I know from table 7.3 (bja3, pg.99) that always standing on 16 v 10 will cost the $100 bettor $2.65 per hour. Is there an easy way to figure out the cost of standing down to TC -1, -2?
It will obviously be a little more, but not all that much. If you look at the graph of the play, the differences between hitting and standing, down to -2, are virtually inconsequential, particularly considering that you're likely to have a minimum bet out at the moment that you stand.
Don
"It will obviously be a little more, but not all that much."
Shouldn't the cost be a little less?
When I say standing down to TC-2, I mean that I will be hitting 16 v10 at anything worse than TC-2, like I'm supposed to. Given that I will be standing in neutral and positive counts and hitting at counts lower than TC-2 where I'm more likely to make the hand, shouldn't it cost a little (or a lot) less than always standing ($2.65 per hr)? I would only be playing the hand wrong at TC-1 and TC-2.
There's something known as "counter's basic strategy" which calls for standing on all 16vT to look less obvious. I don't have a reference for you (Don?), but always standing makes more sense than always hitting, as your incorrect play will always be with a minimum bet out, so the overall cost will be less.
(edited for typo)
Last edited by 21forme; 03-27-2023 at 06:48 AM.
It’s obvious that there isn’t very much probability for TC<=-3, so your cost of wrongfully playing this 16v.10 hand is still about $2.65 per hour. What is more important is the hit/stand critical TC point, though. I always use TC=+1 for hit/stand, but most people just say TC=+0.
Firstly, I don't know how to calculate the exact value for the TC hit/stand decision, but I roughly recall some literature says it is a positive number around zero. Secondly, I simulated this 16v.10 decision millions of times using Phil's simulator and found it was TC=+1. I still need confirmation on this part.
Last edited by BJGenius007; 03-27-2023 at 05:37 PM.
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