playing two hands

Hi Wizard,
I have seen some of the posts about playing two hands but i have a different variation of this question. im trying to take a new angle on the game but am not sure if it is feasible. lets say that you play two hands, but choose the better hand to play each time and surrendered the other. for purposes of this question im trying to calculate including extreme situations, such as if you had a 20 and a 19 against dealer 6, you would surrender the 19. what are the odds that this "better" hand would win. this also brings up another dilema because under these circumstances, do you think it would be better to play a hard 20 or double an 11 against a dealer 6? would it be the same as calculating a player winning one of two hands? which i think would be approx %66. or would it be higher since the better of the two hands would be played each time?
thanks

Quote: sdgambler . . .

'm trying to take a new angle on the game but am not sure if it is feasible. lets say that you play two hands, but choose the better hand to play each time and surrendered the other.

Uh? Blackjack? You will surely be throwing money away on the surrendered hand?

I won't put out hard statistics, but your strategy doesn't seem like a good idea at all. What about cases where you have two hands that should be doubled according to basic strategy? It wouldn't make sense to surrender one hand that has a better-than-even chance of making money. Likewise, in the case where both hands should be surrendered it wouldn't make sense to choose the best-of-the-worst to play. Playing two hands will lower the variance you are exposed to but the cost is more money wagered. To me, it only makes sense to play more than one hand when the count is in your favor, not as an all-the-time tactic.

Quote: sdgambler

Hi Wizard,
I have seen some of the posts about playing two hands but i have a different variation of this question. im trying to take a new angle on the game but am not sure if it is feasible. lets say that you play two hands, but choose the better hand to play each time and surrendered the other. for purposes of this question im trying to calculate including extreme situations, such as if you had a 20 and a 19 against dealer 6, you would surrender the 19. what are the odds that this "better" hand would win. this also brings up another dilema because under these circumstances, do you think it would be better to play a hard 20 or double an 11 against a dealer 6? would it be the same as calculating a player winning one of two hands? which i think would be approx %66. or would it be higher since the better of the two hands would be played each time?
thanks

Each starting triple of two player cards and dealer up-card has an EV. You can look that up, for example, here:

https://wizardofodds.com/games/blackjack/appendix/9/2dh17r4/

Just choose the hand that has the higher EV and play that one and surrender the other one.

Hard T,T vs. 6, stand EV = 0.673275

6,5 vs. 6, double EV = 0.710663

So, in this situation, the best play is to double 6,5 vs. 6 and surrender the 20.

Stanford Wong has similar EV numbers in his appendices to "Professional Blackjack" (1994).

The only catch here (which I don't imagine has any meaningful effect in a multi-deck game) is that you can use the known cards of the other hand to adjust the EVs based on recomputing with those cards removed from the deck. This and $4.50 will buy you a cup of coffee.

i know there are situations where you would not want surrender one of the hands, for example if you had 20 on both. but this is a theoretical question which i will use to adjust strategy accordingly.

If you have two hands of 20, and the dealers has an ACE up, would you surrender 1 hand or just insure 1 hand ?

Quote: sdgambler

i know there are situations where you would not want surrender one of the hands, for example if you had 20 on both. but this is a theoretical question which i will use to adjust strategy accordingly.

You're giving correlation to your two hands when their independent EV's vary against their independent plays...

Essentially, surrendering is worth a 50% loss. However if the option to Stand on a hand brings a return greater than a 50% loss (Say standing 18 v dealer 7) then you're essentially THROWING MONEY AWAY. Each hand should be played for it's own maximum EV/potential. If you do this idea where you surrender 1 hand and keep 1 hand, you're almost always throwing away EV.

Another example:

Hand 1: 10-10
Hand 2: 10-2

Dealer: 7

In your system you would keep 10-10 and surrender 10-2, losing 50% of that hands value and 25% of your overall bet. However, the best play here is to hit the 12 to the dealer 7. As cited by other players, you can go look at the Wizards indexes to see the expected value of each decision (hit/stand/double/surrender/split). You'll make a lot more money (or lose less) in the long run by playing each individual hand according to it's best possible outcome.

Correlating the 2 hands together to make a decision is just an awful idea. Pretend you did this with the person next to you... not sharing the profits. If they get 10-10, you have to surrender your 18 to a dealer 7... Why on EARTH would you do that? Just because they have 10-10? That has very little to do with your expected return of standing on your 18 vs a dealer 7.

Do not do this system, it will lose you a lot of money.

im not trying to figure out the EV here. I understand why ALWAYS surrendering one hand would be a bad idea, but that not what im trying to find out. let me put it this way, lets say there is no money involved, you are dealt two hands and the dealer gets one, but you can only play one of your hands. does this increase your odds of winning the chosen hand assuming you play the hand with higher expected return each time? or would your odds be the same since you could just as easily get two bad hands as two good ones.

You gotta love Romes.
I nominated him for POTUS, there was no second..

Can you play two hands if you only have one hand, just a hypothetical question.
Just for fun, the devil made me do it......

Quote: sdgambler

im not trying to figure out the EV here. I understand why ALWAYS surrendering one hand would be a bad idea, but that not what im trying to find out. let me put it this way, lets say there is no money involved, you are dealt two hands and the dealer gets one, but you can only play one of your hands. does this increase your odds of winning the chosen hand assuming you play the hand with higher expected return each time? or would your odds be the same since you could just as easily get two bad hands as two good ones.

Under the stated conditions that you can only play one of the hands, then of course choosing the better hand will win more of the time. Take the money out of the game and who cares?

I think I’ve got it. You want to play two hands and surrender one. Unless both are good hands, then you keep both. And unless they are both lousy hands, then you surrender both. If you make your surrender decision based on basic strategy then you will lose about twice as much as if you were playing one hand. BUT, you would be “wining” more total hands! If you surrender more hands than basic strategy dictates then you will still win more hands, but lose A LOT more money.

yes i know, but i am trying to figure out what the new odds of winning would be

Basic strategy alone is a losing game. Deviating for the worse rather than for the better will cause that player to lose more while he plays at a bigger disadvantage against the house.