21Revolution
21Revolution
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June 23rd, 2013 at 7:08:21 PM permalink
If the dealer shows a 10, according to the Wiz, he will break 0.229784833. As the count increases, I assume his likelihood of breaking does as well, but how much can you move the needle? Is there a point where you can predict he'll bust 26%? It seems like a high count also increases the likelihood he already has a good hand, so I'd imagine the bust rate doesn't change substantially. If anyone knows where I could find that answer, I'd appreciate it.
Ibeatyouraces
Ibeatyouraces
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June 23rd, 2013 at 7:44:11 PM permalink
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surrender88s
surrender88s
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June 23rd, 2013 at 8:43:17 PM permalink
Yeah, for a 10 to bust, the unseen hole card must be a 2,3,4,5, or 6. As the count gets higher, it is less likely that it will be these cards. So the opposite of what you suggest is true.
"Rule No.1: Never lose money. Rule No.2: Never forget rule No.1." -Warren Buffett on risk/return
hmmm23
hmmm23
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June 26th, 2013 at 1:49:45 PM permalink
Quote: surrender88s

Yeah, for a 10 to bust, the unseen hole card must be a 2,3,4,5, or 6. As the count gets higher, it is less likely that it will be these cards. So the opposite of what you suggest is true.



Cool thread; int questions.

Surrender88, you make a good point, but wouldn't that phenomenon be at least partially offset by the fact that in those fewer instances when the dealer DOES end up with 12 thru 16, he'll bust more often with the higher count than he otherwise would?

Are we sure the "fewer hands from 10 showing to 12 thru 16 total" phenomena is greater than the "but, for those 12 thru 16's he does get, he'll bust more often" phenomenon? (I openly admit to not having a clue).
21Revolution
21Revolution
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June 26th, 2013 at 6:30:07 PM permalink
Quote: hmmm23

Cool thread; int questions.

Surrender88, you make a good point, but wouldn't that phenomenon be at least partially offset by the fact that in those fewer instances when the dealer DOES end up with 12 thru 16, he'll bust more often with the higher count than he otherwise would?

Are we sure the "fewer hands from 10 showing to 12 thru 16 total" phenomena is greater than the "but, for those 12 thru 16's he does get, he'll bust more often" phenomenon? (I openly admit to not having a clue).



Right- I thought in a card counting strategy a higher count says not to hit 16 against a 10, so here again I wasn't sure if it is because the player is too likely to bust, or if the odds of the dealer busting went up as well.
Ibeatyouraces
Ibeatyouraces
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June 26th, 2013 at 6:32:27 PM permalink
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beachbumbabs
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beachbumbabs
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June 26th, 2013 at 6:59:12 PM permalink
Just thinking it through (not sure how to frame the math), regardless of the hole card, a 2 thru 6 will REQUIRE the dealer to take a hit, and therefore a higher probability of busting (assuming dealer hits soft 17). With a 10, more than 1/2 the hole cards will not require a hit. No 2 card combo can bust, so there has to be significantly more risk in taking a hit regardless of whether it results in a win. Or am I missing your point? I think you have to hit 16 against a 10 just because the dealer's odds of not having to take a hit are better than 50/50.
If the House lost every hand, they wouldn't deal the game.
Ibeatyouraces
Ibeatyouraces
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June 26th, 2013 at 7:33:27 PM permalink
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surrender88s
surrender88s
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June 26th, 2013 at 8:28:53 PM permalink
basic strategy and the deviations live the illustrious 18 are correct. It's good to ask questions and learn the theory and math of it, and even to question it- but when I do this I keep coming back to sticking with the time-tested strategies. Just realize you'll run into a lot of people who stick to these ideas without questioning them and they might not be as patient with you.

IF the dealer has a 12-16 (what you're hoping for with his 10 showing), he is more likely to break. But if the count is anywhere above 0, we know that there are more 10's and A's, and quite possibly 7-9's, than 2-6's.

With a 10 up, there's a 8/13 chance of dealer not busting and standing with two cards. Even with the other 5 cards that give dealer a two-card 12-16, there's still a likelihood of breaking that is somewhere around 40%. This gives some amount of balancing effect, but not enough to overcome the difference from the first card being more likely to be a 7 or higher.

But what the numbers and deviations are saying is that, as the count increases, it becomes more likely that higher dealer values,10-A will make hands, and that it is more likely that 2-6 will break. Not sure, but I think it's also more likely that 7-9 will make hands.
"Rule No.1: Never lose money. Rule No.2: Never forget rule No.1." -Warren Buffett on risk/return
hmmm23
hmmm23
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June 28th, 2013 at 2:04:06 AM permalink
Quote: Ibeatyouraces

Even though the dealer will bust less often with a higher up card [with a positive count]



Are we sure a dealer w/ a 10 up will bust less often with a positive count?

True, a dealer with a 10 up will get a two card total of 12 thru 16 less often than with a zero count. But at least partially offsetting that is the fact that when the dealer does get those (fewer) 12 thru 16's, he'll bust them a higher proportion of the time than he would with a zero count.

So I guess my question is, are we sure the impact of # 1 is greater than the impact of # 2?

1. The fewer 12 to 16's the dealer gets due to the high count.
vs
2. The more of those 12 to 16's he busts due to the high count.

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