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Algebraic Indices for Unbalanced Counts
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Algebraic Indices for Unbalanced Count Systems
Fractional Insurance Indexes
©Copyright ETFan December 12, 2010
Did you know the hi-lo index for insurance can be calculated on the back of a (large) envelope? And the right answer for one deck is 17/12, the exact answer for infinite decks is 10/3, and the correct answer for any other number of decks can be found with linear interpolation?
This article attempts to explain and extend the ideas behind Arnold Snyder's 1980/81/82 pamphlet: Algebraic Approximation of Optimum Blackjack Strategies. The extended system can handle unbalanced true counts, such as TKO or C. Membrino's true-counted Red 7, as well as balanced counts, such as hi-lo. The formulas developed will be more complicated than Arnold's, since we have easy access to computers nowadays, freeing us to abandon simplicity in favor of utterly preposterous precision. The system can be used for any strategy decisions where Effects Of Removal (EoRs) are available. In addition, a formula will be given that produces insurance indexes in the form of simple fractions.
Even though a lot of number crunching is involved, the ideas behind the system are fairly simple. If you accept the concepts of proportional deflection and linear transformation of a count as espoused by Peter Griffin in Theory of Blackjack (ToB), then you must grant that the results are exact. Unfortunately, this will not stop debate on the subject, since decks of cards aren't precisely normal in their distribution, making proportional deflection an approximate theory. Therefore, indices generated by simulators may be slightly more accurate. On the other hand, simulators sometimes generate different indexes at different penetrations, and who needs that... ;?
At any rate, from the standpoint of optimizing expectation (as opposed to SCORE) these algebraic indexes are almost certainly "close enough" by any standard. The system cannot provide surrender indexes, because of the way EoR tables have traditionally been layed out.
I'll try to describe the method in such a way that anyone versed in elementary probability theory and Peter Griffin's work can understand. Anyone unfamiliar with ToB will assuredly not fully comprehend a word I've written. In particular, readers should feel comfortable with the material in Appendix A of Chapter 7 and the reasoning in Appendix C of Chapter 5. You must also know how to use the "Effect of Removal" (EoR) tables in Chapter 6.
To illustrate, I'll develop a hi-lo insurance index for 1 deck, and a true-counted Red 7 insurance index for 6 decks.
Exact Insurance EoRs
Very precise insurance indexes are possible because we can derive exact insurance EoRs. An EoR is the amount you need to add to the full deck EV to get the EV for the 51 cards after a particular card is removed. I.e. EV(full deck) + EoR = EV(51 cards with specified removal), or rearranging: EoR = EV(51 cards with specified removal) - EV(full deck). [Note: EV stands for "Expected Value," also known as the expectation, in this case for the insurance bet only.]
Let's find the insurance EoR for a 5. Assume a 1 unit bet ...
There are 16 Ts in a full deck and 36 other cards. If the hole card is a T, you win 2 units, if the hole card is one of the others, you lose 1 unit. EV(full deck) = 16/52 * 2 + 36/52 * (-1) = -1/13
If you remove a 5, there are now 51 cards remaining, still with 16 Ts, but now with 35 others. EV(51 cards with a 5 removed) = 16/51 * 2 + 35/51 * (-1) = -1/17
So EoR(5) = (-1/17) - (-1/13) = 4/221 ~= 1.81% as listed in ToB, Chapter 6. 4/221 is the exact EoR, 1.81% is the decimal approximation.
This same EoR = 4/221 can be used for any number of decks, and any number of cards removed, by using the conversion outlined on the fourth page of Chapter 6, ToB, of 51/(cards remaining). For example, if you remove four 5s from a 6 deck shoe, the total EoR will be: 4*(4/221) * 51/308 = 12/1001
(We're already at the point where a CAS -- Computer Algebra System -- program or calculator would come in handy. Examples of CAS are Mathematica, Maxima, XCas, and calculators like the HP 50g or TI-89. You could also do the fractions by hand or throw caution to the wind and take my numbers on faith.)
Since we know the EV at the top of a 6d shoe is -1/13, we know now that the EV after four 5s are removed is: -1/13 + 12/1001 = -5/77, which you can verify by exact calculation. You should calculate the expectation directly, based on four non-tens removed from a 6d shoe, to convince yourself that -5/77 is the exact answer (ignore the ace upcard, just this once). If you can follow the argument thus far, you have a very good model of how EoRs work for any strategy -- not just for insurance. On the other hand, if you can't calculate the -5/77 EV, you probably should get off now. This article may exceed your level of preparation.
Using reasoning similar to the above, we can show that the insurance EoR for any card other than a T is also 4/221, and that the EoR for a T is -9/221. The EoR for all 52 cards sum to zero.
You can multiply all the insurance EoRs by 221 (a "linear transformation") to get the perfect, balanced, insurance count of 4 4 4 4 4 4 4 4 4 -9 (we follow Griffin's convention of listing tags in numerical order from ace through ten). This is Thorp's old ten count "parameterized as a point count," according to ToB Chapter 4. These tags have correlation 1 with the EoR, so if you know how to calculate a correlation coefficient, you can quickly correlate any set of tags with 4 4 4 4 4 4 4 4 4 -9 to get the an Insurance Correlation (IC) slightly more accurate than what you get using the three digit EoRs listed in ToB.
A Word about Linear Tranformations of Count Systems
A linear transformation is the act of adding a constant to every tag in a count, and/or multiplying every tag by a constant. A little thought should convince you that linear transformations will not degrade the information accessible to an arithmetically adroit counter. For example, suppose all the hi-lo tags were doubled. You of course will adjust your indices and your betting ramp, but at the end of the day, you're no better nor worse off than when you started. Similarly, you could subtract 1 from all the tags, and unbalance the system, but at any given penetration, it's trivial to convert back to hi-lo.
In general a set of tags, tagi for i = 1, 2 ... 52, defines a count system S. If, for constants m and b, each tagi(S) x m + b = tagi(S'), then system S' has linear equivalence with S. The correlation between S and S' will be 1, so the correlation between either count and any third set of numbers (such as EoRs) will be identical.
For example, if S = hi-lo, m = 2 and b = 0.5, we have the count:
S' = -1.5 2.5 2.5 2.5 2.5 2.5 0.5 0.5 0.5 -1.5
which is nothing more than hi-lo in disguise. This new count has an insurance correlation of 76%, same as the insurance correlation for hi-lo. Betting and playing correlations are also identical.
Or if you let m = 3/13 and b = 1/13 and apply those to Thorp's ten count, S = 4 4 4 4 4 4 4 4 4 -9, you generate S' = 1 1 1 1 1 1 1 1 1 -2 -- the well known Noir count. Both counts have an insurance correlation of 100%.
Calculating the One Deck Insurance Index for Hi-lo
Now to find the one deck insurance index for hi-lo, we will perform six steps:
1) Remove the upcard (ace for insurance) as well as any player cards from the shoe.
2) Balance the count with a linear transformation.
3) [A common sense check] Find the effect on each individual card when you increase the TC by 1, using the principle of proportional deflection.
4) Write an equation that finds EV (or EV delta, for strategies other than insurance) based on 3), the EORs, and the starting EV delta.
5) Solve for TC when the EV delta = 0.
6) Convert this TC back to the original count to get your index.
1) Remove an ace from the deck. Now there are 51 cards in the "shoe" and the count is no longer balanced.
2) To balance the count (making it much simpler to apply Griffin's formulas) we need to subtract 1/51 from all the tags, which is a linear transformation. So we have three aces that are tagged -52/51, four each of 2s, 3s, 4s, 5s and 6s tagged 50/51, four each of 7s, 8s and 9s tagged -1/51, and 16 Ts tagged -52/51. If you add the tags all up, they come to zero, and this new count is guaranteed to provide the counter with exactly the same information as a hi-lo counter at any given penetration.
We could multiply all the tags by 51, to make them integers, and it might make life a little easier for a counter using the tags in the real world, but our calculations would actually become slightly more complicated -- and we have no intention of using these tags in the real world.
A subtle point concerning this new "rebalanced" count is how it reveals that the so-called "neutral" cards are now slightly correlated with the count. Because of the removal of the ace, a high count now portends a slightly higher density of 7s, 8s and 9s, and you can see that clearly now, because 7s, 8s, and 9s have small negative tags.
3) For each of the 51 cards, a given True Count (TC) implies that, on average, it's no longer one, single card. The principle of proportional deflection says it's "deflected" from the value of 1, by an amount proportional to the TC and the individual (balanced) count tag. Under this model, cards can assume fractional values in order to represent the "average" situation. The formula for each card is: 1 - TC*Tag*51/y, where y = the sum of (tag squared) for all 51 cards. TC, here, is expressed on a per card basis, so a TC of +1/52 corresponds to a full deck TC of +1. There will still be 51 cards in the deck after the cards are deflected. In this case:
y = 19*(-52/51)^2 + 20*(50/51)^2 + 12*(-1/51)^2 = 1988/51.
If we happen to have a per-card TC of +1/52, the average number of aces would then be:
3*[1 - (1/52)*(-52/51)*51/(1988/51)]
And the average number of Ts would be:
16*[1 - (1/52)*(-52/51)*51/(1988/51)]
And the total number of 2s, 3s, 4s, 5s and 6s, on average:
20*[1 - (1/52)*(50/51)*51/(1988/51)]
And the total number of 7s, 8s and 9s, on average:
12*[1 - (1/52)*(-1/51)*51/(1988/51)]
If you add these all up, you'll find the total number of cards is still 51. It's also fairly easy to prove that it adds up to 51 for any TC you choose to plug in. Finally, if you use any number other than 1988/51 for y, the sum of tags for the 51 cards would no longer equal -51/52 as it must, showing our deflection formula is accurate. (Please take your time to verify each of these statements. You don't have to take my word for anything.)
4) Now the nice thing about the 1d insurance case is we don't have to convert EoRs. If we use our TC formula for the various cards, the number remaining at the end will be 51. So we'll forego multiplying by 51/51 (but we must remember to convert when we tackle 2 or more decks). The number "removed" for each of the 51 cards, is just the expression after the "1 -". E.g. for one given ace, remember the formula was:
1 - TC*(-52/51)*51/(1988/51)
So the amount removed for that ace is:
which will be a negative removal (i.e. an addition) for a positive TC. So combining all this information with the EoRs, and remembering to add the full EoR for the ace removed off the top, we deduce that the total insurance EV, for any given TC is:
EV = -1/13 + 4/221 +
3*TC*(-52/51)*51/(1988/51) *(4/221) +
20*TC*(50/51)*51/(1988/51) *(4/221) +
12*TC*(-1/51)*51/(1988/51) *(4/221) +16*TC*(-52/51)*51/(1988/51) *(-9/221)
This is similar to the example on the fourth page of Chapter 6, ToB, showing how to use the EoR tables, just a tad more tortuous for us mathe-masochists.
5) To find the per-card index for our "rebalanced" count, we set this last expression equal to zero and solve for TC:
TC/(1988/51)/221*[3*(-52)*4 + 20*50*4 + 12*(-1)*4 + 16*(-52)*(-9)] = 1/13 - 4/221
TC*51/1988/221* = 1/17
TC = 497/10608
[As a check, solving:
16*(1 - TC*(-52/51)*51/(1988/51)) * 2 = 51 - 16*(1 - TC*(-52/51)*51/(1988/51)
for TC also returns 497/10608. This last expression just says there are twice as many non-tens as tens in the deck. So you can solve for TC without bothering with EoRs, just from the expression for Ts in 3). But I'm using EoRs to develop a more general system that can handle any strategy -- not just insurance.]
6) Now to get the hi-lo per-card TC from this "rebalanced" count at any penetration, all you need to do, believe it or not, is subtract 1/51. Then to convert to a full deck TC, you multiply this per-card TC by 52. So our surprisingly simple answer is: index = (497/10608 - 1/51) * 52 = 17/12
There are other ways to look at it and think about it, but the blankety-blank mess above is about as simple as it gets for this problem. Luckily we live in the age of computers, and once you codify the six steps in a program or spreadsheet you can immediately harness the system to kick out an index for any strategy decision.
Second Illustration -- Six Deck Insurance Index for Red 7
We follow the same six steps...
1) Remove an ace from the shoe. Now there are 311 cards in the shoe and the count is (still) unbalanced.
2) To balance the count we subtract 13/311 from all the tags, which is a linear transformation. So we have 23 aces that are tagged -324/311, 24 each of 2s, 3s, 4s, 5s and 6s tagged +298/311, 12 red 7s tagged +298/311 and 12 black 7s tagged -13/311, 24 each of 8s and 9s tagged -13/311, and 96 Ts tagged -324/311. If you add the tags all up, they come to zero, and this new count is guaranteed to provide the counter with exactly the same information as a Red 7 true counter at any given penetration.
3) The sum of squares:
y = 119 * (-324/311)^2 + 132 * (298/311)^2 + 60 * (-13/311)^2 = 77892/311
The formula for each card is:
1 - TC*Tag*311/y
The rest of this step is left as an exercise for the interested reader.
4) The amount removed for one of the 23 remaining aces is:
which will be a negative removal (i.e. an addition) for a positive TC. Combining this type of information with the EoRs, and remembering to adjust EoRs by a factor of 51/311 while remembering to add the full EoR for an ace off the top, the total EV for a given TC is:
EV = -1/13 + 4/221*51/311 +
23*TC*(-324/311)*311/(77892/311) *(4/221*51/311) +
132*TC*(298/311)*311/(77892/311) *(4/221*51/311) +
60*TC*(-13/311)*311/(77892/311) *(4/221*51/311) +
5) Set EV equal to zero and solve for TC (A CAS is highly recommended):
TC = 149293/2418336
6) index = (149293/2418336 - 13/311) * 52 = 2015/1944 ~= 1.04
[Note: This is the full deck true counted Red 7 insurance index assuming the normal Initial Running Count (IRC) for 6 decks of -12. People using another IRC will need to adjust the index accordingly.]
Comparison with Arnold Snyder's Algebraic Method
In the case where only the dealer's upcard is removed from one deck the 6 steps can be collapsed to this formula:
index = 52/51 * m * y / i + 52/51 * t
Where m is precisely as the Bishop defined it, namely:
-(Griffin's 11th column) - EoR(upcard).
And y is the same as Snyder's "p" except it is based on sum of squares of the modified tag values, instead of the original tags. And i is the inner product, also based on the modified tags instead of the originals.
Hence, the famous Algebraic Approximation formula:
index = m * p / i + 52/51 * t
is closely related to the procedure above. After calculating a dozen different strategies using Griffin's most recent EoR tables, I can state that indexes calculated with 1) through 6) above rarely differ from Snyder's until you reach the third digit, which is insignificant to the player's hourly win rate.
For example, the one deck insurance index under the '81 system for hi-lo is 1.4332 while the method above yeilds 1.4167 . It's impossible to construct a set of cards with 51 cards or less having a TC that falls between those two numbers. Thus, the two indexes are in practice indistinguishable.
The main value of this new procedure is in expanding the algebraic system to unbalanced true count systems. In addition you get to entertain your friends with insurance indexes in the form of exact fractions!
Toward Perfect Insurance -- A Challenge
Griffin states that "insurance is linear," and we know EoRs can provide exact expectations for insurance bets, and that perfect insurance decisions are possible with the Noir count, for example. So it might seem reasonable to expect our procedure to provide exact insurance indexes, or at least the best possible index for every situation. Unfortunately, for most counts this is only possible when you have different indexes at very deep penetrations.
To see this, imagine one deck dealt down to the last two cards -- one of which is the dealer's hole card. A hi-lo count of zero portends the following probabilities for the two unseen cards: ace-low -- 60/446, ten-low -- 320/446, mid-mid -- 66/446. So an insurance bet of one unit has a positive expectation of +17/223, and a simulator should produce an index of 0 in this extreme situation.
However, I challenge anyone to find an insurance index that's more accurate than the one generated by this algebraic system, for any popular count system throughout a realistic range of penetrations.
Extension to Unbalanced Running Count Systems
Several simulations lead me to this rule of thumb: the best index for a conventional unbalanced count, such as Red 7, is the running count corresponding to the true count index at the most profitable penetration that can actually occur in the game you're facing -- or just under that penetration. The most profitable penetration normally occurs right before the last hand is dealt from the shoe.
For example, for Red 7, since the six deck true counted insurance index is 2015/1944, I would look at the RC where 4.5 decks had been dealt out and calculate: RCindex = 1.5 x 2015/1944 = 1.55, which dovetails nicely with Snyder's recommended index of +2 for shoe games. Again, this is the index when you start with an initial running count of -12, which is necessary to make the final running count equal to zero once all the cards in the shoe are counted.
Simulations might reveal some exceptions to my empirical rule of thumb.
General Linear Insurance Index Formula:
Applying our six steps to the general insurance case and simplifying (don't try this at home) yields:
52 (d s s - 48 d s t - 4 d y + 4 a a + 48 a t - 2 a s + y)
48 (52 d t - d s + a - t)
where d = number of decks, a = ace tag, t = tens tag, s = sum of all tags over one 52-card deck, and y = sum of squares of tags over one deck.
For example, for Red 7, 6 deck case, d = 6, a = -1, t = -1, s = 2, and y = 42. So index =
52*(6*2*2 - 48*6*2*(-1) - 4*6*42 + 4*(-1)*(-1) + 48*(-1)*(-1) - 2*(-1)*2 + 42)
.....all divided by.....
48*(52*6*(-1) - 6*2 + (-1) - (-1))
= 2015/1944 as above.
It is assumed, for unbalanced counts, that an IRC of -s*d is used, so true count division can work normally.
Changing the 52 in the numerator to 26 gives the half deck index; changing it to 13 gives a quarter deck index; and changing it to 1 returns a per-card index.
Some interesting conclusions can be drawn from the General Insurance Index Formula. For example, you can use it to show that when a = t (as in hi-lo or Red 7), interpolation by reciprocal of the number decks (1/d) is possible. Interpolation can also work perfectly in some other cases, e.g. for any perfect insurance count, such as 0 0 0 0 0 0 0 0 0 1. a = 0, t = 1, s = 16, y = 16, and index = -52/3, for any d save d = 1/36 (which is unlikely).
The GLII formula works for insurance, but for other strategies I revert to the six steps above. A formula would get pretty hairy.
Here is an input string for the formula that you can copy and paste for use in some CAS programs:
52*(d*s*s - 48*d*s*t - 4*d*y + 4*a*a + 48*a*t - 2*a*s + y)/(48*(52*d*t - d*s + a - t))
[Hint: It may be useful to paste this into a text file and recopy from there, to rid the string of unwanted html control codes]
Here is a formula for the infinite deck case:
52 (s s - 48 s t - 4 y)
index = --------------------------
48 (52 t - s)
And an input string for infinite deck:
52*(s*s - 48*s*t - 4*y)/(48*(52*t - s))
Here are some insurance indexes:
Thorp Noir decks hi-lo Red 7 Zen (4/-9) (1/-2) 1 17/12 -65/108 923/309 52/3 0 2 19/8 247/648 2587/621 52/3 0 3 97/36 689/972 1417/311 52/3 0 4 137/48 377/432 1183/249 52/3 0 5 59/20 1573/1620 7579/1557 52/3 0 6 217/72 2015/1944 3081/623 52/3 0 7 257/84 13/12 10907/2181 52/3 0 8 99/32 2899/2592 12571/2493 52/3 0 infinite 10/3 221/162 16/3 52/3 0
With the General Linear Insurance Index Formula, all the problems of the world can be solved!
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