I am playing around with the index generators for multiparameter counts from www.bjmath.com. I was wondering if anyone can explain how I calculate when to start taking insurance? The spreadsheets don't have this oh-so-important index number on them. I'm no math-phobe, so hit me with whatever it takes. Thanks!
~Magic Man
Also take insurance at TC=+1 if you have A8 or A9 but do not yet take even money if you have A10 (BJ) unless you spot a couple non tens around you.
Puiu
Why should one go against the advice and buy insurance at +2, no matter how many decks of cards used? This is only matemathically correct for one deck games.
An other JP tip off...
A point count version of the ten count counts tens as -9 and all other cards as +4. Insurance is profitable when the ratio of tens to non-tens is 1 to 2, which is the reverse of the payout.
There are 16 tens in a single deck and normally 36 other cards. When 4 non-tens are removed this is just right.
Now, and this is how algebraic approximation indexes are done genrally too, the point count version of the ten count has the equivalent of having a single deck index that is +4*4*52/48---your calculator or mine? Use that as the perfect index for this point count version of the ten count for any number of decks.
To convert this index to an index that you can use, you first have to to divide by the correlation coefficient between your count and this point count--your count is not so good as this ten count and will tend to be more "out there" to find the point where insurance becomes profitable. You also have a point count that has smaller values than -9 and such, so you multiply by the square root of the ratio of the sum of squares of your count and this ten count, which is perfect for insurance. The result is the perfect insurance index (your calculator or mine?)(which is in blue above) multiplied by the sum of squares of your point count, and divided by the inner product of the "pointed" ten count and the count you actually use.
You now have the infinite deck index--but since you don't have to play shoes that big....well we move on to the single deck. There is a known card removed from the pack: the dealer's ace. For the perfect count for insurance it has the count value of +4. If we see other player's cards we would have other values to include. That "precounts" our index in 2 ways: we should adjust for how the perfect count values the ace, and we should adjust for how our count values the ace.
So we subtract from the infinite deck index we would have to shoot for if conditions got really bad, by taking the perfect index number (in blueabove) and subtracting, using n for the number of known cards we are dealing with 4*51/(52-n). 4 is the value given in the perfect index count for the ace. Now you may or may not count the ace in your count, but if you do your adjusted index needs to have that value multiplied by 51/(52-n) added and the inner product has to be done without including the value of the known card (or cards). So instead of finding the inner product with the card values and then using the inner product of the tens value 4 times, you find the inner product with 4 times this and subtract the product of the ace value your count uses and +4, the value this perfect count uses.
That is your one deck insurance index. You interpolate to get the insurance indexes for other numbers of decks. An example, that more or less shows how it is done for all numbers of decks, is that the double deck index is halfway between the one and infinite deck indexes. A 4 deck index is 1/4 the way toward the one deck index from the infinite deck index.
Now this works if you, instead of the perfect insurance count values, used "effects of removal," from many sources, such as Theory of Blackjack, by Peter Griffin, that are only near perfect, for each decision, and used the minus of his full deck favorability numbers for the perfect count index above. That "52 and n stuff," above is how you adjust for things like soft double indexes where you know 3 exact cards that are removed, and the same with pair splits. You then use the total effects of removal and the n, for the number of cards known. Don't forget to adjust the inner product too.
Full details of algebraic approximation for other indexes can be found best in Arnold Snyder's paper that is still sold by rge I believe and is even somewhat superior to later equations to find such indexes from effects of removal data.
Here's a method you can use that doesn't require the use of simulators. Although my description of this process is rather lengthy, with a little practice you can do the necessary calculations in 5-10 minutes. I'll use the Hi-Lo system and the single-deck game for this example, but this technique will work with just about any system and number of decks.
First, find the earliest point in the deck (or shoe) at which the insurance wager becomes a potentially profitable venture for the player. In the single-deck game, that point is reached when the first 5 cards dealt from the top of the deck are Non-tens. For the six-deck shoe, the earliest point occurs when the first 25 cards dealt from the top of the shoe are Non-tens (incredibly unlikely).
For the single-deck game, compute the average RC and TC for the first 5 cards when we know that they're all Non-tens and we know that at least one of them is an ace (the dealer's up-card). How do we do that? We know that, along with the dealer's ace, the other four cards came from the 36-card set of Non-tens in the single deck. We set the dealer's Ace aside for the moment and concentrate on the remaining 35 cards. What is the average tag-value of each card in the 35-card subset? Using the tag-values for Hi-Lo, we add them up, then divide by 35:
4 + 4 + 4 + 4 + 4 + 0 + 0 + 0 - 3 = 17
17/35 = .4857
So the average tag-value of the cards in the subset of 35 Non-tens is .4857. We can use this figure to compute the average RC and TC of the first 5 cards dealt from the deck:
(4 * .4857) - 1 = .9428[RC]
.9428 * 52/47 = 1.0431[TC]
Notice that we've taken into account the 5th card (the dealer's Ace up-card), shown in the RC calculation with a tag-value of -1. Since this card is the only one of the five whose rank is known with certainty we can use its true tag-value in the calculations. Were it not for the presence of at least one ace (the dealer's up-card), the dealer wouldn't even offer the insurance wager. There may or may not be additional aces appearing in the other 4 cards; their occasional presence there was taken into account when we computed the average tag-value (.4857).
Let's compute the TC for the next two potentially profitable insurance scenarios:
When the first 6 cards dealt from the deck are Non-tens:
(5 * .4857) - 1 = 1.4285[RC]
1.4285 * 52/46 = 1.6148[TC]
When the first 7 cards dealt from the deck are Non-tens:
((6 * .4857) - 1 = 1.9142[RC]
1.9142 * 52/45 = 2.2120[TC]
Now, repeat these calculations at three consecutive points somewhere around the maximum penetration you expect to be able to play. I'll use a few points at mid-deck for this example:
When 26 cards have been dealt and no more than 7 of them are Tens:
(18 * .4857) - 7 - 1 = .7426[RC]
.7426 * 52/26 = 1.4852[TC]
When 27 cards have been dealt and no more than 7 of them are Tens:
(19 * .4857) - 7 - 1 = 1.2283[RC]
1.2283 * 52/25 = 2.5549[TC]
When 28 cards have been played and no more than 7 of them are Tens:
(20 * .4857) - 7 - 1 = 1.714[RC]
1.714 * 52/24 = 3.714[TC]
I determined that the maximum number of Tens allowable in the three subsets of played cards was 7 by looking at each subset's associated number of unplayed cards. With 28 played cards, there is a subset of 24 unplayed cards. Looking at those 24 unplayed cards, insurance is profitable when at least 9 of the 24 are Tens. Subtracting those 9 from the original 16 Tens in the deck leaves us with a maximum of 7 cards in the 28 played cards.
You now have 6 TCs calculated at opposite ends of the playable portion of the deck. You could, if you felt it necessary, add all 6 TCs together, then divide the total by 6 to get an average TC:
1.0431 + 1.6148 + 2.2120 + 1.0804 + 1.4852 + 2.5549 + 3.714 = 13.7044
13.7044 / 6 = 2.2841
Round off the 2.2841 to +2 for your Hi-Lo, single-deck insurance index.
If you care to bother with it, you can perform these calculations at several other points throughout the deck. You'll find the TCs slowly rising (in groups of three) the deeper you go into the deck. The biggest reason for this is the decreasing denominator in the calculations to convert RC to TC. For example, going from 52/47 to 52/46 isn't nearly as big a leap as going from 52/25 to 52/24.
During the final calculation to determine an average for the 6 TCs, I inadvertently tossed in a 7th figure:
1.0431 + 1.6148 + 2.2120 + 1.0804+ 1.4852 + 2.5549 + 3.714 = 13.7044
13.7044 / 6 = 2.2841
Obviously, the correct calculations should be:
1.0431 + 1.6148 + 2.2120 + 1.4852 + 2.5549 + 3.714 = 12.624
12.624 / 6 = 2.104
