Surrender when playing two hands

[ElementX]

When playing two hands, would it be advisable to sometimes go against basic strategy and surrender the poorer hand, depending upon the player's advantage of each of the two hands. For example, say the dealer is showing a 4 up, your first hand is a hard 16 (average player advantage of -19.4%) and your second hand totals 8 (average player advantage of +8.4%). This would give an average overall advantage of -5.5%. Would it be good to surrender the 16 and then play out the 8? What if you had to play the 8 first and the 16 second? Would you surrender the 16 depending on the outcome of the first hand? This is not considering the situation of the count. I am wondering if there is a more optimal set of indices for two-hand play. Okay.. my mind has definitely shut down for the night!

ElementX

 

[phantom007]

I would not surrender 16 vs. Dealer 4 in your example. As a general rule, only surrender hands that you will lose more than 75% of the time. This is usually 16 vs. 9,10,+/-A, and 15 vs. 10, and ONLY at neutral counts and above.

Certainly, there are other conditions where surrender might be advantageous, such as 14 vs. 10 at higher TC's.....however, against mid-to-low # cards, not so. High TC should cause one to Stand on the 16, DD on the 8, and most likely, the dealer will bust out.

 

[The Mayor]

Surrender those hands it is proper to surrender, and no others.

>For example, say the dealer is showing a 4 up, your first hand is a hard 16 (average player advantage of -19.4%) and your second hand totals 8 (average player advantage of +8.4%). This would give an average overall advantage of -5.5%. Would it be good to surrender the 16 and then play out the 8?

Neither of these hands is proper to surrender under ANY conditions I am aware of (with the single exception of hole card play, when you know the dealer has a 6 or 7 burried, it is proper to surrender the 16).

Surrender: 16 vs. T,A, 15 vs. T all the time.
In H17 games, also surrender 17 vs. A and 15 vs. A.

--Mayor

 

[zengrifter]

Re: Surrender DISAGREE

I have to disagree with Mayor, though I cannot state my position in math-terms, ElemX has a germ of a good idea that has infrequent application, and is supported by recent 'out-of-box' thinking related to insurance-bets. The strategy that ElemX has suggested will REDUCE VARIANCE but I would not do it in the instant example - I WOULD do it (perhaps) if I held a 16 and a 20 vs 8 for example. Yes/no? zg

see this post and JGrossjean's Beyond Counting for further (similar) treatment -
(Dead link: http://www.cardcounter.com/main.pl?read=126)

 

[ElementX]

Re: Surrender DISAGREE

ZG- you give a very good example. I would likely surrender a 16 vs. 8 if I also had a 20 and the TC was substantially positive. My example was not very illustrative and less clear-cut. Likewise I am unable to explain it mathematically- when I employ this tactic in the casino, it seems to save money on individual hands (when dealer gets a 19 or 20 after having an up card of 5 or 6), but I don't know what the long-term effect is. Certainly the reduced variance you mentioned causes the bankroll to fluctuate less.

ElementX

 

[The Mayor]

Re: Surrender DISAGREE

The index for surrender for 16. vs. 8 is about 5 (+/- depending on the game). In other words, take the hit!

 

[steve]

to ElementX: is this what you are really asking

Hello ElementX,
I don't think anyone answered your question. I think what you are really asking is this: "Is there a strategy that would treat the two hands as one outcome by summing the Ev of both and adopting a strategy that seeks to optimize the overall Ev"? Is that your real question?

For this question to have any merit, the two events would have to be interdependent, because if you look at each hand individually, assuming them to be independent, then the hit/stand decisions are obvious and don't vary due to the presence of the other hand.

In your example of 16 v 4, the player will lose 19% when standing, but will lose 50% when surrendering. Obviously, you want to stand.

But, what if the two events are interdependent in some way? Given that the two hands are dealt during the same round, does this make them interdependent in some way when they wouldn't be interdependent if they were dealt in two separate rounds?

I think that what you are getting at is this: if I play two hands instead of one and my goal is to have the highest Ev per round, instead of per hand, would I modify my play?

Hmmmmm, verrrrryyyyy interesting. I have to think about this.

Let me make sure I see where you are really going with this: "If I 'know' I am going to win one hand and I 'know' I am going to lose the other hand, should I surrender the loser and end up with a win of .5 my bet or take the push"? Obviously, you don't 'know' the outcome of the hands, but you expect it.

My initial feeling is that the two events are independent and there is no reason to vary your play. But, I need to do some reading on the interdependency (or lack thereof) of two hands played during the same round.

Regards,
Steve

 

[ElementX]

Yes

Steve,

You hit the nail right on the head with your interpretation. If you try to take the two hands and try to maximize the total OVERALL EV, would there be a divergence from normal play, especially when surrender is concerned? When playing two hands, the player should be concerned with the overall result, rather than the result of each hand individually. Here is another scenario: Dealer has a 2 up, you have two 16s- would you surrender one of them? I don't know how much literature is out there analyzing play of two or more hands. Any further thoughts on this would be greatly appreciated. Thanks for your reply Steve.

ElementX