Quote Originally Posted by seriousplayer View Post
Don

What do you think about this count for perfect insurance instead of the Unbalanced Ten Count you are talking about?

A=1
2=1
3=1
4=1
5=1
6=1
7=1
8=1
9=1
T=-1

IC of 100%

It's true that this count could be used for perfect insurance. However, it wouldn't be as simple. The point of tagging non-tens as -1 and tens as +2 is that when there are twice as many non-tens as tens the running count is 0 and the probability of drawing a ten is 1/3. Since insurance pays 2 to 1 it is an even bet at this count. Now let's say insurance paid 3 to 1. If tens were tagged at +3 and non-tens were tagged at -1 then insurance would also be an even bet if running count is 0. If this was the case then the initial running count would be 16*3*(number of decks) - 36*1*(number of decks) = 12*(number of decks). Since this number is greater than 0, insurance would be positive EV at or near the start of a shoe if insurance paid 3 to 1.

Choosing tags like above reduces to something pretty simple.

The reason the count you listed could be used to perfectly gauge insurance is that it tags all of the ranks and tags all non-tens with the same tag. The initial running count for your count would be 16*1*(number of decks) - 36*1*(number of decks) = -20*(number of decks). Insurance would be an even bet when (number of tens)/((number of tens)+(number of non-tens)) = 1/3.

Doing some algebra:
3*(number of tens) = (number of tens) +(number of non-tens)
2*(number of tens) = (number of non-tens)
(number of tens) = 1/2*(number of non-tens)
(running count) = (number of tens) - (number of non-tens) = (number of tens) - 2*(number of tens)
(running count) = -(number of tens)
(total cards) = (number of tens) + (number of non-tens) = (number of tens) + 2*(number of tens)
(total cards) = 3*(number of tens)

(true count) = 52*(running count)/(total cards)
in above case:
(true count) = 52*(-(number of tens))/(3*(number of tens)) -17 1/3

Insurance for your count is an even bet when true count = -17 1/3. It is positive EV when true count is greater than -17 1/3 and negative EV when less than -17 1/3. In order to gauge insurance with 100% accuracy you'd need to make sure to count exactly the number of cards remaining to be dealt and calculate the true count as above, but it could be done.

Hope this helps.

k_c