Quote Originally Posted by k_c View Post
I don't know if I will explain this very well.

I try to test my methods by trying them on a problem with an obvious answer. Use insurance count (tag non-tens as -1 and tens as +2) (sign of tag relative to what remains in shoe) as the basis for a question.

For this count this is data listing min RC values for single deck where Prob(ten) >= 1/3, which we should agree on.
Code:
Count tags {-1,-1,-1,-1,-1,-1,-1,-1,-1,2}
Decks: 1
Insurance Data (without regard to hand comp)
No subgroup (removals) are defined

**** Player hand: x-x ****
Cards   RC      TC ref

48      0       0.00
47      1       1.11
46      2       2.26
45      0       0.00
44      1       1.18
43      2       2.42
42      0       0.00
41      1       1.27
40      2       2.60
39      0       0.00
38      1       1.37
37      2       2.81
36      0       0.00
35      1       1.49
34      2       3.06
33      0       0.00
32      1       1.63
31      2       3.35
30      0       0.00
29      1       1.79
28      2       3.71
27      0       0.00
26      1       2.00
25      2       4.16
24      0       0.00
23      1       2.26
22      2       4.73
21      0       0.00
20      1       2.60
19      2       5.47
18      0       0.00
17      1       3.06
16      2       6.50
15      0       0.00
14      1       3.71
13      2       8.00
12      0       0.00
11      1       4.73
10      2       10.40
9       0       0.00
8       1       6.50
7       2       14.86
6       0       0.00
5       1       10.40
4       2       26.00
3       0       0.00
2       1       26.00
1       2       104.00
Using TC as the index metric, obvious answer is buy insurance when TC > 0. Also acceptable, buy insurance when TC >= 0

For any of the above data points I can get this answer by interpolating to find RC when prob(ten) = 1/3, using 27, 26, 25 cards remaining as example.

27 cards (no interpolation necessary since prob(ten) = 1/3
RC = 0, prob(ten) = 1/3
TC = 52*0/27 = 0

26 cards single deck insurance (insurance count)
RC = +1, prob(ten) = 9/26
RC = -2, prob(ten) = 8/26
interpolate: RC = 0, prob(ten) = (8 2/3)/26
TC = 52*0/26 = 0

25 cards single deck insurance (insurance count)
RC = +2, prob(ten) = 9/25
RC = -1, prob(ten) = 8/25
interpolate: RC = 0, prob(ten) = (8 1/3)/25
TC = 52*0/25 = 0

This is my question:
How would your method of finding TC index work on this data set?

Thanks,
k_c
I'll give you an idea of ??what I do. Maybe this can help you.
Let's use as an example a deck of 52 cards and the unbalanced count of tens (1 1 1 1 1 1 1 1 1 -2).
1) Remove an ace from the pack (which is the ace the dealer receives)
2) Let's assume that we are going to play until there are 10 cards left (CL)
3) Two nested loops are needed, one loop that goes through all the possible RC from a minimum to a maximum.
It doesn't matter if you don't know the limits, it can be between -50 and +50. Then the ones that do not correspond
will be discarded (they are the ones whose probability is equal to zero).
The other loop will go through the different depths between 51 (52 -1) cards and 11 cards (10 have been left out).
4) Within these two nested loops you are going to calculate, through a combinatorial analysis, the probability and the
expected value for each combination of RC and depth and the values ??obtained are going to be accumulated into two arrays.
The index of the array is going to be the TC which in turn you get from doing (RC / depth * 52).
array1 (TC) += probability * EV
array2 (TC) += probability
5) You are going to get the index from array1. To do this, you have to go through that array from the lowest value of TC
until you find the TC in which the expected value becomes positive. That point corresponds to the searched index.
It's actually a little more elaborate but for now I don't want to complicate it for you.
The other array is used to know the frequency of each TC.

Hope this helps.

Sincerely,
Cac