 The following gets into how combinatorial analysis could be used to analyze any counting system and its limitations.

## Accounting for all possible subsets

In a full shoe there is only 1 possible shoe composition. For example a full single deck's composition consists of (ranks 2 through ace) {4,4,4,4,4,4,4,4,4,16,4}. Suppose 1 card is now randomly dealt from a full single deck leaving 51 cards remaining. There would be 10 possible shoe compositions that could result from this. The probability of each subset would be 1/13 if the card dealt was a non-ten (2 through 9 or ace) and 4/13 if it was a ten value card (ten, jack, queen, or king.) If 2 cards were randomly dealt there would be 55 possible subsets, each with its own probability. If 3 cards were randomly dealt there would be 220 subsets and so on. The maximum number of possible subsets occurs when half of the number of cards in the shoe remains. For a single deck this is when 26 cards have been randomly dealt. This results in 1868755 possible subsets, each with its own probability of occurrence. Listed here are the number of possible distinct random compositions for 1 or 2 decks when a given number of cards remain to be dealt and the cards that have been previously dealt have been exposed and are known. As can be seen, the number of possible subsets for a 2 deck shoe can be much larger than for a single deck with the maximum number of 375268773 ocurring at 52 cards remaining (half shoe.) As the number of starting decks for a full shoe increases, the number of possible subsets will tend to grow exponentially to a number that could not be stored in a 32-bit integer. If stored in a 64-bit integer then data may be displayed for more than 2 decks but I'm not sure of the maximum number of starting decks before an overflow of a 64-bit integer would occur for such data.

## Computing overall EV for a given number of cards remaining when all dealt cards may be any random card

It can be seen from above that randomly removing a given number of cards from a full shoe results in a number of possible subset compositions and each subset composition has its own probability of occurrence. Let's define EV_cardsRemoved = sum of the computed overall EV of each subset multiplied by its probability of occurrence. The overall EV of each individual subset composition is dependent upon player's strategy. If player strategy is to use full shoe composition dependent basic strategy it turns out that EV_cardsRemoved is equal to full shoe (basic strategy) EV (= +0.000247914 for a single deck) when all possible random combinations for any given number of cards removed is considered. This makes sense and I have shown it to be true for the resultant subsets of 1, 2, 3, and 4 cards randomly removed from a single deck (10, 55, 220, and 715 subsets respectively.) If best strategy is used instead of basic strategy then EV_cardsRemoved is greater than full shoe (basic strategy) EV. Below are the (composition dependent) EVs for single deck with rules of s17, bj pays 3:2, full peek, double any 2 cards, 1 split allowed for any pair, 1 card to split aces, no double after split, no surrender. Listed are EVs for full shoe and 1, 2, 3, and 4 cards randomly removed and exposed:
full shoe (0 cards removed, 52 cards remain): 0.000247914
(1 card removed, 51 cards remain): 0.000499114
(2 cards removed, 50 cards remain): 0.000738031
(3 cards removed, 49 cards remain): 0.000997096
(4 cards removed, 48 cards remain): 0.00126127
The difference between the above values and the full shoe EV is the maximum overall gain possible (for any counting system) given 1, 2, 3, or 4 cards have been randomly dealt and exposed from a single deck. Additionally it would be possible to enumerate EV for any counting system for all possible running counts (and their probabilities) for each cards removed/cards remaining value, including any optional side count data that may be chosen to be included, by parsing each subset relative to the counting system being analyzed. Above data is for up to 4 random cards removed from a single deck and there are 715 subsets possible when 4 cards are removed. If 26 random cards are removed from a single deck there are 1,868,755 subsets possible. Out of the 1868755 possible subsets, listed below are the number of possible random subsets for a running count of 0 for 4 different counting systems for a single deck with 26 cards removed/26 cards remaining. In each case an EV for a running count of 0 could be computed using basic strategy, optimal strategy, or any other enumerated strategy as could EV for any other possible running count. The uncounted ranks (ranks with a zero tag) are not side counted in any of this data.

```	Number of subsets where running count = 0, single deck
HiLo (tags 2 through ace) = {-1,-1,-1,-1,-1,0,0,0,1,1} 112695
Tarzan (tags 2 through ace) = {-1,-1,-1,-1,0,0,0,0,1,0} 117115
Wong Halves (doubled) (tags 2 through ace) = {-1,-2,-2,-3,-2,-1,0,1,2,2} 54685
Ace-Five (tags 2 through ace) = {0,0,0,-1,0,0,0,0,0,1} 364179
```

It may seem that the only limitation to computing the maximum overall EV gain possible and also EV for each running count of a counting system for any number of random cards removed would be the great amount of computing time needed to process a large number of subset compositions. This would be true if a given number of cards were exposed and burned (from a full shoe) prior to a round being dealt. However, there is a complication. In a normal blackjack game some subsets may not be possible since a round must be completed before a new round can begin and the next few random cards to be dealt at the beginning of a new round may not be sufficient to complete a blackjack round. For example in a heads up game of 1 player versus dealer a new round with a subset composition of 48 cards remaining where 4 fives have been removed is not possible. This would correspond to a hand where player is dealt 5-5 and dealer is also dealt 5-5. Even if player adopts a strategy of standing on 5-5, dealer must draw to his 5-5, so a new round with this subset is not possible in the normal course of a blackjack game of 1 player versus dealer. (However, player's strategy for 5-5 would not be to stand anyway.) In fact there are many subsets consisting of 4 removals, 48 cards remaining that could not occur in the normal course of such a game. The 48 card subsets that are possible to begin a new round in a heads up game will consist mostly of compositions where cards that benefit player are predominantly removed, whereas the subsets that are not possible consist of removals that predominantly favor player. Therefore the overall EV for 48 cards remaining using basic strategy when the 4 dealt cards must comprise a blackjack round, unlike the case where all random subset compositions are possible, is less than the full shoe (basic strategy) EV for a heads up game. In general when any number of random cards are exposed and removed and a new round is dealt, EV using basic strategy is constant but when the exposed and removed cards must have completed a finished blackjack round, EV using basic strategy is variable.

Now suppose a dummy card can be placed at some point in the shoe and the only purpose of this card is to force a reshuffle if it is encountered. Further, suppose this dummy card is placed after the fifth card in a full shoe. In that case the dummy card will not be encountered in a heads up game if the first round of the shoe consists of exactly 4 cards, and a second round will be dealt. As was stated above, the possible 48 card subsets resulting from 4 cards removed that can complete a round in a heads up game will yield an overall EV that is less than the full shoe EV for the second round. Note that it is possible for 5, 6, 7,...., maxNumCards to be dealt after the dummy card is encountered (where maxNumCards represents the greatest number of cards that might be needed to complete a round) and some of these subsets may be favorable to player if a second round is dealt. However, the presence of the dummy card, called a cut card causes any favorable subsets to be shuffled away. This type of reduction in overall EV has been referred to as the cut card effect. This example shows how a loss of EV occurs to player in a heads up game, with player playing all rounds, using total dependent basic strategy when a cut card is placed after the fifth card and cards are dealt from a full single deck. Player's EV for the first round is always the total dependent basic strategy EV for this game. However, if the cut card is not encountered a second round will be dealt where player's EV is much less than total dependent basic strategy EV. In this example the cut card will not be encountered when the round consists of exactly 4 cards. In other words both player and dealer must not draw any cards for a second round to occur. In general the overall EV for a clump of random cards using total dependent basic strategy as above is the same basic full shoe EV as above for the first round dealt whether a cut card is encountered or not. Any potential loss of EV due to cut card effect occurs in either the next or a subsequent round.

So what happens when the cut card is in another position other than after the fifth card? In order to try understand this let's consider another scenario. Suppose the timing of a reshuffle is not determined by a cut card but instead occurs after exactly 2 rounds (still a heads up game) have been dealt. In that case there would be no cut card to force the shuffling away of possible favorable subsets after the first round. Although some second round subsets would be unfavorable to player, others would be favorable and on balance no EV is lost for the second round like when a cut card can cause the shuffling away of some favorable subsets. Therefore if the cut card is sufficiently deep that there is no possiblity that it will be encountered in the first round then neither the first round nor the second will have a cut card effect. (There is simulation data in the next section below (Computing overall EV when all dealt cards must comprise a completed blackjack round,) that supports this conclusion.) In general, player can eliminate cut card effect altogether simply by making the last round of a shoe the first round there is any possiblity that the cut card will be encountered by wonging out following this round (whether or not the cut card is actually encountered.) If player is not counting cards, wonging out at at this point would be the best overall strategy. However, if player is counting cards he may determine that it may be worth the risk of encountering the cut card because it is sufficiently deep that it may not be encountered and if not a possible favorable count could occur and be exploited. Also if a favorable count already exists playing at least one more round would be best even if the cut card could have been encountered during the previous round but it turned out that it didn't actually appear, allowing the dealing of at least one more round.

## Computing overall EV when all dealt cards must comprise a completed blackjack round

It was shown earlier that when any number of cards are randomly removed and exposed from a shoe a basic strategy player is playing at a fixed EV regardless of the number of cards removed and a player using optimal strategy is playing at an EV greater than the basic strategy EV. Optimal strategy increases as more cards are seen while basic strategy EV remains fixed because basic strategy is a fixed strategy. This is in the case where all random possibilities may occur. However, in a normal blackjack game a round does not end just because a number of random cards have been exposed. For the round to end enough cards to complete 1 or more players' hands given the strategy of each player and also dealer's hand must be dealt and the overall EV for the next round cannot be computed until this happens. Therefore in order to determine all of the possible subsets and their probabilities we'd need to go through all of the possible player(s)/dealer drawing sequences. The possible subsets would be dependent on player strategy and also the possibility of encountering a cut card. Whereas in the first case basic strategy EV is fixed, in the second basic strategy EV is variable. In the second case the only time basic strategy EV is equal to full shoe basic strategy EV is when a random clump of cards is the starting point whereas in the first case basic strategy EV is always equal to the full shoe basic strategy EV. In the second case the number of subsets is limited to ones where the removals can comprise a full blackjack round where each blackjack round could be completed in differing ways, which would affect the probabilities of subsets for ensuing rounds. Data from the link in the above example shows that exact overall EV can be computed by combinatorial analysis given a heads up game of 1 player versus dealer and a cut card placed after the fifth card where total dependent basic strategy is used. (Composition dependent basic strategy could have been used instead to illustrate the effect.) In the example dealer hits soft 17. EV using basic strategy for the first round is -.2051%. EV for the second round, which can occur only if the cut card is not encountered in the first round is -1.4661% and a second round can occur about 17.01% of the time. These figures are derived by going through all of the possible drawing sequences that can comprise a full round with a cut card that follows the fifth card from the top of a full single deck. Overall EV is figured using basic strategy for each set of removals weighted by their probability of occurrence. As you can see, EVs for the first and second round are quite different.

It may be possible to compile all of the drawing sequences and their probabilities for any cut card position. Such a compilation would depend on number of players as well as the strategy employed by each player. Therefore overall EV for the case where all cards removed needs to have completed 1 (or more) blackjack round(s) is for a very specific condition. For example overall EV for 2 players when the cut card follows the fifth card and player uses total dependent basic strategy, as in the example, is -.2051% whereas for 1 player it is -.4545%. This is because the cut card will always be encountered in the first round for 2 players and there will never be a second round. With 1 player a second round will be dealt (at a greatly reduced EV) about 17% of the time. Rather than trying to come up with an algorithm for drawing sequences I have decided to run several simulations for the specific condition of 1 player using composition dependent basic strategy versus dealer with a cut card placed at varying positions in order to explore how overall EV can be affected by cut card position. Modest simulations of a reshuffle of 10,000,000 shoes each with a cut card following cards 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15 from the top were run recording separate results for rounds 1 and 2 as well as total results using composition dependent basic strategy. The default set of rules used in my composition dependent programs is used (dealer stands on soft 17, blackjack pays 3:2, full peek, double any 2 cards, all pairs can be split once, no double after splitting, 1 card to split aces, no surrender.) When the cut card follows card 4 from the top no second round is possible and the EV of the sim should agree with basic strategy EV. When the cut card follows cards 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 from the top a second round is possible but EV of the first round should also agree with basic strategy EV. It is the EV of the second round that is of interest. The result of the sim where the cut card followed the fifth card was that a a first round was dealt 10,000,000 times followed by a second round 1,765,056 times. When there was no second round a shuffle occurred after the first round otherwise a shuffle occurred after the second round. Sim EV for round 1 was 0.0000051 = .00051% (actual computed CD basic strategy EV is 0.0002479 = .02479%.) Sim EV for round 2 was -0.0117010 = -1.17010%, which shows how the cut card affects round 2 EV. The overall EV of the sim is the combined EV of rounds 1 and 2 (-0.0017511 = -.17511%.) The result of the sim where the cut card followed the sixth card was that a first round was dealt 10,000,000 times followed by a second round 5,740,191 times. Sim EV for round 1 was 0.0003043 = .03043% (actual computed CD basic strategy EV is 0.0002479 = .02479%.) Sim EV for round 2 was -0.0047776 = -.47776%, which shows that cut card effect when the cut card follows the sixth card is less than when it follows the fifth card. The overall EV of the sim is the combined EV of rounds 1 and 2 (-0.0015490 = -.15490%.) As the cut card is moved so that it follows the seventh, eighth, etc. card the respective sims show that in general EV for round 2 approaches a value that is pretty close to the known computed EV of composition dependent basic strategy. Also the overall EV for all rounds starts decreasing for the sims where the cut card follows more the the eighth card position. This is because if more than 8 cards can be dealt in a heads up game it is possible that more than 2 rounds can be dealt, causing a new cut card effect for the next round. When the cut card followed the fifteenth card there were 10,000,000 rounds for both the first and second round meaning a second round was always dealt while the trend continued that EV for the second round is reasonably close to basic strategy EV. Actually it appears that EV for the second round tends to slightly exceed basic strategy EV when the cut card follows the 10th, 11th, 12th, 13th, and 14th cards possibly indicating the existence of a floating advantage for the second round. Floating advantage is a term used to describe the tendency of the increasing of player EV as the number of remaining cards decreases. In any event it appears that if the cut card follows no less than the 15th card a player in a heads up game using basic strategy is guaranteed 2 rounds where EV is at minimum the basic strategy EV with no cut card effect for either round. Also possibly player's EV may be at least basic strategy EV if the cut card follows no less than the 10th card due to floating advantage if the trend of the sims is correct (although there is probably insufficient data to conclude this.) These sims were generated using the random number generator of C++. They are intended to show trends and not necessarily super accurate results. The parameter to best measure the effectiveness of the sims is EV for round 1, which is known to be +.02479% from combinatorial analysis. There were 12 sims of 10,000,000 shuffles so there were 120,000,000 round 1 rounds and the average EV was +.024708%, which turned out to be very close to the actual value. (Basic strategy EV for the listed rules is very close to 0, which is an even game.) The results of these sims are listed here.

## Computing EVs and strategies for individual hands relative to a given counting system using combinatorial analysis

Above it was shown that overall EV is not a good candidate to be computed using combinatorial analysis. Overall EV is only computed at the end of a finished blackjack round. This eliminates some random composition subsets that can occur before the round ends and changes the probabilities of the subsets that remain. However, computing EVs for an individual hand does not require that a round be completed and all random subsets and their probabilities can be used to do this. There may be a lot of subsets to compute, but with enough computing time an individual hand can be analyzed relative to any given count. This is an example of a player hand of T-6 dealt from a single deck with 4 cards removed/48 cards remaining using the HiLo counting system. The data shows that a stand, hit, and double EV can be computed for each possible HiLo running count for a player hand of T-6 versus each up card. Note that even for a running count (RC) of +3 (which would correspond to a true count of more than +3 since 45 cards remain to be dealt at the point the RC is recorded) the EV for hit is greater than EV for stand for an up card of 10. This may surprise a HiLo player using index plays generated from simulation software, where standing would be recommended for a count greater than 0 for a single deck. The EVs are computed by going through all of the possible subsets, computing a HiLo running count for each, and computing a weighted average for each possible RC. There are some subsets where standing on T-6 versus an up card of 10 would be better than hitting, but on balance hitting is the best option at any HiLo count with 4 random cards removed and exposed (in addition to the cards in player's hand (T-6) and dealer's up card.) The results take into consideration the composition of player's hand (T-6) plus dealer's up card whereas simulation software most likely would only account for the removal of the up card. This is like assigning player a hand total of hard 16, removing an up card, proceeding by removing additional cards by simulation, and compiling results for not only 4 additional cards removed but any number of cards removed. Simulation software most likely would not account for player's hand composition and would treat all numbers of cards removed with equal weight.

The number of cards parameter is essential in computing the probability of each individual subset which is used to weight results obtained by combinatorial analysis in order to get the displayed EVs. It can be shown by example that the number of cards parameter can affect EVs and strategies to a very noticeable extent. Suppose the Tarzan Count is being used. The tags for this count are (2 through ace) {-1,-1,-1,-1,0,0,0,0,+1,0}. (The +/- of the tag here is defined as relative to what remains in the shoe. If the +/- is defined as what has been removed then the sign of the tags would be reversed. However the +/- of the tags is defined does not change results in any way.) Suppose further that the ranks tagged as 0 are exactly side counted and it is known that no 0 tagged ranks at all remain to be dealt and that the current running count (RC) is 0. So far all that is know with this information is that the number of ranks with tag -1 is equal to the number of ranks with tag +1 and there are no ranks with tag = 0 present, a condition that could occur with many differing numbers of cards remaining to be dealt. Suppose player has a hand of hard 16 versus a dealer up card of 10 and needs to make a decision of what to do with RC = 0 and no 0 tagged cards remaining to be dealt. The EVs/strategy depends upon number of cards remaining. In order to display this we will look at compositions such that the condition in question is encountered with a varying number of cards. Inputs will be made into my combinatorial analyzer that uses an exact shoe composition as input to compute EVs. 3 shoe compositions with the same running count/side count info will be considered: 2 cards remaining, 24 cards remaining, and 16 cards remaining. Only in the case of 2 cards remaining is an exact EV computed. The other 2 cases use a representative composition. A representative composition is one and only one composition used to compute (approximate) EVs for a given running count, number of cards, and side counted cards used to simply get a reasonable approximation of EV by doing only 1 calculation. Actual EVs would be a composite computation of many compositions that satisfy the conditions of a given running count, number of cards, and side counted cards. The probability of drawing each rank in a representative composition should be exactly equal or very close to the probability of drawing each rank computed from a composite of all possible compositions that satisfy the above conditions. The first set of screen shots shows input of a composition of 3 tens, 1 six, and either a 2, 3, 4, or 5 depending upon the particular screen shot and a player hand input of 6-10. The desired composition is input by removing all but the above 5 cards from a single deck. When the analyzer computes it deals the player cards (6-10) and cycles through all of the possible up cards, one of which is a 10. When the up card is 10, 2 tens and one six are used leaving one 10 and either a 2, 3, 4, or 5 (depending upon screenshot) remaining to be dealt. The probability that the last low card is a 2, 3, 4, or 5 is 1/4 = 25% for each. Therefore the EV for hitting is 1/4 * (-100% - 100% - 50% + 0%) = -62.5%. The EV for standing is 0% (even bet) no matter what the rank of the last low card is, so standing is much better than hitting and of course doubling is much worse than hitting still. Screenshots for hard 16 versus up card of 10 - Tarzan RC = 0, 2 cards remain to be dealt with no 6, 7, 8, 9, ace remaining. Next we'll look at the case for hard 16 versus up card of 10 when Tarzan RC = 0 and 24 cards remain to be dealt with no 6, 7, 8, 9, ace remaining. 1 deck is input into the combinatorial analyzer. We remove 1 each of ranks 2, 3, 4, and 5 and remove 2 of rank 10. We remove all but one 6 and remove all 7, 8, 9, and ace and input 6-10 for player's hand. When player hand of 6-T is dealt from this composition player has hard 16 and when up card is 10 any cards drawn are from a composition of {3,3,3,3,0,0,0,0,12,0} (2 through ace.) Here EV for standing is -38.28% and EV for hitting is -36.56%, so hitting is the better option. Screenshot for hard 16 versus up card of 10 - Tarzan RC = 0, 24 cards remain to be dealt with no 6, 7, 8, 9, ace remaining. We can see that standing is by far the best option for 16 versus an up card of 10 for a Tarzan running count of 0 with zero 6, 7, 8, 9, ace possible when 2 cards remain and hitting is the best option for the same condition with the exception that 24 cards remain. It follows that there should be some number of cards between 2 and 24 where hitting and standing are approximately equal options. Screenshot for hard 16 versus up card of 10 - Tarzan RC = 0, 16 cards remain to be dealt with no 6, 7, 8, 9, ace remaining. (Here I created the composition starting with 2 decks but it could have been created starting with 1 deck as well.) This shows that EV for hitting is slightly better than EV for standing for this condition but the EVs for hitting and standing are very close to equal.

The point of the above was to show an example of the effect that number of cards remaining to be dealt has for a given counting system. Inputs were made into my combinatorial analyzer that uses input of an exact composition. An exact result was obtained for the case of 2 cards remaining by computing results for all possibilities. Approximate results for the other 2 cases were obtained by using representative subsets. However, it may be possible to get exact results for all cases by computing rank probabilities in a way that is count dependent.

## What all (rank tagged) counting systems have in common

All counting systems separate the 10 possible ranks into 2 or more groups and assign each group a differing tag value. We'll look at 5 different count systems as examples. Each counting system has an initial running count (IRC) = (sum of (cards per deck for each group)*(tag of group))*(number of decks used)

```Hilo:
group 1 {2,3,4,5,6} tag = -1, group 1 cards per deck = 20
group 2 {7,8,9} tag = 0, group 2 cards per deck = 12
group 3 {T,A} tag = +1, group 3 cards per deck = 20
IRC = ((-1)*20+(0)*12+(+1)*20)*(number of decks used) = (0)*(number of decks used) = 0
KO:
group 1 {2,3,4,5,6,7} tag = -1, group 1 cards per deck = 24
group 2 {8,9} tag = 0, group 2 cards per deck = 8
group 3 {T,A} tag = +1, group 3 cards per deck = 20
IRC = ((-1)*24+(0)*8+(+1)*20)*(number of decks used) = (-4)*(number of decks used)
Insurance Count:
group 1 {A,2,3,4,5,6,7,8,9} tag = -1, group 1 cards per deck = 36
group 2 {T} tag = +2, group 2 cards per deck = 16
IRC = ((-1)*36+(+2)*16)*(number of decks used) = (-4)*(number of decks used)
Wong Halves (doubled):
group 1 {2,7} tag = -1, group 1 cards per deck = 8
group 2 {3,4,6} tag = -2, group 2 cards per deck = 12
group 3 {5} tag = -3, group 3 cards per deck = 4
group 4 {8} tag = 0, group 4 cards per deck = 4
group 5 {9} tag = +1, group 5 cards per deck = 4
group 6 {T,A} tag = +2, group 6 cards per deck = 20
IRC = ((-1)*8+(-2)*12+(-3)*4)+(0)*4+(+1)*4+(+2)*20)*(number of decks used) = (0)*(number of decks used) = 0
Tarzan:
group 1 {2,3,4,5} tag = -1, group 1 cards per deck = 16
group 2 {6,7,8,9,A} tag = 0, group 2 cards per deck = 20
group 3 {T} tag = +1, group 3 cards per deck = 16
IRC = ((-1)*16+(0)*20+(+1)*16)*(number of decks used) = (0)*(number of decks used) = 0
```

Assuming we start with a full shoe of n number of decks consisting of 52*n number of cards, the probability of drawing any non-ten rank (A,2,3,4,5,6,7,8,9) = 1/13 = .0769231 and the probability of drawing a ten value card = 4/13 = .307692. We will randomly remove exactly half of the cards in the shoe so that the number of cards remaining = 26*n and look at only the subsets where running count (RC) = IRC/2. It turns out that for this condition for any counting system that the probabilities for drawing any rank exactly equal the probabilities of drawing any rank for a full shoe. Also, half shoe with RC = IRC/2 is the only condition where the rank probabilities are the same as they are for a full shoe. I will use a program I've written that computes rank probabilities for any counting system to display this for the above counting systems for 1, 2, and 8 decks. The program computes the probability of each possible counting system subset which is then used to compute the probability of drawing each rank given what has been input. It can be seen that for a half shoe and a running count equal to half the count's initial running count that the rank probabilities for any counting system equal those of a full shoe for any number of decks. For any other number of cards other than half shoe (= 26*n cards) or any other RC other than IRC/2 for any counting system the rank probabilities computed by the program are not identical to a full shoe's rank probabilities. There is a special case that may be of some interest that occurs for counts such as HiLo and the Tarzan Count. What these counts have in common is that there are the same number of cards equally tagged as positive and negative and the IRC for them = 0. For systems such as these when half shoe (= 26*n cards) remains the probability for drawing any rank with a 0 tag = 1/13 for any possible running count (and this is only true when 26*n cards remain.)

The output below may look redundant but the program goes through all of the possible count subsets to compute the rank probabilities and they all compute to the same thing for each of the input conditions. The point is that this will be true for any counting system and any number of decks. The only variable outputs are the number of count subsets and the probability of the input running count.

```HiLo:
1 deck, RC = (0)/2 = 0, cards remaining = 1*26 = 26
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 7
Prob of running count 0 with above removals from 1 deck: 0.124165

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
2 decks, RC = (0)/2 = 0, cards remaining = 2*26 = 52
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 13
Prob of running count 0 with above removals from 2 decks: 0.0885002

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
8 decks, RC = (0)/2 = 0, cards remaining = 8*26 = 208
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 49
Prob of running count 0 with above removals from 8 decks: 0.0445147

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
KO:
1 deck, RC = (-4)/2 = -2, cards remaining = 1*26 = 26
Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): -4
Running count: -2
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 5
Prob of running count -2 with above removals from 1 deck: 0.11886

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
2 decks, RC = (-8)/2 = -4, cards remaining = 2*26 = 52
Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): -8
Running count: -4
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 9
Prob of running count -4 with above removals from 2 decks: 0.0846975

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
8 decks, RC = (-32)/2 = -16, cards remaining = 8*26 = 208
Count tags {1,-1,-1,-1,-1,-1,-1,0,0,1}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): -32
Running count: -16
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 33
Prob of running count -16 with above removals from 8 decks: 0.0425947

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
Insurance Count:
1 deck, RC = (-4)/2 = -2, cards remaining = 1*26 = 26
Count tags {-1,-1,-1,-1,-1,-1,-1,-1,-1,2}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): -4
Running count: -2
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 1
Prob of running count -2 with above removals from 1 deck: 0.235516

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
2 decks, RC = (-8)/2 = -4, cards remaining = 2*26 = 52
Count tags {-1,-1,-1,-1,-1,-1,-1,-1,-1,2}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): -8
Running count: -4
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 1
Prob of running count -4 with above removals from 2 decks: 0.168019

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
8 decks, RC = (-32)/2 = -16, cards remaining = 8*26 = 208
Count tags {-1,-1,-1,-1,-1,-1,-1,-1,-1,2}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): -32
Running count: -16
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 1
Prob of running count -16 with above removals from 8 decks: 0.084571

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
Wong Halves (doubled):
1 deck, RC = (0)/2 = 0, cards remaining = 1*26 = 26
Count tags {2,-1,-2,-2,-3,-2,-1,0,1,2}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 281
Prob of running count 0 with above removals from 1 deck: 0.0591537

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
2 decks, RC = (0)/2 = 0, cards remaining = 2*26 = 52
Count tags {2,-1,-2,-2,-3,-2,-1,0,1,2}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 3091
Prob of running count 0 with above removals from 2 decks: 0.0421775

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
8 decks, RC = (0)/2 = 0, cards remaining = 8*26 = 208
Count tags {2,-1,-2,-2,-3,-2,-1,0,1,2}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 582905
Prob of running count 0 with above removals from 8 decks: 0.0212199

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
Tarzan:
1 deck, RC = (0)/2 = 0, cards remaining = 1*26 = 26
Count tags {0,-1,-1,-1,-1,0,0,0,0,1}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 11
Prob of running count 0 with above removals from 1 deck: 0.138595

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
2 decks, RC = (0)/2 = 0, cards remaining = 2*26 = 52
Count tags {0,-1,-1,-1,-1,0,0,0,0,1}
Decks: 2
Cards remaining: 52
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Number of subsets for above conditions: 21
Prob of running count 0 with above removals from 2 decks: 0.0988675

p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231
p 0.0769231  p 0.0769231  p 0.0769231  p 0.0769231  p 0.307692
8 decks, RC = (0)/2 = 0, cards remaining = 8*26 = 208
Count tags {0,-1,-1,-1,-1,0,0,0,0,1}
Decks: 8
Cards remaining: 208
Initial running count (full shoe): 0
Running count: 0
Subgroup removals: None
Specific removals (1 - 10): {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}