Surrender after split
I can't imagine it's actually +EV for the player but the rest of the rules are:
Dealer hits S17
Double only on 10 and 11
Double after Splits allowed
Resplitting to 3 times allowed
Aces split only once
Insurance pays 11/5
7 Card Charlie
Insurance pays 11/5
What's the best way to attack this version? Which hands benefit most from late late surrender rule?
Quote: MarcoPoloI was surprised to find this rule at one of the online casinos I play at. This seems like a major advantage for the player?(snip)
Not as big of an advantage as you would think, as it shaves off about 1/650,000 from the house edge for a six deck*** game (according to MGP...###).
***: Just a guess, but I don't think more or less decks would matter "too much".
###: Search and download MGP's Blackjack Combinatorial Analyzer (as it has the Surrender after Split option that you need)
Note: There are other good ones out there (eg >> http://www.bjstrat.net/cgi-bin/cdca.cgi ) but this one has more options.
Quote:(snip)What's the best way to attack this version?
Surrender these HARD hands:
15 vs 10 or A
16 vs 9, 10 or A
17 vs A (since the dealer hits on soft 17).
Surrender these hands after Splitting Aces, if you can't hit split aces (standard rule for most games):
A-A to A-5 vs 8, 9, 10 or A
A-6 vs A
Quote:(snip)Which hands benefit most from late late surrender rule?
16 vs 10 or A. benefit the most
Note: After splitting A's, surrendering A-A to A-5 vs 10 or A, benefit the most.
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Update (about 410pm)
Even though you didn't ask, you would take insurance if you were playing 2 or 3 hands per game, and none of your hands had a 10-value card in it (only checked for 6-deck, when insurance pays 11/5)
Also, even though I don't think it is a major advantage, I would have thought that the reduction in house edge would have been better than 1/650,000, but I am too lazy to work it out my self, sorry (it would probably take me over two hours to work out, since I am slow at doing this sort of stuff manually).
Quote: ksdjdjNot as big of an advantage as you would think, as it shaves off about 1/650,000 from the house edge for a six deck*** game (according to MGP...###).
[snip]
Also, even though I don't think it is a major advantage, I would have thought that the reduction in house edge would have been better than 1/650,000, but I am too lazy to work it out my self, sorry (it would probably take me over two hours to work out, since I am slow at doing this sort of stuff manually).
I don't think that surrendering after you have split aces is the intent of the rule. If you cannot hit a split ace, then surrendering an 'A-(2-5) vs Anything' is a really weird rule.
Other than split aces, the only time you are going to takes advantage of "surrender after splitting" is vs 10,A or occasionally vs 9. The only pair you will split against these dealer cards is 88. And its hard to make 16 vs A or 16 vs 10 after splitting 8's because
- if you draw to a split 8 and get another 8 to make 16, you re-split rather than surrender.
- if you draw to a split 8, you Stand if you draw Ace or 10 or 9, and you double if you draw 2 or 3. The only way to make a multicard 16 is to draw a 4,5,6 or 7, and then get another card to exactly make 16.
-and, oh yeah, when dealer peeks with a 10 or A showing, he sometimes has BJ
Quote: MarcoPoloI was surprised to find this rule at one of the online casinos I play at. This seems like a major advantage for the player?
I can't imagine it's actually +EV for the player but the rest of the rules are:
Dealer hits S17
Double only on 10 and 11
Double after Splits allowed
Resplitting to 3 times allowed
Aces split only once
Insurance pays 11/5
7 Card Charlie
Insurance pays 11/5
What's the best way to attack this version? Which hands benefit most from late late surrender rule?
After splitting aces, the player would surrender 12-17 vs A, and 12-16 vs 8-10, as well as the usual surrenders.
However, the rule allowing surrender on two cards after a split only improves the base game by about 0.01% (my estimate.)
The seven-card Charlie increases the player's EV by about 0.01%.
The effects of these two favorable rules are small compared to the decrease in player EV with H17 (-0.22%) and double on only 10-11 (-0.18%.)
See https://wizardofodds.com/games/blackjack/rule-variations/ for the Wizard's effects of rule variations.
I thought maybe splittable hands could be played more aggressively
For what it's worth though, I forgot to mention it's single deck and it appears the rules are outdated on their site as I can double on any two cards as well
I guess I have to wonder if there are other incorrect rules listed but playing through some hands, I didn't see any duplicated cards so it would appear it is single deck though I wasn't able to test the resplitting rule
Quote: MarcoPolo(snip)
For what it's worth though, I forgot to mention it's single deck(snip)
Please can someone correct me if I am wrong, but If you are paid 11/5 for insurance, then I think the overall reduction in house edge is between 0.08% and 0.09%, for when you play this strategy below:
. Play one hand per game
. Only take insurance, when you have "no 10-value cards in your hand"
Note: The reduction in house edge would be higher for 2 or 3 handed games (someone good at this sort of thing may be able to work out what that amount is for you)
Note 2: Insurance options are assumed as follows, (a) you can either insure for 1/2 your initial bet, or (b) you can choose NOT to take any insurance.
Using the Wizard's house edge calculator, your single-deck game would have a house edge of 0.31067% -0.01% (for surrender after split) -0.01% (for 7-card charlies) - 0.0806% (for insurance) = 0.21%.
However, if your game is double on any two cards instead of double on 10-11, the house edge would be -0.09%, for a player advantage.
Can you see other players' cards before your insurance decision? As ksdjdj pointed out, your insurance EV would improve. (For one other player, the house edge would decrease by 0.10% instead of by 0.08%. And for two other players, the house edge would decrease by 0.12%.)
I get a decrease of 0.145%.Quote: ChesterDogksdjdj's point about insurance is correct. In the first hand of single-deck, if you are dealt two non-tens against the dealer's ace, your insurance bet has an EV of 4.49%. This decreases the house edge by 0.0806%.
4 * 16 * 15 / (52 * 51 * 50) * 1/5 = 0.00145
Maybe wrong
Quote: Ace2I get a decrease of 0.145%.
4 * 16 * 15 / (52 * 51 * 50) * 1/5 = 0.00145
Maybe wrong
I don't know, but I did it this way:
Start with 16 "10-valued" cards, and 36 cards that are "not 10-valued"
The chance for both of the two cards in your hand to not be a "10-valued card and the dealer showing an Ace," i get this:
36 x 35 x 4 / (52 x 51 x 50) = 3.80... %
EV of insurance bet (when the above occurs):
(16/49) x 3.2 = 104.48...% RTP = + 4.48...% EV
Multiply the above chance and EV figures together
3.80...% x 4.48...% = 0.170...%
Then divide that figure above by 2 (because the insurance bet is half the initial bet).
= 0.085...%
Therefore the reduction in the "initial bet house edge " is 0.085%
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Again, I could be wrong.
Also, I get a slightly different "house edge reduction" figure to ChesterDog's figure of 0.0806% (but we do seem to get the same EV of about 4.49%, so I don't know why we have a different figure overall).
4 * 36 * 35 / (52 * 51 * 50) * 1/5 = 0.0076
Quote: ksdjdjI don't know, but I did it this way:
Start with 16 "10-valued" cards, and 36 cards that are "not 10-valued"
The chance for both of the two cards in your hand to not be a "10-valued card and the dealer showing an Ace," i get this:
36 x 35 x 4 / (52 x 51 x 50) = 3.80... %
EV of insurance bet (when the above occurs):
(16/49) x 3.2 = 104.48...% RTP = + 4.48...% EV
Multiply the above chance and EV figures together
3.80...% x 4.48...% = 0.170...%
Then divide that figure above by 2 (because the insurance bet is half the initial bet).
= 0.085...%
Therefore the reduction in the "initial bet house edge " is 0.085%
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Again, I could be wrong.
Also, I get a slightly different "house edge reduction" figure to ChesterDog's figure of 0.0806% (but we do seem to get the same EV of about 4.49%, so I don't know why we have a different figure overall).
I used your method with one small difference. Instead of "36 x 35 x 4 / (52 x 51 x 50)," I did "35 x 34 x 4 / (52 x 51 x 50)" because the dealer's ace is one of the 36 non-tens.
That would account for the small difference in our results.
Quote: gordonm888(snip)
I don't think that surrendering after you have split aces is the intent of the rule. If you cannot hit a split ace, then surrendering an 'A-(2-5) vs Anything' is a really weird rule. (snip)
After doing some practice play, I can confirm you are correct about "not being able to surrender after splitting aces".
Quote: ChesterDogI used your method with one small difference. Instead of "36 x 35 x 4 / (52 x 51 x 50)," I did "35 x 34 x 4 / (52 x 51 x 50)" because the dealer's ace is one of the 36 non-tens.
That would account for the small difference in our results.
Edit (about 1240 am): Whether your answer is exact or just an approximation, that make sense to me ***
***: I think I get the reasoning behind your proof, but I don't know the reasoning behind Ace2's proof (even though his answer is fairly close to your answer).
Quote: Ace2It’s been a rough night. My calculation was for 2 face cards to player. Now I get
4 * 36 * 35 / (52 * 51 * 50) * 1/5 = 0.0076
If I was forced to guess about the two answers, I would either say :
(a) "that one method is a quick^^^ way to get an approximation and the other way gets you the exact answer" or
(b) "they are both equally valid ways to get an approximation easily ^^^" (?)
^^^: quick or easy, when compared to working it out the "exact answer way"?
Again, just a guess.
Quote: ChesterDogksdjdj's point about insurance is correct. In the first hand of single-deck, if you are dealt two non-tens against the dealer's ace, your insurance bet has an EV of 4.49%. This decreases the house edge by 0.0806%.
Using the Wizard's house edge calculator, your single-deck game would have a house edge of 0.31067% -0.01% (for surrender after split) -0.01% (for 7-card charlies) - 0.0806% (for insurance) = 0.21%.
However, if your game is double on any two cards instead of double on 10-11, the house edge would be -0.09%, for a player advantage.
Can you see other players' cards before your insurance decision? As ksdjdj pointed out, your insurance EV would improve. (For one other player, the house edge would decrease by 0.10% instead of by 0.08%. And for two other players, the house edge would decrease by 0.12%.)
This isn't a live game so no other cards and can only play one hand at a time. That is very nice to hear though that the game has a player advantage
On the first hand of a single deck game: If you have two "non-Ten" cards vs Ace you should NOT take insurance!
Given: 52 cards to start, and you see 3 cards that are not tens (two in your hand and the dealer's Ace):
Thus, 49 cards are left and all 16 "tens" are still left. Insurance pays 2:1, so to have a positive EV the number of remaining Tens must be greater than 1/3 of the remaining cards. And 16 x 3 =48, so there must be less than 48 remaining cards for the Insurance wager to be +EV.
Quote: gordonm888I am having a hard time following the insurance decision discussion.
On the first hand of a single deck game: If you have two "non-Ten" cards vs Ace you should NOT take insurance!
Given: 52 cards to start, and you see 3 cards that are not tens (two in your hand and the dealer's Ace):
Thus, 49 cards are left and all 16 "tens" are still left. Insurance pays 2:1, so to have a positive EV the number of remaining Tens must be greater than 1/3 of the remaining cards. And 16 x 3 =48, so there must be less than 48 remaining cards for the Insurance wager to be +EV.
You are correct for when insurance pays 2:1. However, in MarcoPolo's special 1-deck blackjack game, insurance pays 11:5.
