I knew a 10 was coming so I surrendered

A

assume_R

Well-Known Member
This was the last hand of the shoe (in the game from spain with no ten's, only faces), and the TC was +7. I had 2 hands of my max, and I got a 14 and a 13 vs a 10 upcard. I signaled to surrender my 14, but the dealer thought I had wanted to hit, so accidentally pulled a jack out of the shoe. I said that I had wanted to surrender, and she said sorry and let me surrender. So the jack would have gone to either my 13 or as her hit card.

This is what I thought, and what I did, and I don't actually know if it was correct. So obviously I couldn't hit, or I would have busted. Therefore, my options were to stand or surrender. If her hole card was a 2-6, I would have won (since she would take a hit and bust). If her hole card was 7, 8, 9, J, Q, K, I would have lost. Since the TC was +7, I figured there was a lower chance of her having a 2-6. In addition, the 0 variance of surrender appealed to me with a max bet out. So I surrendered. Now I don't know if it was correct.

****************************

Time to do some math.

Maximizing EV:

Surrendering
EV(surrender) = -0.50

Standing
X = P(win) = P(2) + P(3) + P(4) + P(5) + P(6)
P(lose) = P(7) + P(8) + P(9) + P(J) + P(Q) + P(K)
P(win) + P(lose) = 1
EV(stand) = P(win) - P(lose) = X - (1 - X) = 2X - 1

EV(stand) >? EV(surrender)
2X - 1 > -0.5
2X > 0.5
X > .25
So if P(win) > 0.25, I should have stood.

With a TC of +7 was P(2) + P(3) + P(4) + P(5) + P(6) greater than 0.25? Probably... I suppose I should have stood.


Using Variance:

Var(surrender) = 0.0

Var(stand) = P(win) * (1 - EV(stand))^2 + P(lose) * (-1 - EV(stand))^2
Var(stand) = P(win) * (1 - P(win) + P(lose))^2 + P(lose) * (-1 - P(win) + P(lose))^2
Var(stand) = X * (1 - X + (1-X) )^2 + (1-X) * (-1 - X + (1-X) )^2
Var(stand) = X * (1 - X + (1-X) )^2 + (1-X) * (-1 - X + (1-X) )^2
Var(stand) = 4X * (1 - X)

Var(stand) = 4 * P(win) * P(lose)

So we have:
Standing: EV = P(win) - P(lose), Var = 4 * P(win) * P(lose)
Surrendering: EV = -0.50, Var = 0.0

With a TC of +7, I suppose I need to calculate P(win) for a bunch of different TC +7 shoes, and see what the median P(win) is or something, right?
 
B

BJgenius007

Well-Known Member
assume_R said:
This was the last hand of the shoe (in the game from spain with no ten's, only faces), and the TC was +7. I had 2 hands of my max, and I got a 14 and a 13 vs a 10 upcard. I signaled to surrender my 14, but the dealer thought I had wanted to hit, so accidentally pulled a jack out of the shoe. I said that I had wanted to surrender, and she said sorry and let me surrender. So the jack would have gone to either my 13 or as her hit card.

This is what I thought, and what I did, and I don't actually know if it was correct. So obviously I couldn't hit, or I would have busted. Therefore, my options were to stand or surrender. If her hole card was a 2-6, I would have won (since she would take a hit and bust). If her hole card was 7, 8, 9, J, Q, K, I would have lost. Since the TC was +7, I figured there was a lower chance of her having a 2-6. In addition, the 0 variance of surrender appealed to me with a max bet out. So I surrendered. Now I don't know if it was correct.

****************************

Time to do some math.

Maximizing EV:

Surrendering
EV(surrender) = -0.50

Standing
X = P(win) = P(2) + P(3) + P(4) + P(5) + P(6)
P(lose) = P(7) + P(8) + P(9) + P(J) + P(Q) + P(K)
P(win) + P(lose) = 1
EV(stand) = P(win) - P(lose) = X - (1 - X) = 2X - 1

EV(stand) >? EV(surrender)
2X - 1 > -0.5
2X > 0.5
X > .25
So if P(win) > 0.25, I should have stood.

With a TC of +7 was P(2) + P(3) + P(4) + P(5) + P(6) greater than 0.25? Probably... I suppose I should have stood.


Using Variance:

Var(surrender) = 0.0

Var(stand) = P(win) * (1 - EV(stand))^2 + P(lose) * (-1 - EV(stand))^2
Var(stand) = P(win) * (1 - P(win) + P(lose))^2 + P(lose) * (-1 - P(win) + P(lose))^2
Var(stand) = X * (1 - X + (1-X) )^2 + (1-X) * (-1 - X + (1-X) )^2
Var(stand) = X * (1 - X + (1-X) )^2 + (1-X) * (-1 - X + (1-X) )^2
Var(stand) = 4X * (1 - X)

Var(stand) = 4 * P(win) * P(lose)

So we have:
Standing: EV = P(win) - P(lose), Var = 4 * P(win) * P(lose)
Surrendering: EV = -0.50, Var = 0.0

With a TC of +7, I suppose I need to calculate P(win) for a bunch of different TC +7 shoes, and see what the median P(win) is or something, right?
Click to expand...
If you have to calculate every scenario, you won't have time to play. That is why you buy the books. Somebody else has calculated the indexes for you.

I am using Omega II, the indexes for surrendering

13 vs. 10 is +6.5
14 vs. 10 is +3

Since +7 is higher than either number, I will surrender both hands, either. (However, considering margin of error, if you are not so confident on your counting skill or you often miss one or two, you should use Basic Strategy to play the 13 vs. 10 hand, that is hit.)
 
A

assume_R

Well-Known Member
Firstly, the dealer already checked for an ace, so therefore you don't include the probability of an ace since you assume that the hole card is not an ace by the time this decision has to be made.

Secondly, yes, I know that the books have already written about most games, but not this game for how I play it. I wasn't planning on being able to calculate this on the fly in the future, but I want to know the process so I can calculate it on my own after the fact (or hopefully before) for games in which the books haven't been written yet. The indices for these types of counts and games are very much unpublished. I want to know the process more than the actual result.
 
M

MangoJ

Well-Known Member
Yeah you are right, I forgot that the hole card was checked before.

Then P(holecard, peeked) = P(holecard, non-peeked) / (1 - P(ace, non-peeked) )

The non-peeked probabilities should be derived from the TC-implied card distribution. Frankly I'm not sure how to do that.


@Books: The indices given are for information of TC only. Obviously when you know the next dealt card is a ten, things may change.
 
K

k_c

Well-Known Member
assume_R said:
This was the last hand of the shoe (in the game from spain with no ten's, only faces), and the TC was +7. I had 2 hands of my max, and I got a 14 and a 13 vs a 10 upcard. I signaled to surrender my 14, but the dealer thought I had wanted to hit, so accidentally pulled a jack out of the shoe. I said that I had wanted to surrender, and she said sorry and let me surrender. So the jack would have gone to either my 13 or as her hit card.

This is what I thought, and what I did, and I don't actually know if it was correct. So obviously I couldn't hit, or I would have busted. Therefore, my options were to stand or surrender. If her hole card was a 2-6, I would have won (since she would take a hit and bust). If her hole card was 7, 8, 9, J, Q, K, I would have lost. Since the TC was +7, I figured there was a lower chance of her having a 2-6. In addition, the 0 variance of surrender appealed to me with a max bet out. So I surrendered. Now I don't know if it was correct.

****************************

Time to do some math.

Maximizing EV:

Surrendering
EV(surrender) = -0.50

Standing
X = P(win) = P(2) + P(3) + P(4) + P(5) + P(6)
P(lose) = P(7) + P(8) + P(9) + P(J) + P(Q) + P(K)
P(win) + P(lose) = 1
EV(stand) = P(win) - P(lose) = X - (1 - X) = 2X - 1

EV(stand) >? EV(surrender)
2X - 1 > -0.5
2X > 0.5
X > .25
So if P(win) > 0.25, I should have stood.

With a TC of +7 was P(2) + P(3) + P(4) + P(5) + P(6) greater than 0.25? Probably... I suppose I should have stood.


Using Variance:

Var(surrender) = 0.0

Var(stand) = P(win) * (1 - EV(stand))^2 + P(lose) * (-1 - EV(stand))^2
Var(stand) = P(win) * (1 - P(win) + P(lose))^2 + P(lose) * (-1 - P(win) + P(lose))^2
Var(stand) = X * (1 - X + (1-X) )^2 + (1-X) * (-1 - X + (1-X) )^2
Var(stand) = X * (1 - X + (1-X) )^2 + (1-X) * (-1 - X + (1-X) )^2
Var(stand) = 4X * (1 - X)

Var(stand) = 4 * P(win) * P(lose)

So we have:
Standing: EV = P(win) - P(lose), Var = 4 * P(win) * P(lose)
Surrendering: EV = -0.50, Var = 0.0

With a TC of +7, I suppose I need to calculate P(win) for a bunch of different TC +7 shoes, and see what the median P(win) is or something, right?
Click to expand...
I can give a ballpark answer but it's based on several assumptions. Chances are one or more of these assumptions may not apply but here goes.
1) 6 decks
2) You are using HiLo adapted to Spanish 21
3) Since Sp 21 is absent 24 tens, initial running count = -24
4) 2 decks remain
5) Approx average Sp 21 comp of 2 decks {8,8,8,8,8,8,8,8,24,8} (2-ace)
6) Start with average comp, remove 2 ea 2-6 and add 8 tens and 2 aces
7) All of this is to create a somewhat reasonable composition that could show if there is any decision that is definitely better. If the answer is too close to call it won't be of much use.

The stand EV in the image wouldn't apply because next card = T.
Since dealer's next card is a ten and hole card can't be an ace, standing has following probs:
29/83 sure win (dealer hole card is 2,3,4,5, or 6
54/83 sure loss (dealer hole card is 7,8,9, or T)
Stand EV = -25/83 = -30.12%

Late surrender EV = -50%

Note: Initial running count metric is wrong in the image because it is geared to a standard deck rather then a Spanish 21 deck. True count in image needs to be divided by 52/48 since there are 52 cards in a standard deck and 48 cards in a Spanish 21 deck. (Also hit and double EVs wouldn't apply.)

I appears to me that standing is clearly the best play (even if some assumptions may not apply.) In this case since you know dealer's hit card is a ten, the lower the count the greater stand EV will be and the higher the count the better the surrender option becomes until at some plus count it will be best.
 

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FLASH1296

FLASH1296

Well-Known Member
BJgenius007 ,

Your comments re: Advanced Omega Late Surrender Indices were
well-meant buy they do not apply in this case; as the mathematics
of Spanish21 are so very far removed from those of blackjack.

Thinking empirically requires one to avoid mixing apples and oranges.

:eyepatch:

p.s. I like A.O. II, although it is often referenced as an out-moded count.
 
A

assume_R

Well-Known Member
K_c it seems you used the same calculation as me but used a general "tc 7" distribution of remaining 2-6's to get P(win). How did you pick that distribution?
 
zengrifter

zengrifter

Banned
FLASH1296 said:
BJgenius007 ,

Your comments re: Advanced Omega Late Surrender Indices were well-meant but they do not apply in this case; as the mathematics
of Spanish21 are so very far removed from those of blackjack.

Thinking empirically requires one to avoid mixing apples and oranges.

:eyepatch:

p.s. I like A.O. II, although it is often referenced as an out-moded count.
Click to expand...
That is what we call a BJgenius-ism.
And yes the AO2 is definitely out-moded. zg
 
B

BJgenius007

Well-Known Member
FLASH1296 said:
BJgenius007 ,

Your comments re: Advanced Omega Late Surrender Indices were
well-meant buy they do not apply in this case; as the mathematics
of Spanish21 are so very far removed from those of blackjack.

Thinking empirically requires one to avoid mixing apples and oranges.

:eyepatch:

p.s. I like A.O. II, although it is often referenced as an out-moded count.
Click to expand...
First, my bad. Don't know it is Spanish 21.

Regarding A.O. II, I don't mind doing extra calculation because I am a genius.

I can see newer counting system have all kinds of short cuts to minimize the numbers people need to memorize. But this is what I have done in one second or a blinking of the eyes:

So in this example, I have running count +61, remaining deck 2.5, and I have 14 against dealer up card 8.

+61 / 5 = +12 (my TC)

Memory access the hit/stay 14 row: 13 13 11 7.5 11
Retrieve tag: 13
Decision: hit (if surrender is not available)

Memory access surrender 14 row: 11 6 3 5
Retrieve tag: 11
Decision: surrender

Signal to dealer: surrender

I memorize numbers by row. And it may sound strange. But I found the newer systems are more difficult for me. I prefer to memorize the 6 rows for stick hands (row 14, 15, 16 for both surrender and hit/stay). When I see my hands and the up card, I immediately "see" the two rows I am about to use to make my play decision.
 
FLASH1296

FLASH1296

Well-Known Member
You surrendered the 14 and contemplated the play of the 13.

I use +9 for the LS index for 14 vs Face

I use +13 for the LS index for 13 vs Face

The indices above assume abnormal distribution of card ranks, adjusted for the accelerated TEN-density.

In this case you know that ALL dealer stiffs will 'bust'.

As such I think that you do well to ignore the index as such.

There is no possibility of the dealer's final hand being a 3 card winner or a push.
Ergo you win whenever the dealer flips up a 2, 3, 4, 5, or 6 and lose otherwise.

That is 7 to 5 losing odds — 42% chances.

We know that you need 25% winning chances in order to play on without surrendering.

So … does the increased ten-density indicate that our winning chances drop below 25% ?

Using a single deck as our model, (in accordance with the True Count Theorem),
we can create a more accurate picture for the outcome of this hand,
only after we know the tags being used to generate the +7 True Count.

I think that the correct (expectation-maximizing) decision may have been
to surrender the 13 and to hit the 14; although "over the table" I would
have factored in my sense of Risk Aversion (as these were max' bets)
and surrendered both hands.
 
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A

assume_R

Well-Known Member
Flash, the only problem with your analysis is I only knew all dealer stiffs would bust on the 13 vs 10 not the 14 vs 10 in your analysis.
 
K

k_c

Well-Known Member
assume_R said:
K_c it seems you used the same calculation as me but used a general "tc 7" distribution of remaining 2-6's to get P(win). How did you pick that distribution?
Click to expand...
It's just a way of approximating a reasonably representative composition. I first remove cards in their starting ratio. So for a standard deck remove 1 each non-ten per 4 tens. This maintains the initial shoe rank ratios. After that I leave the neutral ranks (HiLo 7,8,9) as is and then remove low and add high cards in equal numbers to increase count and the opposite to decrease count. This works best at exactly 1/2 shoe. From the programs I've written enumerating count subsets I've found that at midshoe the probability of drawing a neutral card is exactly equal to 1/13 for any running count given that nothing else has been specifically removed. Midshoe (and full shoe) are the only penetrations this is exactly true but the variation from 1/13 is small for most penetrations.

To get the spanish 21 comp I started with a 6 deck spanish 21 shoe {24,24,24,24,24,24,24,24,72,24} and removed cards to {8,8,8,8,8,8,8,8,24,8}. Then I removed 2 each (2-6) and added 8 tens and 2 aces. Since in spanish 21 there are more low cards than high I possibly should have only added 6 tens but I was going to deal a dealer up card of ten and also a player hand of T-3 so 2 more tens would be removed in the calculation. It was just an approximate composition to try and get a general idea of what to possibly expect.
 
Automatic Monkey

Automatic Monkey

Banned
k_c said:
It's just a way of approximating a reasonably representative composition. I first remove cards in their starting ratio. So for a standard deck remove 1 each non-ten per 4 tens. This maintains the initial shoe rank ratios. After that I leave the neutral ranks (HiLo 7,8,9) as is and then remove low and add high cards in equal numbers to increase count and the opposite to decrease count. This works best at exactly 1/2 shoe. From the programs I've written enumerating count subsets I've found that at midshoe the probability of drawing a neutral card is exactly equal to 1/13 for any running count given that nothing else has been specifically removed. Midshoe (and full shoe) are the only penetrations this is exactly true but the variation from 1/13 is small for most penetrations.

To get the spanish 21 comp I started with a 6 deck spanish 21 shoe {24,24,24,24,24,24,24,24,72,24} and removed cards to {8,8,8,8,8,8,8,8,24,8}. Then I removed 2 each (2-6) and added 8 tens and 2 aces. Since in spanish 21 there are more low cards than high I possibly should have only added 6 tens but I was going to deal a dealer up card of ten and also a player hand of T-3 so 2 more tens would be removed in the calculation. It was just an approximate composition to try and get a general idea of what to possibly expect.
Click to expand...
I've found that synthesizing a count by removing an equal number of high/low cards works well enough to get a usable answer. For more accuracy I use the MRI function on CVData because that provides the counts with the distributions you'd see in the live game.
 
T

The Chaperone

Well-Known Member
If the 10 is going to be burned, then you hit obviously. If it's not burned, it's better to stand, because the dealer is very likely to bust (a little less than 4/11 of the time since the count is high). This assumes no one else is acting behind you.
 
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