All the cheap camo plays in the world will not cover the fact that you increased from 1 unit to 10, 12 or 15 units when the count jumped to +4. I will disagree with the experts, except to the extent that cheap camo plays generally gets them to let their guard down UNTIL you announce your real intent. Of course it does matter what the game min is ($5 or $50), and also what the size of the store is, as to what will GRAB their attention. For cheap camo, not hitting hard 12 vs 12/13 will both please the ploppies at the table and assure the educated eye (not necessarily the one in the sky) that you are not a pro player. Always hitting your hard 12s will usually get the table's attention and alert the educated eye that you don't make this most common mistake (other ones are A/7 vs 9, 10 and A, which I do correctly announcing that I read it in the book if anyone gripes).
Doubling hard 12 question
- Thread starter Blue Efficacy
- Start date
Sucker said:
I believe he's saying that the next card ONLY is the one that's not an ace or a face; and once he takes that card, the deck becomes normal again; and the dealer has a normal chance of busting. The cdca on that website is a VERY valuable tool, but there doesn't seem to be a way to make it give the specific information that the OP is looking for.
It will let you take the aces and faces out of the ENTIRE deck, but if that were the case; it would be INSANE to double.
It will let you take the aces and faces out of the ENTIRE deck, but if that were the case; it would be INSANE to double.
Conditional EV = (Unconditional EV - prob(T)*EV(if T is drawn)/(1-prob(T))
Example: hand of 10-2 versus 5, 6 decks full shoe
Unconditional double EV = -.3772
Prob(T) given hand of T-2 v 5 = 95/309
Conditional EV = (-.3772 -95/309*(-2))/(1-95/309) = .3432 = +34.32%
k_c said:
You can compute the conditional EV of not drawing a ten from the program's output:
Conditional EV = (Unconditional EV - prob(T)*EV(if T is drawn)/(1-prob(T))
Example: hand of 10-2 versus 5, 6 decks full shoe
Unconditional double EV = -.3772
Prob(T) given hand of T-2 v 5 = 95/309
Conditional EV = (-.3772 -95/309*(-2))/(1-95/309) = .3432 = +34.32%
Conditional EV = (Unconditional EV - prob(T)*EV(if T is drawn)/(1-prob(T))
Example: hand of 10-2 versus 5, 6 decks full shoe
Unconditional double EV = -.3772
Prob(T) given hand of T-2 v 5 = 95/309
Conditional EV = (-.3772 -95/309*(-2))/(1-95/309) = .3432 = +34.32%
k_c said:
Conditional EV = (Unconditional EV - prob(T)*EV(if T is drawn)/(1-prob(T))
Example: hand of 10-2 versus 5, 6 decks full shoe
Unconditional double EV = -.3772
Prob(T) given hand of T-2 v 5 = 95/309
Conditional EV = (-.3772 -95/309*(-2))/(1-95/309) = .3432 = +34.32%
Example: hand of 10-2 versus 5, 6 decks full shoe
Unconditional double EV = -.3772
Prob(T) given hand of T-2 v 5 = 95/309
Conditional EV = (-.3772 -95/309*(-2))/(1-95/309) = .3432 = +34.32%
From your post it appears to me that maybe you have forgotten to include the fact that the next card cannot be a ten, but it also cannot be an ACE. Perhaps this explains the discrepancy?
A simpler way of figuring this problem and USING the program from that website:
1/8 of the time you will get a deuce for a total of 14; EV = -15.85
1/8 of the time you will get a three for a total of 15; EV = -15.85
1/8 of the time you will get a four for a total of 16; EV = -15.85
1/8 of the time you will get a five for a total of 17; EV = - 4.07
1/8 of the time you will get a six for a total of 18; EV = +19.85
1/8 of the time you will get a seven for a total of 19; EV = +43.69
1/8 of the time you will get an eight for a total of 20; EV = +66.86
1/8 of the time you will get a nine for a total of 21; EV = +89.08
Add these up and divide by 8, and you get an average EV of 20.9%; which is much closer to the 20.5% figure that MY software came up with. (I was using 8 decks rather than 6).
Either way; it's definitely a no-brainer of a double-down if you choose to take it!
Sucker said:
I'm not sure why your figure of 34.32% differs from the figure of 20.5% that I originally quoted. Could you please define the terms in this formula so I can check your work against mine? (Conditional EV, Unconditional EV, prob(T))
From your post it appears to me that maybe you have forgotten to include the fact that the next card cannot be a ten, but it also cannot be an ACE. Perhaps this explains the discrepancy?
A simpler way of figuring this problem and USING the program from that website:
1/8 of the time you will get a deuce for a total of 14; EV = -15.85
1/8 of the time you will get a three for a total of 15; EV = -15.85
1/8 of the time you will get a four for a total of 16; EV = -15.85
1/8 of the time you will get a five for a total of 17; EV = - 4.07
1/8 of the time you will get a six for a total of 18; EV = +19.85
1/8 of the time you will get a seven for a total of 19; EV = +43.69
1/8 of the time you will get an eight for a total of 20; EV = +66.86
1/8 of the time you will get a nine for a total of 21; EV = +89.08
Add these up and divide by 8, and you get an average EV of 20.9%; which is much closer to the 20.5% figure that MY software came up with. (I was using 8 decks rather than 6).
Either way; it's definitely a no-brainer of a double-down if you choose to take it!
From your post it appears to me that maybe you have forgotten to include the fact that the next card cannot be a ten, but it also cannot be an ACE. Perhaps this explains the discrepancy?
A simpler way of figuring this problem and USING the program from that website:
1/8 of the time you will get a deuce for a total of 14; EV = -15.85
1/8 of the time you will get a three for a total of 15; EV = -15.85
1/8 of the time you will get a four for a total of 16; EV = -15.85
1/8 of the time you will get a five for a total of 17; EV = - 4.07
1/8 of the time you will get a six for a total of 18; EV = +19.85
1/8 of the time you will get a seven for a total of 19; EV = +43.69
1/8 of the time you will get an eight for a total of 20; EV = +66.86
1/8 of the time you will get a nine for a total of 21; EV = +89.08
Add these up and divide by 8, and you get an average EV of 20.9%; which is much closer to the 20.5% figure that MY software came up with. (I was using 8 decks rather than 6).
Either way; it's definitely a no-brainer of a double-down if you choose to take it!
Code:
[u]Hand EV Weight[/u]
T-2-1 -.3248 24/214
T-2-2 -.3220 23/214
T-2-3 -.3212 24/214
T-2-4 -.3204 24/214
T-2-5 -.0787 23/214
T-2-6 +.4060 24/214
T-2-7 +.8804 24/214
T-2-8 +1.3412 24/214
T-2-9 +1.7828 24/214
(1.7828+1.3412+.8804+.4060−.3204−.3212−.3248)×24/214
+(−.3220−.0787)×23/214 = .3432
k_c said:
The way I did it is a shortcut using conditional probabilities. If you itemize the possibilities weighted by their probability you get the same answer. (Figures are for dealer stands on soft 17, btw.) EV is for drawing 1 card and standing but remember bet is now doubled.
Code:
[u]Hand EV Weight[/u]
T-2-1 -.3248 24/214
T-2-2 -.3220 23/214
T-2-3 -.3212 24/214
T-2-4 -.3204 24/214
T-2-5 -.0787 23/214
T-2-6 +.4060 24/214
T-2-7 +.8804 24/214
T-2-8 +1.3412 24/214
T-2-9 +1.7828 24/214
(1.7828+1.3412+.8804+.4060−.3204−.3212−.3248)×24/214
+(−.3220−.0787)×23/214 = .3432
Sucker said:
Ok - I get the drift. Good work on your part! But I see you HAVE included the ace as a possibility, which the OP excluded in the problem. Without going through the work myself, I'm going to guess that without the possibility of an ace the answer WILL be very close to a 41.8% IBA; which agrees perfectly with the 20.9% DBA that my simulation program came up with.
Revised calculation to include ace is pretty much in the same ballpark as your sim.
(1.7828+1.3412+.8804+.4060−.3204−.3212)×24/190+(−.3220−.0787)×23/190 = +.4276
k_c said:
OK, sorry I didn't notice that OP's condition was no ace as well as no ten could be drawn.
Revised calculation to include ace is pretty much in the same ballpark as your sim.
(1.7828+1.3412+.8804+.4060−.3204−.3212)×24/190+(−.3220−.0787)×23/190 = +.4276
Revised calculation to include ace is pretty much in the same ballpark as your sim.
(1.7828+1.3412+.8804+.4060−.3204−.3212)×24/190+(−.3220−.0787)×23/190 = +.4276
1) Figure unconditional EV where any card may be drawn
UEV = -.3772 for doubling T-2 v 5 with no special condition
2) Figure EV for the case of an ace and case of a ten being drawn when doubling
EV(T-2-A) = -.3248; stand EV(T-2-A) v 5 = -.1624, doubled = -.3248
EV(T-2-T) = -2; (double EV for busting = -2)
3) Figure probability of drawing an ace and probability of drawing a ten given T,2,5 have been removed from a full 6 deck shoe
prob(A) = 24/309
prob(T) = 95/309
4) Compute EV on the condition that no ace or ten can be drawn on the next card
CEV = (UEV - prob(A)*EV(T-2-A) - prob(T)*EV(T-2-T))/(1 - (prob(A)+prob(T))
CEV = (-.3772-24/309*(-.3248)-95/309*(-2))/(1-(24/309+95/309))
CEV = (−.3772+24÷309×.3248+95÷309×2)÷(1−119÷309) = +.4276
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