have been a bit past the OP though =p.