dbldwn23
dbldwn23
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September 26th, 2012 at 2:06:08 PM permalink
My first post so please dont flame me:) Assume the count is neutral what is the difference between a 6 vs 2 showing for the dealer. Yes, if he has a 16 anything 6 or over will bust him, but the 12 has the A-4 which will have no effect and we will be right back to the 16. I am sure I am miscalculating somewhere, can somebody please explain the difference. Thank you for your time.
bigfoot66
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September 26th, 2012 at 2:08:19 PM permalink
The blackjack page at wizard of odds has a chart that shows the likelyhood of dealer busting with a given up card, is that what you wanted?
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Ibeatyouraces
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September 26th, 2012 at 2:12:08 PM permalink
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dbldwn23
dbldwn23
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September 26th, 2012 at 2:30:13 PM permalink
I understand the dealers % of busting with any up card showing. I dont understand why we are so afraid of a 2 and not a 6. Yes with a 12 he can hit more than once, but we dont care if he hits once, twice, 3 times as long as he busts (the A-4 are useless cards dont hurt or help). I just trying to understand this more clearly. Thanks again to any input.
bigfoot66
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September 26th, 2012 at 2:33:20 PM permalink
Either card is only half a hand. We want the dealer to bust, and to do that at some point his hand must be a 12, 13, 14, 15, or 16. He is more likely to start with a 12-16 with a 6 showing than a 2.
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sodawater
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September 26th, 2012 at 2:35:24 PM permalink
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Last edited by: sodawater on Oct 1, 2018
Ibeatyouraces
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September 26th, 2012 at 2:36:37 PM permalink
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ChesterDog
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September 26th, 2012 at 2:43:59 PM permalink
From here I copied the Wizard's dealer probabilities for an 8-deck shoe with the dealer standing on soft 17 for dealer's up cards of 2 and 6. Here they are:


P(dealer) 2 6
21 0.12 0.10
20 0.12 0.10
19 0.13 0.11
18 0.13 0.11
17 0.14 0.17
bust 0.35 0.42

The dealer is more likely to get 21, 20, 19, and 18 with a 2 up than he is with a 6 up. And the dealer is more likely to get 17 or bust with a 6 up than with a 2 up.

Using these values, one can calculate the EV of these various player (standing) totals:

EV(player) 2 6
21 0.88 0.90
20 0.64 0.70
19 0.39 0.50
18 0.12 0.28
17 -0.15 0.01
<17 -0.29 -0.15

The EV of each hand of the player is greater vs a 6 than vs a 2.
Ibeatyouraces
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September 26th, 2012 at 2:49:18 PM permalink
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dbldwn23
dbldwn23
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September 28th, 2012 at 4:14:47 AM permalink
I really appreciate everyone's feedback. With a 16 you have 5 cards that will make a hand 8 that will bust, with a 12 you have the same 5 cards that will make a hand 4 that will bust and 4 that will have no effect just bringing you closer to 16. Yes there are less bust cards with a 12 but 4 neutral cards bringing you closer to 16 and busting on the following card with now more bust cards available to you. I am sorry it is taking me so long to grasp this situation.
Ibeatyouraces
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September 28th, 2012 at 6:31:31 AM permalink
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dbldwn23
dbldwn23
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September 28th, 2012 at 12:08:37 PM permalink
With the 6 you have A-5 to help make a possible hand and with the 2 you have a 5-9 to make a poosible hand. Both have 5 cards that could put the dealer in position of making 17-21. I am not trying to be difficult I am just trying to understand why I don't see this.
dwheatley
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September 28th, 2012 at 12:47:32 PM permalink
You seem to be assuming the dealer has a 10 under, which he probably doesn't. You have to pretend the dealer's undercard isn't even there, it's a random card coming out next.

One advantage for the 2: The 2 has more ways to become a 7-11 than a 6 does. Those small hands can then hit 10s for a pat hand.

There are others. The probability tables are available for your perusal. If you don't get it, then I doubt any more explaining will help you. Work it out for yourself in Excel.
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LonesomeGambler
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September 28th, 2012 at 12:51:25 PM permalink
An easy (and accurate) answer that hopefully should give you some relief:

Whenever asking "why do we play this hand this way?" with regards to basic strategy, the answer is not, "because if x = y, then z is a likely outcome," it's, "because computer simulations of billions of hands have proven that this is the play that will produce the highest mathematical expectation." Trying to understand each and every basic strategy play from an intuitive perspective is a fool's errand, as the answer is quite often not at all intuitive (nor does it need to be for it be the correct play). Life is short, so if you want my two cents: learn the BS charts and save the time spent trying to "understand" basic strategy for something more fun!
FleaStiff
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September 28th, 2012 at 1:55:55 PM permalink
Quote: dbldwn23

I am sorry it is taking me so long to grasp this situation.

That's okay. Its taking me even longer and I'm nowhere near grasping it yet.
MonkeyMonkey
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September 28th, 2012 at 4:11:27 PM permalink
Quote: LonesomeGambler

Life is short, so if you want my two cents: learn the BS charts and save the time spent trying to "understand" basic strategy for something more fun!



Best answer yet. Everytime I deal bj I hear a bj "pro" explaining why a certain hand is played the way it is (stuff like you have to assume the hole card is a 10, etc.) and it's nearly always wrong. That's how these myths get perpetuated.
Ibeatyouraces
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September 28th, 2012 at 4:26:04 PM permalink
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dbldwn23
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September 29th, 2012 at 5:01:49 AM permalink
Thank you for all of your input. I will just drop it and trust the percentages and millions and millions of computer simulations. Now back to the tables here in MN. I don't think we have many players from MN.
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