Table Games math question for all you card counters out there

If the dealer is showing a 10, and you have 12-16 as your hand. What is the rough estimate percent chance you'll win your hand. I know it depends on how you hit out, and what the dealer has underneath. But just on your experience how often do you win these types of hands. I'm thinking its about 40% of the time.

 

I'm confused by your question. I'm assuming your saying that you have a hard 12-16. Each of those hands has a different probability to win. Are you asking what the chances to win are if you play proper basic strategy, or are you asking about whether the True Count changes what the probabilities are?

Seems to me this question has nothing to do about Card Counting.

 

Dutch is correct. Card counting really only has to do with how much you bet...very rarely does it impact how one plays a hand.

But the math on this is easy. Assuming a brand new deck of 52 cards, and assuming your 15 (let's say) consists of a 10 & 5, then we know that there are exactly 14 tens, 4 each of the Aces through fours, 3 fives, and 4 each of sixes through nines.
So, hitting your 15? There are 26 cards that bust you, 4 that give you 16, and 19 that give you 17 - 21.
Let's consider the Ace a bust card, since you would hit the 16 vs. 10. That gives you a 30 to 19 chance of busting, or 61.22%.

For the dealer, his hole card will be a 7 or better 22 of 44 (you know he has no ace!)...so, half the time.

For easy math, he will have a 2-card hand half of the time, and you will bust 60% of the time before even seeing his hole-card.

The 40% of the time when you don't bust, you have an even chance of winning, losing or pushing...meaning the odds of any given pat hand are the same for you as they are for the dealer.
But, with a 15, there are only 7 cards giving you a 20 or 21. For the dealer, there are 14 cards that give him a 20. and there are multiple 2-hit card combinations that give him a 20 or 21.
But for perspective, let's look at his 20 only. He is twice as likely to have 20 as you are to have 20 or 21...assuming you didn't bust in the first place...which you will do 60% of the time.
So, for basic easy math purposes, you will lose half of the 40% of the hands for which you do not bust...so 20% wins & 80% losses.

That's just for 15, but similar math can be figured for the other crap hands vs a dealer 10...with your hard 12 being the best, obviously.

 

Basic strategy will say to either hit or surrender if allowed. If someone is supposed to be counting in this scenario, what are you proposing the true count is? How many decks in the shoe, surrender allowed, H17, S17?

 

On a stiff hand, you wont win unless the dealer busts. With a dealer card of 10, the dealer busts 23% of the time.

 

Now - for the sake of putting this to rest...what would happen if you stand with your hard 15 vs. a dealer 10?
There are 26 hole cards that give the dealer a 17-20 (14 tens, 12 7s through 9s). That's a 26 out of 45 chance (no ace) that he has a pat hand and you lose...58% loss.
There are 19 crap hole-cards he could have. For each of them them (except 5 & 6) there are at least 18 cards giving him a 17+.
Let's look at a dealer 14. He can hit any of four 2s, 3s, 6s & 7s + two 5s...that's 18 of 48. The same is true for the other crap hands.
So, 38% of his crap hands, he will still make a hand, beating your 15. 62% of those, he will bust.
58%, you lose without a dealer hit. Of the other 42%, you would win 62% of those - or 26%.
That's 12% wins and 88% losses...vs 20% wins & 80% losses if you hit.
So - yeah - always hit your hard 15 vs. a dealer 10.

 

http://wizardofodds.com/games/blackjack/appendix/1/

Look up your hand. Look up dealer's up card. There you go.

 

thanks for the input guys, especially your analysis topcard. It was exactly what I was looking for.