I'm a newbie to card counting and am wondering if someone can explain the disparity between standing on 16 v. dealer 10 (zero or any + running count) and standing on 16 v. dealer 9 (true count of +5 or higher) in the Illustrious 18. Seems to me that the odds of busting on 16 are the same regardless of whether the dealer has 10 or 9. I know that there is more to the story though.
JT
Your probability of busting when you hit hard 16 is 8/13 (infinite deck approximation), or, 8n/(13n-1), where n is the number of decks used. The 1 represents the dealer's upcard, which is known. The formula assumes a player's generic 16, rather than specifically 10, 6; 9, 7; 5, ,5 , 5, A; etc.
To beat the dealer, you must avoid busting and either the dealer must bust or your hand must beat the dealer's non-busted hand. The dealer is more likely to have an almost unbeatable two-card 20 if his upcard is a 10 than if it is a 9. A dealer's 19 is a very strong hand, but 20 is even stronger. The dealer is also less likely to bust with an upcard of 10 than an upcard of 9. This is the rest of the story.
> I'm a newbie to card counting and am wondering if
> someone can explain the disparity between standing on
> 16 v. dealer 10 (zero or any + running count) and
> standing on 16 v. dealer 9 (true count of +5 or
> higher) in the Illustrious 18. Seems to me that the
> odds of busting on 16 are the same regardless of
> whether the dealer has 10 or 9. I know that there is
> more to the story though.
> JT
> Your probability of busting when you hit hard 16 is
> 8/13 (infinite deck approximation), or, 8n/(13n-1),
> where n is the number of decks used.
You don't mean that. You mean 32n/(52n-1), which, of course, isn't the same thing.
Don
> Your probability of busting when you hit hard 16 is
> 8/13 (infinite deck approximation), or, 8 n /(13 n
> -1), where n is the number of decks used. The 1
> represents the dealer's upcard, which is known. The
> formula assumes a player's generic 16, rather than
> specifically 10, 6; 9, 7; 5, ,5 , 5, A; etc.
> To beat the dealer, you must avoid busting and
> either the dealer must bust or your hand must beat the
> dealer's non-busted hand. The dealer is more likely to
> have an almost unbeatable two-card 20 if his upcard is
> a 10 than if it is a 9. A dealer's 19 is a very strong
> hand, but 20 is even stronger. The dealer is also less
> likely to bust with an upcard of 10 than an upcard of
> 9. This is the rest of the story.
Hitting with 16 vs 10 is borderline, though hitting with 16 vs 9 is a MUST unless the count is extremely high.
I'm not sure your response explains this issue.
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