ricardo
ricardo
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February 1st, 2011 at 11:20:31 PM permalink
In blackjack game, the dealer is forced to draw to 17. The player is also strive to get to 17 and above than stop according to the basic strategy. Both the dealer and the play has same chance to be burst. The problem is, the player has inherited disadvantage that they have to draw first, burst first and out of the game first. To overcome this disadvantage, I think that in the game that offer late surrender, we should be more aggressive to use the surrendering. We should surrender all bad hand with hard 5,6,12,13,14,15,16, and the dealer has good card a ten showing, especially the true count is +1 and above; We should stay when the dealer has a bad(burst) card such as 2,3,4,5,6, showing, to let the dealer burst; after all it only cost 50% of the bet to surrender. When the dealer burst, we gain 100% of the bet. At the end, the player win. Am I right?
P90
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February 2nd, 2011 at 12:02:07 AM permalink
No. Basic Strategy prescribes when to surrender, exactly.
Here are the charts: https://wizardofodds.com/blackjack/strategy/4deck.html

There are also a few surrender indexes that prescribe additional surrenders, like 15vA at TC+1, 15v9 at TC+2, 14v10 at TC+3. But you shouldn't even bother thinking about counting until you have at least basic strategy down cold.
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weaselman
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February 2nd, 2011 at 5:46:23 AM permalink
Quote: ricardo

The player is also strive to get to 17 and above than stop according to the basic strategy.


I think, you have the answer to your own question :)
Basic strategy is called "basic" not because it is incorrect, but because it is the basis of the optimal play. It tells you exactly what is the best play in each situation (the only thing it doesn't take into account is the changes in deck composition due to previous cards out - that's what CDEs and indexes are for, but that's more like "fine tuning"). You surrender when the expected value of your hand against the dealer's is less than -0.5 (i.e., roughly speaking, you more than twice as likely to lose than to win) otherwise you play the hand, because in the long run you'll lose less this way.
"When two people always agree one of them is unnecessary"
ricardo
ricardo
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February 2nd, 2011 at 7:23:54 AM permalink
Both of you are correct, of course, that is what the book says. Unfortunately, it did not answer the question. My question is, if the probability of bursting are the same, that means that the probability of the player or the dealer to have the bad hand(hard 5,6,12,13,14,15,16) are the same, other things are equal, using the surrender strategy, player should win. The real trouble is, when dealer's showing card is 2,3,4,5,6, we don't know the hole card is a sure ten or not. When the true count is 1 or above, the probability of the hole care would more than likely a ten. I know I may be wrong, but how to prove I am wrong convincingly. By the way, how to calculate the expected value?
P90
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February 2nd, 2011 at 7:35:52 AM permalink
Quote: ricardo

My question is, if the probability of bursting are the same, that means that the probability of the player or the dealer to have the bad hand(hard 5,6,12,13,14,15,16) are the same, other things are equal, using the surrender strategy, player should win.


I'm afraid I don't understand what you mean here, at all.


Quote: ricardo

The real trouble is, when dealer's showing card is 2,3,4,5,6, we don't know the hole card is a sure ten or not. When the true count is 1 or above, the probability of the hole care would more than likely a ten.


If you meant "more likely than not", then no. There are 16 tens and 36 other cards in a fresh deck. At TC+1 in ace-neutral counts there are 16 tens and 35 other cards.

Seriously, don't throw words like "true count" around until you got all the basic math, if you are even using the term correctly, which you seem not to be.
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Wizard
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February 2nd, 2011 at 7:48:57 AM permalink
I think the question being asked is for a situation like 16 against a 10, where the player would otherwise (incorrectly) stand, what would be the dealer bust probability for surrender to be the best play? In other words, a situation where the player's only hope otherwise is for the dealer to bust. The answer to that question is the player should surrender if the dealer's bust probability is less than 25%.
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P90
P90
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February 2nd, 2011 at 8:05:11 AM permalink
It appears he's rather talking about surrendering all bad hands, even down to 5 and 6, at plus counts, against all 7+ dealer upcards.


Anyway, to OP, a good set of tables for quick analysis can be found here: http://www.beatblackjack.org/tables.html. The tables look fixed, but they are interactive and allow for changing the exact card distribution.
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mkl654321
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February 3rd, 2011 at 1:31:12 AM permalink
If the dealer burst, wouldn't he spatter all over everyone and maybe pieces of him would land in your drink? Yuck!
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
irbylamer
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February 3rd, 2011 at 5:06:23 AM permalink
The option to surrender you Blackjack hand is not always available when playing Blackjack games in a casino. This is mainly due to the fact that if the player utilizes the surrender strategy properly, then the casino's built-in mathematical advantage is reduced. The decision on when to surrender in Blackjack depends on the number of decks used, whether early or late surrender is offered and if you are counting cards or not. For more information you may check this link
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