Originally Posted by
Secretariat
The 88vT situation is just one example.
So 88vT at TC0 with a ratio of 789/Tens lower than 25% will probably happen more often than TC+11
This is something of a check for your hypothesis that num_789/num_T = .25 more often than TC = 11. It's one example from single deck for 33 cards remaining with 8-8 and T removed to account for hand of 8-8 vs ten.
First I record all subsets and parse them according to num_A, num_789, num_T. Note that if you have this info along with number of cards remaining, RC can be determined (RC = 2 * (num_A + num_T) + num_789 - cards remaining). (RC could be input instead of computing. It would make no difference.) Then num_A, num_789, and num_T are input and number of subsets for this condition and the probability are output from recorded data.
Code:
Decks: 1
Count system: HiLo
Cards remaining: 33
Cards removed (number of cards ranks 1 to 10):
0, 0, 0, 0, 0, 0, 0, 2, 0, 1
Total subsets: 529774
Display info for given num aces, num 7-8-9, num tens
given 33 cards remain with above removals
Enter number of aces: 3
Enter number of 7-8-9: 3
Enter number of tens: 12
HiLo RC: 0
Total Prob RC = 0: 0.130432
Number of subsets (0, 3, 3, 12): 1089
Probability (0, 3, 3, 12): 0.00101134
Next I use my count enumeration program to compute RC = 7 with 33 cards remaining and 8-8, T removed. This computes to a TC of about 11.
Code:
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 1
Cards remaining: 33
Initial running count (full shoe): 0
Running count: 7
Specific removals (1 - 10): {0,0,0,0,0,0,0,2,0,1}
Subgroup removals: None
Number of subsets for above conditions: 5
Prob of running count 7 with above removals from 1 deck: 4.51675e-003
p[1] 0.10524 p[2] 0.057553 p[3] 0.057553 p[4] 0.057553 p[5] 0.057553
p[6] 0.057553 p[7] 0.084941 p[8] 0.04247 p[9] 0.084941 p[10] 0.39465
In the first case probability = 0.00101134.
In the second case probability = 0.00451675.
This one example would show prob(TC=11) > prob(num_789/num_T=.25)
Hope this helps,
k_c
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