When the exact number of each rank of a blackjack shoe are known, an exact calculation of EVs can be made by using the probabilities of drawing any rank for different shoe states. When a player counts cards he knows 2 things: the running count of his particular counting system and the (approximate) number of cards remaining to be dealt. This is less information than is known for the case when the exact numbers of each rank are known. A counter may have access to additional information if he side counts individual ranks or groups of ranks.
Probability of drawing a given rank for an exactly known shoe composition
Computing the probability of drawing a given rank when the exact shoe composition is known is straight forward. Since the number of each rank present in the shoe is known the total number of cards is also known and is equal to the sum of the numbers of each rank present. The probability of drawing any given rank is the number of that rank present divided by the total number of cards. After one of a given rank is drawn, the number of that rank present is decreased by 1 and also the total number of cards present is decreased by 1. After one of a given rank is drawn, the probability of drawing another of the same rank is less than it was before the rank was drawn and the probability of drawing any other rank (as long as the number of this other rank is greater than 0) is more than it was prior to the given rank being drawn.
Probability of drawing a given rank for a given counting system
I have written 2 programs to better analyze counting systems. One takes as input number of decks and number of cards and is capable of outputting every random subset and its probability for the input number of decks/number of cards. The limitation of this program is that there can be an extremely large number of random subsets. The other program, which I will describe in this section, takes as input the tags of any counting system, number of decks, optional side counts of subgroups with the same tag value, number of cards remaining to be dealt, and number of cards of each rank that are known to be specifically removed from the shoe. It outputs the number of random subsets for the given input, probability of the running count that was input, and the probability of drawing each rank in the shoe (for the given input.) There are fewer subsets as cards are grouped by tag value and there are the same number of groups as there are different tag values in the counting system. A complicated count like Wong Halves has a maximum of 582,905 subsets for 8 decks. A simpler count such as HiLo has a maximum of 49 subsets for 8 decks. The limitation of the program is that the subsets cannot be directly input into a combinatorial analyzer that uses an exact shoe composition as input. In its present form this program could be used to analyze when to take insurance for any counting system. (It was used to output the insurance data for HiLo and KO counting systems on this website.) I'll use an example of computing insurance EV to show how the program works. Whenever the probability of drawing a ten value card is greater than 1/3 = .3333333..., insurance EV is positive. Below is output for Wong Halves for the condition of 39 cards remaining dealt from 2 decks with 1 ace, 1 two, and 1 ten specifically removed for running counts of +6, +5 and +4. (No ranks are side counted.) Since the program requires that tag values be entered as whole numbers the Wong Halves tag values are doubled and would exactly correspond to a Wong Halves running count of +3, +2.5, and +2. The specific removals account for a dealer up card of ace and a player hand of 10-2. A player hand of 10-7 would yield the same result relative to insurance EV since in Wong Halves the tag of the 2 and 7 are the same. (Note a ten and ace also have the same tag but for insurance purposes these ranks are not equivalent.) At a running count of +6 probability of drawing a ten = .340475 and insurance EV = 3*.340475-1 = +2.1425%. At a running count of +5 probability of drawing a ten = .335719 and insurance EV = 3*.335719-1 = +0.7157%. At a running count of +4 probability of drawing a ten = .330972 and insurance EV = 3*.330972-1 = -0.7084%. This indicates that with 39 cards remaining dealt from 2 decks insurance is positive EV for a running count greater than or equal to +5 using Wong Halves (doubled) for a hand of T-2 or T-7 given up card is an ace. Notice that all that is considered is running count. (We would have the option of later categorizing results by true count = 52*(running count)/(number of cards remaining) after all data has been enumerated.)
RC = +6 Count tags {2,-1,-2,-2,-3,-2,-1,0,1,2} Decks: 2 Cards remaining: 39 Initial running count (full shoe): 0 Running count: 6 Subgroup removals: None Specific removals (1 - 10): {1, 1, 0, 0, 0, 0, 0, 0, 0, 1} ....computing, please wait Number of subsets for above conditions: 2585 Prob of running count 6 with above removals from 2 decks: 0.0323179 p[1] 0.0768814 p[2] 0.0657327 p[3] 0.0708072 p[4] 0.0708072 p[5] 0.0662731 p[6] 0.0708072 p[7] 0.0751231 p[8] 0.0794413 p[9] 0.083652 p[10] 0.340475 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: RC = +5 Count tags {2,-1,-2,-2,-3,-2,-1,0,1,2} Decks: 2 Cards remaining: 39 Initial running count (full shoe): 0 Running count: 5 Subgroup removals: None Specific removals (1 - 10): {1, 1, 0, 0, 0, 0, 0, 0, 0, 1} ....computing, please wait Number of subsets for above conditions: 2624 Prob of running count 5 with above removals from 2 decks: 0.0350046 p[1] 0.0758076 p[2] 0.0662668 p[3] 0.0719834 p[4] 0.0719834 p[5] 0.0680028 p[6] 0.0719834 p[7] 0.0757335 p[8] 0.0794591 p[9] 0.0830607 p[10] 0.335719 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: RC = +4 Count tags {2,-1,-2,-2,-3,-2,-1,0,1,2} Decks: 2 Cards remaining: 39 Initial running count (full shoe): 0 Running count: 4 Subgroup removals: None Specific removals (1 - 10): {1, 1, 0, 0, 0, 0, 0, 0, 0, 1} ....computing, please wait Number of subsets for above conditions: 2655 Prob of running count 4 with above removals from 2 decks: 0.0374578 p[1] 0.0747357 p[2] 0.0667957 p[3] 0.073159 p[4] 0.073159 p[5] 0.0697436 p[6] 0.073159 p[7] 0.0763379 p[8] 0.0794711 p[9] 0.0824667 p[10] 0.330972 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:
As you can see the program computes the probability of drawing each rank for a condition that is count dependent. In order to compute insurance EV all that is needed is the probability of drawing a ten value card. However, it would be possible to replace the computation of rank probabilities in my combinatorial analyzer that uses an exact shoe composition with one where rank probabilities are count dependent like in the above example and the result would be a combinatorial analyzer that outputs EVs relative to a given counting system. Any such combinatorial analyzer would use all possible random count subsets. The EVs of any individual hand could be computed for any counting system defined condition similar to the ones defined in the insurance example. Overall EV would not be as good a candidate for computation since it requires a completed blackjack round before it can be computed. This means that some random subsets could be eliminated and the probabilites of the ones that remain could be changed. Therefore the best approach using combinatorial analysis would probably be to get individual hand strategies by combinatorial analysis and to get overall EV by simulation of these strategies.