Advantage of the Insurance bet?

k_c

Well-Known Member
#2
tensplitter said:
What's the house or player edge of the insurance bet by itself at the true counts of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 with a 6 deck shoe?
I can tell you how to get a reasonable idea of value of insurance by using the program on my website. I am too lazy to do it for you but I can show what to do if you're interested. The trouble with math is that most just don't care to deal with it. Also I wouldn't be surprised if QFIT had some sort of data listed somewhere.

The program on my website computes composition dependent EV for any shoe composition. It also computes insurance EV at the same time. For insurance it computes 2 values - insure all and insure only hands where EV is positve.

Relative to a count the only problem is to input a reasonably representative composition. Since you're interested in 6 decks, the first step is to input 6 for
number of decks. Mid-shoe is the best place to compute values, so the next step is the set shoe composition to 12 for all non-tens and 48 for tens. Mid-shoe is good because I have independently shown at that point probability of drawing a neutral Hi-Lo count card (7,8,9) at that point equals exactly 1/13 for any running count.

Next step is to vary composition relative to Hi-Lo.

For a 0 count nothing needs to be done. Just click compute. Insure all EV displays -.2844%, Since there are no +EV opportunites for a composition of
12,12,12,12,12,12,12,12,48,12 (2-A) insure +EV displays .0000%.

To create non-zero counts change low card compositions (2-6) and high card compositions by equal amounts. i.e. to create a running count of +2 remove 1 low card and add 1 high card (T-A). Total number of cards will always be kept equal to 156. It doesn't matter which low cards are removed because relative to the insurance calculation they are all equivalent. However the insurance calculation is different depending upon whether the high card added is a ten or an ace. So the calculation needs to be done twice - once with 49 tens and 12 aces and once with 48 tens and 13 aces to account for both possibilites. The probability that the high card added was a ten = 48/60 and the probability it was an ace = 12/60 since the starting high card composition is 48 tens and 12 aces, so multiply insurance EV with a ten added by 48/60 and add to insurance EV with an ace added multiplied by 12/60. Result is insurance EV for running count of +2 with 156 cards (3 decks remaining,) evaluating to a H-Lo true count of +.67. For reference this comes out to -.2179% for insure all EV and .0000% for insure +EV.

Next compute EV for 2 low cards removed and 2 high cards added. This is a running count of +4 with 156 cards (3 decks remaining) evaluating to a Hi-Lo true count of +1.33. I will not do this calculation but will explain waht is involved. It never matters which low cards are removed but now there are 3 possible combinations of high cards that can be added - TT, TA, AA and 3 calculations need to be done. Prob(TT) = 48/60*47/59, Prob(TA) = 48/60*12/59*2, Prob(AA) = 12/60*11/59. Sum the 3 calculation results for each of the high card compositions multiplied by their probabilities.

Whenever compositions are chamged in multiples of 5 low removed : 5 high added then only 1 calculation is needed. This is because the correct ratio of tens to aces can be maintained by simply adding 4 tens and 1 ace, 8 tens and 2 aces, etc.

What you will end up with is a table like the following. I listed end results for
TC=+3.33 and TC=+6.67 because they involve changing composition by multiples of 5 cards and require only 1 calculation.

You could use a program such as Excel to do the probabilites for cases of composition changes that are not in multiples of 5 because that could be a pain in the neck to do by hand.

Online program to compute blackjack EV

Insurance EV at various Hi-Lo true counts for 6 decks at pen of 156 cards
0 Insure all -.2844%, Insure only when +EV .0000%
+.67 Insure all -.2179%, Insure only when +EV .0000%
+1.33 compute yourself
+2 compute yourself
+2.67 compute yourself
+3.33 Insure all +.0269%, Insure only when +EV +.0369%
+4 compute yourself
+4.67 compute yourself
+5.33 compute yourself
+6.67 Insure all +.3763%, Insure only when +EV +.3763%
 

blackriver

Well-Known Member
#3
k_c said:
I can tell you how to get a reasonable idea of value of insurance by using the program on my website. I am too lazy to do it for you but I can show what to do if you're interested. The trouble with math is that most just don't care to deal with it. Also I wouldn't be surprised if QFIT had some sort of data listed somewhere.

The program on my website computes composition dependent EV for any shoe composition. It also computes insurance EV at the same time. For insurance it computes 2 values - insure all and insure only hands where EV is positve.

Relative to a count the only problem is to input a reasonably representative composition. Since you're interested in 6 decks, the first step is to input 6 for
number of decks. Mid-shoe is the best place to compute values, so the next step is the set shoe composition to 12 for all non-tens and 48 for tens. Mid-shoe is good because I have independently shown at that point probability of drawing a neutral Hi-Lo count card (7,8,9) at that point equals exactly 1/13 for any running count.

Next step is to vary composition relative to Hi-Lo.

For a 0 count nothing needs to be done. Just click compute. Insure all EV displays -.2844%, Since there are no +EV opportunites for a composition of
12,12,12,12,12,12,12,12,48,12 (2-A) insure +EV displays .0000%.

To create non-zero counts change low card compositions (2-6) and high card compositions by equal amounts. i.e. to create a running count of +2 remove 1 low card and add 1 high card (T-A). Total number of cards will always be kept equal to 156. It doesn't matter which low cards are removed because relative to the insurance calculation they are all equivalent. However the insurance calculation is different depending upon whether the high card added is a ten or an ace. So the calculation needs to be done twice - once with 49 tens and 12 aces and once with 48 tens and 13 aces to account for both possibilites. The probability that the high card added was a ten = 48/60 and the probability it was an ace = 12/60 since the starting high card composition is 48 tens and 12 aces, so multiply insurance EV with a ten added by 48/60 and add to insurance EV with an ace added multiplied by 12/60. Result is insurance EV for running count of +2 with 156 cards (3 decks remaining,) evaluating to a H-Lo true count of +.67. For reference this comes out to -.2179% for insure all EV and .0000% for insure +EV.

Next compute EV for 2 low cards removed and 2 high cards added. This is a running count of +4 with 156 cards (3 decks remaining) evaluating to a Hi-Lo true count of +1.33. I will not do this calculation but will explain waht is involved. It never matters which low cards are removed but now there are 3 possible combinations of high cards that can be added - TT, TA, AA and 3 calculations need to be done. Prob(TT) = 48/60*47/59, Prob(TA) = 48/60*12/59*2, Prob(AA) = 12/60*11/59. Sum the 3 calculation results for each of the high card compositions multiplied by their probabilities.

Whenever compositions are chamged in multiples of 5 low removed : 5 high added then only 1 calculation is needed. This is because the correct ratio of tens to aces can be maintained by simply adding 4 tens and 1 ace, 8 tens and 2 aces, etc.

What you will end up with is a table like the following. I listed end results for
TC=+3.33 and TC=+6.67 because they involve changing composition by multiples of 5 cards and require only 1 calculation.

You could use a program such as Excel to do the probabilites for cases of composition changes that are not in multiples of 5 because that could be a pain in the neck to do by hand.

Online program to compute blackjack EV

Insurance EV at various Hi-Lo true counts for 6 decks at pen of 156 cards
0 Insure all -.2844%, Insure only when +EV .0000%
+.67 Insure all -.2179%, Insure only when +EV .0000%
+1.33 compute yourself
+2 compute yourself
+2.67 compute yourself
+3.33 Insure all +.0269%, Insure only when +EV +.0369%
+4 compute yourself
+4.67 compute yourself
+5.33 compute yourself
+6.67 Insure all +.3763%, Insure only when +EV +.3763%
this is the longest post ive ever seen that starts out with "im too lazy"
 

London Colin

Well-Known Member
#4
A couple of points

First of all, I got very confused looking at k_c's figures until the penny eventually dropped and I realised that the insurance expectations being calculated are the contributions of insurance towards the overall EV of the original bet, not the EVs of the actual insurance bets themselves.

So for example, I would calculate the OTT (therefore TC=0) insurance EV for 6D as follows -

(+2 * 96/311) - (1 * 215/311) = -7.4%

And the insurance EV of an average hand (before we know the dealer is going to get an ace) is therefore -7.4/13 = -0.569%.

k_c's calculator, however, gives -.2844%, half the above figure, which I think is beacuse it is being related to the initial bet, which is twice the size of the insurance bet. (N.B. The figure quoted for the 3D composition is actually what the calculator gives for the full 6D.)



Secondly, assuming I've understood all that correctly, isn't the most straightforward approach to this just to use the insurance EORs to give an average change in expectation per TC (much like the 0.5% we quote for the regular advantage)?

The EORs are quoted in TOBJ: +1.81% for the non-tens and -4.07% for the tens. So the average for Hi-Lo is (6*1.81 + 4*4.07)/10 = 2.7%

So with an OTT EV of -7.4%, we can compute the EV for any TC as:
-7.4 + TC*2.7

(at least I think we can; I may have totally misunderstood some of the theory behind all this.)

And to get the figures in the same terms as k_c's calculator, they would then have to be divided by 26. (13 and then 2)

For a TC of .67, this would give: (-7.4 + .67 * 2.7)/26 = -.215%
For a TC of 3.33, it would give : (-7.4 + 3.33 * 2.7)/26 = +.062%
For a TC of 6.67, it would give : (-7.4 + 6.67 * 2.7)/26 = +.408%

(which look to be in the same neighbourhood as k_c's figures)
 

k_c

Well-Known Member
#5
blackriver said:
this is the longest post ive ever seen that starts out with "im too lazy"
It is simple enough to get insurance EVs for any shoe composition from my online program. Overall insurance EVs or insurance EV for any 2 card hand can be displayed by simply inputting decks, shoe composition, and optionally a 2 card hand and just clicking 'Compute'. The harder thing to do is to use the data to relate to a given counting system in a systematic way.
 

k_c

Well-Known Member
#6
London Colin said:
First of all, I got very confused looking at k_c's figures until the penny eventually dropped and I realised that the insurance expectations being calculated are the contributions of insurance towards the overall EV of the original bet, not the EVs of the actual insurance bets themselves.

So for example, I would calculate the OTT (therefore TC=0) insurance EV for 6D as follows -

(+2 * 96/311) - (1 * 215/311) = -7.4%

And the insurance EV of an average hand (before we know the dealer is going to get an ace) is therefore -7.4/13 = -0.569%.

k_c's calculator, however, gives -.2844%, half the above figure, which I think is beacuse it is being related to the initial bet, which is twice the size of the insurance bet. (N.B. The figure quoted for the 3D composition is actually what the calculator gives for the full 6D.)
Yes it's overall insurance EV based on half of initial bet. The program's calculation depends upon how many cards are optionally input into player's hand when 'Compute' is clicked. If number of player cards is 0 or 1 then overall EV is displayed. If 2 or more player cards are input then EVs for that particular hand are displayed.

London Colin said:
Secondly, assuming I've understood all that correctly, isn't the most straightforward approach to this just to use the insurance EORs to give an average change in expectation per TC (much like the 0.5% we quote for the regular advantage)?

The EORs are quoted in TOBJ: +1.81% for the non-tens and -4.07% for the tens. So the average for Hi-Lo is (6*1.81 + 4*4.07)/10 = 2.7%

So with an OTT EV of -7.4%, we can compute the EV for any TC as:
-7.4 + TC*2.7

(at least I think we can; I may have totally misunderstood some of the theory behind all this.)

And to get the figures in the same terms as k_c's calculator, they would then have to be divided by 26. (13 and then 2)

For a TC of .67, this would give: (-7.4 + .67 * 2.7)/26 = -.215%
For a TC of 3.33, it would give : (-7.4 + 3.33 * 2.7)/26 = +.062%
For a TC of 6.67, it would give : (-7.4 + 6.67 * 2.7)/26 = +.408%

(which look to be in the same neighbourhood as k_c's figures)
The program outputs exact calculations. There is nothing wrong with using EORs to get a reasonable value if that's what suitable to you.
 

London Colin

Well-Known Member
#7
k_c said:
The program outputs exact calculations. There is nothing wrong with using EORs to get a reasonable value if that's what suitable to you.
The program's exact calculations are for a specific shoe composition. Does the linear nature of insurance mean that the mid-shoe composition you chose (combined with the method of generating TCs by adding and removing cards) gives exact answers for each given TC at any depth? (Where the exact answer would presumably equate to the average of all the different advantages of every possible composition that gives a particular TC.)

My assumption was that your method gives an approximation for the general case, by looking at the specific, balanced mid-shoe case. But I'm a little out of my depth on this.

Similarly, I'm not sure whether my method actually gives an exact answer, because of the linear nature of insurance, or whether it gives an approximation of the overall, average advantage associated with any TC.
 

k_c

Well-Known Member
#8
London Colin said:
The program's exact calculations are for a specific shoe composition. Does the linear nature of insurance mean that the mid-shoe composition you chose (combined with the method of generating TCs by adding and removing cards) gives exact answers for each given TC at any depth? (Where the exact answer would presumably equate to the average of all the different advantages of every possible composition that gives a particular TC.)

My assumption was that your method gives an approximation for the general case, by looking at the specific, balanced mid-shoe case. But I'm a little out of my depth on this.

Similarly, I'm not sure whether my method actually gives an exact answer, because of the linear nature of insurance, or whether it gives an approximation of the overall, average advantage associated with any TC.
I chose mid-shoe because at that point the method I used does equate to the average Hi-Lo composition for any running count. This is because that mid-shoe is the only shoe depth (other than full shoe) where the probability of drawing a neutral card (7,8,9) is exactly equal to 1/13 given no other information other than count and depth.. This can be seen from a separate program I have that computes the probabilites of drawing each rank based on count/pen input. At other depths the prob of drawing a (7,8,9) can vary some as running count varies. Below 6 decks, 156 cards, RC=+10 and 6 decks 104 cards, RC=0. With 156 cards remaining RC can be any possible value and prob(7,8,9) remains = 1/13. This is not true at any other depth. For this reason I think that mid-shoe is the best place to look for a reasonable average for the least amount of trouble.

Code:
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 6
Cards remaining: 156
Initial running count (full shoe): 0
Running count: 10
Specific removals
        A: 0
        2: 0
        3: 0
        4: 0
        5: 0
        6: 0
        7: 0
        8: 0
        9: 0
        T: 0

Number of subsets for 156 cards: 7501
Prob of running count 10 from 6 decks: 0.0224398

p[1] 0.0833333  p[2] 0.0705128  p[3] 0.0705128  p[4] 0.0705128  p[5] 0.0705128
p[6] 0.0705128  p[7] 0.0769231  p[8] 0.0769231  p[9] 0.0769231  p[10] 0.333333

Press x or X to exit program (it may take some time to close,)
any other key to enter more data for same count tags/decks:


Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 6
Cards remaining: 104
Initial running count (full shoe): 0
Running count: 0
Specific removals
        A: 0
        2: 0
        3: 0
        4: 0
        5: 0
        6: 0
        7: 0
        8: 0
        9: 0
        T: 0

Number of subsets for 104 cards: 5037
Prob of running count 0 from 6 decks: 0.0544645

p[1] 0.076886  p[2] 0.076886  p[3] 0.076886  p[4] 0.076886  p[5] 0.076886
p[6] 0.076886  p[7] 0.0770465  p[8] 0.0770465  p[9] 0.0770465  p[10] 0.307544

Press x or X to exit program (it may take some time to close,)
any other key to enter more data for same count tags/decks:
Sorry this is all I can post right now as I need to leave for work.
 

k_c

Well-Known Member
#9
London Colin said:
The program's exact calculations are for a specific shoe composition. Does the linear nature of insurance mean that the mid-shoe composition you chose (combined with the method of generating TCs by adding and removing cards) gives exact answers for each given TC at any depth? (Where the exact answer would presumably equate to the average of all the different advantages of every possible composition that gives a particular TC.)

My assumption was that your method gives an approximation for the general case, by looking at the specific, balanced mid-shoe case. But I'm a little out of my depth on this.

Similarly, I'm not sure whether my method actually gives an exact answer, because of the linear nature of insurance, or whether it gives an approximation of the overall, average advantage associated with any TC.
Sorry my previous post was probably not a very satisfactory answer to your question but maybe the data could at least show where I'm coming from.

Exact answers can be obtained at mid-shoe. If the same method is used at a pen other than mid-shoe then it can't be considered exact. This is because it turns out that mid-shoe is the only pen where the prob of a neutral card (7,8,9) equals exactly 1/13 for any running count so when you alter the shoe by changing the number of low (2-6) and high (T-A) by the same number of cards then you wind up with the exact average Hi-Lo running. Composition changes are integral numbers.

With other pen levels when you look at the probabilities of drawing each rank for various running counts then you will be dealing with shoe compositions that do not consist of integral numbers. For example the data I listed for 104 cards remaining of 6 decks @ RC=0 shows prob of drawing a 7 = .0770465. 104*.0770465 = 8.012836, which is not an integral number and cannot be represented exactly in the program on my website.

Hopefully at least that may shed some additional light
 

London Colin

Well-Known Member
#10
k_c said:
Sorry my previous post was probably not a very satisfactory answer to your question but maybe the data could at least show where I'm coming from.
No problem. I think I got the gist of what you were saying. (And apologies to tensplitter if I'm hijacking the thread a little.)


k_c said:
Exact answers can be obtained at mid-shoe. If the same method is used at a pen other than mid-shoe then it can't be considered exact.
As ever, there is much scope for ambiguity in both the asking and answering of questions. In this case, I think it all hinges on the meaning of 'exact'.

If I understand you right, you are asserting that the answer given for any Hi-Lo TC at the mid-shoe point will be exact for that depth in the shoe, but you are not asserting that this answer will also be the exact average for all the multitude of shoe compositions, at varying depths, which can give rise to that TC. (Though I wonder if this might actually be the case, given the symmetry of the situation.)

Have I got that about right?


But the point I was trying to make is that a method based on EORs may actually be more accurate (or at least no less accurate), as well as being a good deal simpler.

It's fair to say that I only have a partial grasp of some of the theory behind all this, essentially what the layman can pick up from reading the TOBJ and skipping over the difficult bits:), so please bear with me while I try to lay out some of the factors influencing my thinking (and let me know if you spot any errors in my assumptions) -

Insurance is linear, which means that if you use a point count that is 100% correlated with the insurance EORs, you can compute the exact advantage from the count, for any given deck composition. (Whereas when we apply the same technique to something like the value of a blackjack hand, we are making an 'assumption of linearity' which we know to be false. It does however give good estimates unless you get down to a very small remnant of cards remaining to be dealt.)

The Hi-Lo count is only loosely correlated with insurance. That means that you can't calculate the exact advantage for any specific situation if you only know the Hi-Lo TC. You can only get an estimate.

But what effect, if any, does this lack of correlation have on the accuracy overall? i.e. Can we say that the computed estimate will in fact be equal to the average of all the exact values it is the estimate for? [And if not, why not?]

----------------

By the way, I spotted a couple of errors in my original post -

It seems the insurance EORs are based on a full 52-card deck (i.e. against a notional dealer ace). As a result, the mean figure against which the EORs are applied is -7.69, not -7.4 (regardless of the number of decks in play).

Basing calculations on the single-deck EORs, as I did, is the practical approach, but for supreme accuracy you would have to use the EORs for the number of decks in play, as they don't quite scale precisely.

-----------------

Hope everyone is now as confused as I am. :grin:
 

k_c

Well-Known Member
#11
London Colin said:
No problem. I think I got the gist of what you were saying. (And apologies to tensplitter if I'm hijacking the thread a little.)



As ever, there is much scope for ambiguity in both the asking and answering of questions. In this case, I think it all hinges on the meaning of 'exact'.

If I understand you right, you are asserting that the answer given for any Hi-Lo TC at the mid-shoe point will be exact for that depth in the shoe, but you are not asserting that this answer will also be the exact average for all the multitude of shoe compositions, at varying depths, which can give rise to that TC. (Though I wonder if this might actually be the case, given the symmetry of the situation.)

Have I got that about right?


But the point I was trying to make is that a method based on EORs may actually be more accurate (or at least no less accurate), as well as being a good deal simpler.

It's fair to say that I only have a partial grasp of some of the theory behind all this, essentially what the layman can pick up from reading the TOBJ and skipping over the difficult bits:), so please bear with me while I try to lay out some of the factors influencing my thinking (and let me know if you spot any errors in my assumptions) -

Insurance is linear, which means that if you use a point count that is 100% correlated with the insurance EORs, you can compute the exact advantage from the count, for any given deck composition. (Whereas when we apply the same technique to something like the value of a blackjack hand, we are making an 'assumption of linearity' which we know to be false. It does however give good estimates unless you get down to a very small remnant of cards remaining to be dealt.)

The Hi-Lo count is only loosely correlated with insurance. That means that you can't calculate the exact advantage for any specific situation if you only know the Hi-Lo TC. You can only get an estimate.

But what effect, if any, does this lack of correlation have on the accuracy overall? i.e. Can we say that the computed estimate will in fact be equal to the average of all the exact values it is the estimate for? [And if not, why not?]

----------------

By the way, I spotted a couple of errors in my original post -

It seems the insurance EORs are based on a full 52-card deck (i.e. against a notional dealer ace). As a result, the mean figure against which the EORs are applied is -7.69, not -7.4 (regardless of the number of decks in play).

Basing calculations on the single-deck EORs, as I did, is the practical approach, but for supreme accuracy you would have to use the EORs for the number of decks in play, as they don't quite scale precisely.

-----------------

Hope everyone is now as confused as I am. :grin:
To me exact means computing EVs strictly dependent upon shoe composition by going through all of the possibilities. Counting happens after the fact. Counting systems try and estimate what the actual EV is as well as they can.

Relative to a counting system exact can mean something a bit different. You'll notice that my counting system subset enumeration program identifies a certain number of subsets for the pen that has been input. There are some subsets that compute to RC=0. There are subsets that compute to RC=+-1, RC=+-2, etc. When RC and pen are input there are a certain number of subsets that compute to that particular RC. Each subset has its own probability of occurrence. Each subset is weighted according to its probability and it's possible to compute the probabilities of drawing each rank. What you then have is shoe composition based upon a counting system. Since the shoe rank probabilities are based upon a weighted average of multiple subsets for the most part the number of each rank will not compute to a whole integer.

In the case of insurance all we need to know is whether the probability of drawing a ten is greater than 1/3. My old enumeration program computed a count based insurance EV but I'm not sure how well it works. Computing EV based on count subsets has an interesting phenomena. Normally when you remove a rank from a shoe the probabilities of drawing the other ranks after the removal increase. Example if an ace is removed from a full single deck probability of drawing a ten increases from 16/52 to 16/51. Compare this to a 2 card Hi-Lo running count of zero at the end of a single deck. The 2 possible subsets are L-H and M-M. Probabilities of drawing each rank are -
p[1] 0.0858369 p[2] 0.0858369 p[3] 0.0858369 p[4] 0.0858369 p[5] 0.0858369
p[6] 0.0858369 p[7] 0.0472103 p[8] 0.0472103 p[9] 0.0472103 p[10] 0.343348
If an ace is drawn then that means count = -1 and so the remaining card must necessarily be a low card. Probability of drawing a ten has decreased from 0.343348 to 0.0 even though a non-ten was drawn. If a medium card is drawn then that means the remaining card must also must be medium. Probability of drawing a 7,8,9 has gone from 0.0472103 (edit) to either 3/11 or 4/11 depending upon which medium card has been specifically removed.

My old subset enumeration program gets ins EV=.03% for RC=+10, cards left = 156 and ins EV=+.32% for RC=+20, cards left = 156. These are count based calculations. The first value is in agreement with my other method but the second is a little lower. However, like I said I'm not sure about the output. It's possible count based calculations may turn out a little differemtly because of the counting dynamics I tried to describe above as cards are removed and the various possibilities computed.

So to try and answer your question about mid-shoe, we can represent the starting composition by changing composition by whole numbers of cards at mid-shoe but maybe we may be neglecting some maybe minor count dynamics as cards are drawn to the various possibilities after that.

Just trying to describe my experience in looking at relating to a count. :grin:
 

London Colin

Well-Known Member
#12
k_c said:
To me exact means computing EVs strictly dependent upon shoe composition by going through all of the possibilities. Counting happens after the fact. Counting systems try and estimate what the actual EV is as well as they can.
That's not entirely true. In the case of insurance, a counting system can be as exact as you choose to make it. Though obviously if you actually wanted to employ an insurance count at the tables, practical considerations (integer tags, TC rounding, and so on) would mean that you would still be dealing with estimates.

So for instance, to use an example from TOBJ, if you see the dealer's ace and three other non-tens dealt from a single deck game, you know that the insurance EV is exactly -7.69 + (4*1.81)*51/48 (which equals 0).

But, thinking some more about the question of what happens when the counting system has tags which exaggerate (or even reverse the sign) of the values of the certain denominations, as happens when you apply Hi-Lo to insurance, I think there must necessarily be errors in the answers produced for each TC.

I was speculating that some kind of cancellation effect might get rid of the errors, but I think that could only apply overall (e.g., if the EV for a TC +1 is understimated by a certain amount, pehaps the EV for -1 is correspondingly overestimated.)

Hope that makes some kind of sense.:)
 

The Chaperone

Well-Known Member
#13
Actually if you wanted to do strictly an insurance count there is no need to estimate at all. You just start your running count at -4n where n is the number of decks in the game. Each non-ten will be counted as +1 and each ten will be counted as -2. If the RC is at 0, the insurance bet is break even. If it is positive, then you have an edge betting insurance.
 

k_c

Well-Known Member
#14
London Colin said:
That's not entirely true. In the case of insurance, a counting system can be as exact as you choose to make it. Though obviously if you actually wanted to employ an insurance count at the tables, practical considerations (integer tags, TC rounding, and so on) would mean that you would still be dealing with estimates.

So for instance, to use an example from TOBJ, if you see the dealer's ace and three other non-tens dealt from a single deck game, you know that the insurance EV is exactly -7.69 + (4*1.81)*51/48 (which equals 0).

But, thinking some more about the question of what happens when the counting system has tags which exaggerate (or even reverse the sign) of the values of the certain denominations, as happens when you apply Hi-Lo to insurance, I think there must necessarily be errors in the answers produced for each TC.

I was speculating that some kind of cancellation effect might get rid of the errors, but I think that could only apply overall (e.g., if the EV for a TC +1 is understimated by a certain amount, pehaps the EV for -1 is correspondingly overestimated.)

Hope that makes some kind of sense.:)
By exact I mean the starting shoe composition can be anything and is without regard to any counting system. When the cards are dealt there are 55 possible 2 card player hands and 10 possible dealer up cards. This is true for both insurance EV and stand/double/hit/split/surrender. All that is needed for insurance is to compute the probability of each 2 card hand times probability of up card of ace times 1/2 (relates to 1/2 initial bet) times (3*probability of drawing a ten after 2 card hand and ace are removed - 1) and sum the total. This is insurance EV for insuring every hand. To get EV for insuring only +EV only sum when the calculation is positive. There are shortcuts that can be taken because for insurance purposes all non-tens-aces are equivalent but going through all of the possibilities will get the final result as well. Shoe composition can be anything means each rank present before dealing is 0,1,2,3,.....,whole number up to number of cards of rank in a full shoe.

As Chaperone notes there is one count that can identify with 100% accuracy when insurance EV is positive, 0, or negative. This is because it identifies if probability of drawing a ten is > 1/3, 0, or < 1/3. There is no other count that can do this.

I think what you're saying about EORs makes sense when starting with a full shoe. When given less than a full shoe EORs may be different than full shoe EORs.
 

London Colin

Well-Known Member
#15
k_c said:
As Chaperone notes there is one count that can identify with 100% accuracy when insurance EV is positive, 0, or negative. This is because it identifies if probability of drawing a ten is > 1/3, 0, or < 1/3. There is no other count that can do this.

I think what you're saying about EORs makes sense when starting with a full shoe. When given less than a full shoe EORs may be different than full shoe EORs.
The EORs are a count, just not a very practical one. They can not only tell you when the probability of drawing a ten is > 1/3; they also tell you the exact advantage of any composition.

The starting point is, by definition, the full shoe. You count all the cards that have been removed to leave you with whatever composition you are currently faced with.
 

k_c

Well-Known Member
#16
London Colin said:
The EORs are a count, just not a very practical one. They can not only tell you when the probability of drawing a ten is > 1/3; they also tell you the exact advantage of any composition.

The starting point is, by definition, the full shoe. You count all the cards that have been removed to leave you with whatever composition you are currently faced with.
If the EORs can tell you when probability of drawing a ten is less than, equal to, or greater than 1/3 then that equates to the count in Chaperone's post and we are saying the same thing. :) The insurance count is 100% effective for insurance but otherwise isn't so good.
 

London Colin

Well-Known Member
#17
k_c said:
If the EORs can tell you when probability of drawing a ten is less than, equal to, or greater than 1/3 then that equates to the count in Chaperone's post and we are saying the same thing. :) The insurance count is 100% effective for insurance but otherwise isn't so good.
Indeed. I did refer to all this in my earlier posts: The fact that insurance is 'linear' means that EORs, and the counting methods which we derive from them, have the capability of being 100% accurate, whereas for other, non-linear aspects of blackjack they can only ever be approximations.

This is all just me quoting Griffin in TOBJ; I'm not claiming to have discovered anything.

I was initially confused about what this linearity might mean when we apply a count like Hi-Lo to the task of calculating insurance EV. While the answer it would produce for any given composition would obviously not be accurate, I thought the errors might cancel out, giving exact figures for each TC overall. But I now see that can't be the case.
 
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