3,3 VS 2 index DD NDAS

[aceside]

Thanks for the update! How would you revise the calculation to account for 6,6?

I don’t understand what Don is talking about, but here is what I would calculate. In this part, get ride of the 6,6 pair contribution to the hard 12 hand.
”Furthermore, the probability that you are dealt a 12 is 0.00375+0.00094+0.00082+0.00094+0.00041 = 0.00686

Hence, the expected true count conditioned on the event that you are dealt a hard 12 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094)(2/(101/52))+(0.00041)(3/(101/52)))/0.00686 = 0.779.”

You only count the probability of getting a hard 12, excluding a 6,6 pair. Let me comment a little further. Don has threatened to ban me off from this website a few times!

[JohnGalt007]

"How would you revise the calculation to account for 6,6?"

You wouldn't, especially if aceside tells you it's wrong. It is, of course, perfectly correct. I don't care for your notation, though, which is clumsy and confusing. For 6,6 vs. 4, why not the much clearer and intuitive:
8/104 x 7/103 x 8/102 = 448/1092624 = 0.00041?

Don

Thanks Don! I appreciate it. I had assumed my work was correct, but I'm always open to revisions if they're needed. I used combinations in my formulas above so I can more easily generalize the same process to other hands. Writing out the work in fraction form leaves fewer clues as to how to do that.

[JohnGalt007]

I don’t understand what Don is talking about, but here is what I would calculate. In this part, get ride of the 6,6 pair contribution to the hard 12 hand.
”Furthermore, the probability that you are dealt a 12 is 0.00375+0.00094+0.00082+0.00094+0.00041 = 0.00686

Hence, the expected true count conditioned on the event that you are dealt a hard 12 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094)(2/(101/52))+(0.00041)(3/(101/52)))/0.00686 = 0.779.”

Hi Aceside, I think the issue here is that we only need to treat 6-6 separately from the other combinations that yield a hard 12 total for the purposes of deciding what to do with that hand (i.e. calculating the conditional EV of hitting/standing/doubling down/splitting with 6-6, versus calculating the conditional EV of hitting/standing/doubling down with 2-T, 3-9, 4-8, or 5-7). We don't need to do that for the purposes of calculating the probability of being dealt a hard 12 and how that changes the true count.

[aceside]

I’ll take a look at this part tonight.

[DSchles]

We'll all rest easier once you do!

Don

P.S. I don't/can't ban anyone from this site. That's Norm's domain. But I have been known to make suggestions!

[21forme]

P.S. I don't/can't ban anyone from this site. That's Norm's domain. But I have been known to make suggestions!

I was thinking that, too.

Come on Norm. This month is aceside's 6 year anniversary on this forum. Nothing has changed. His troll tactic of posting incorrect information hasn't changed in all that time. Please get rid of him permanently.

[aceside]

Apparently someone disagrees with me. Still, let me post my calculation. A 2-card hard-12 includes only these four hands:

T, 2;
9, 3;
8, 4;
7, 5.

The probability that you are dealt a hard-12 vs. 4 is 0.00375+0.00094+0.00082+0.00094 = 0.00645.

Hence, the expected true count conditioned on the event that you are dealt a hard-12 vs. 4 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094) (2/(101/52)))/0.00645 = +0.726.

This is slightly off from John’s result.

[JohnGalt007]

Apparently someone disagrees with me. Still, let me post my calculation. A 2-card hard-12 includes only these four hands:

T, 2;
9, 3;
8, 4;
7, 5.

The probability that you are dealt a hard-12 vs. 4 is 0.00375+0.00094+0.00082+0.00094 = 0.00645.

Hence, the expected true count conditioned on the event that you are dealt a hard-12 vs. 4 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094) (2/(101/52)))/0.00645 = +0.726.

This is slightly off from John’s result.

Hi aceside, I'm aware that this is the calculation assuming that we don't count 6-6 as a hard 12. My question is, why would that be a valid assumption? We "count 6-6 separately" from other totals of 12 solely for the purposes of evaluating the conditional expectation of the playing strategies available to the player, because we can split a pair of sixes whereas we cannot split a 2-T, 3-9, 4-8, or 5-7. None of that matters for the purposes of calculating the average true count conditioning on being dealt a hard 12. The discrepancy in understanding lies in the definition of "hard 12" for the purposes of calculating the average true count.

[Freightman]

Hi aceside, I'm aware that this is the calculation assuming that we don't count 6-6 as a hard 12. My question is, why would that be a valid assumption? We "count 6-6 separately" from other totals of 12 solely for the purposes of evaluating the conditional expectation of the playing strategies available to the player, because we can split a pair of sixes whereas we cannot split a 2-T, 3-9, 4-8, or 5-7. None of that matters for the purposes of calculating the average true count conditioning on being dealt a hard 12. The discrepancy in understanding lies in the definition of "hard 12" for the purposes of calculating the average true count.

I have to agree. For whatever reason, I seem to have a plethora of 66 v2 hands. Though I avoid NDAS games, hitting vs splitting is not an unusual play. As stated above, “it is a conditional expectation of the playing strategies available”

It comes out to - what is the index.

[DSchles]

It comes out to - what is the index.

Well, then, to be fair, the index for 6,6, vs. 2 is very different from standing on hard 12 vs. 2. There's probably more to this discussion than meets the eye.

Don

[Zenfighter]

Hi, friends!

Enjoyed this thread. Thank you, all. I couldn’t resist this time to add my contribution, too.

roliin wrote:

In table 26.2 of Don/Gron's book, the index is a 10. Does this mean you would not hit a 3,3 VS 2 @ +10? If that's what the sims say that's what they say.

2 Decks, H17, NDAS, 3,3 v 2

Hitting	-0.142295496
Splitting	-0.202137284
Frequency	0.000410022

What the simulated index is telling us, basically, is that to overcome the difference (6%) in gains for action 1, we need to be patient and avoid splitting. Nothing strange, therefore, with the appearance of such a high index.


John Galt wrote:

Hence, the expected true count conditioned on the event that you are dealt a hard 12 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094)(2/(101/52))+(0.00041)(3/(101/52)))/0.00686 = 0.779.

Note that I calculated the running count with the tags from the Hi-Lo system; a more accurate calculation would require the effects of removal for each card, not the approximations afforded by the tags.

	2 Decks, H17, 12 v 4
	Hitting	Standing
12v4	-0.206889222	-0.195748581
	Hitting	Standing	Frequencies
T,2	-0.204205449	-0.206124526	0.003748774
9,3	-0.219573436	-0.195903587	0.000937193
8,4	-0.214138826	-0.191941785	0.000820044
7,5	-0.203561198	-0.178661647	0.000937193
		Total	0.006443205
	Hitting	Splitting	Frequency
6,6	-0.202729595	-0.069138850	0.000410022

Here we can see the 6,6 index playing in another league, and thus it must be kept separate. A mandatory split!


With accurate EORs for 2 decks, h17, ndas.

Generic: 12 v 4 Stand if TC >= -0.157476

Composition-dependent:

Hand	EORs Indices	Occurrence	Product
T,2	0.622041	0.003748774	0.002331891
9,3	-0.557348	0.000937193	-0.000522343
8,4	-0.448992	0.000820044	-0.000368193
7,5	-0.615712	0.000937193	-0.000577041
	Totals	0.006443205	0.000864313
		Weighted Index	0.134143418

The full-deck favorability, m2, for carrying out the action (hitting) equals -0.371119 (see BJA3, page 519). Nevertheless, here the CD index for T,2 vs. 4 behaves itself like the head of operations, and determines the final fate of the index. In other words: The majority of the time, 58%, your index to stand will be 0.7 by rounding up. On the other hand, for the rest of the two-card holdings, 42% of the time, your index to stand will be negative, around -0.5. What this sort of inner contradiction is telling us is that the use of composition-dependent indices, totally worthless when dealing with shoes, is not a bad idea at all for a DD enthusiast, or for others addicted to saving pennies.

Enjoy!

Sincerely,

Zenfighter

[DSchles]

Readers: It isn't often that Zenfighter graces our pages with his presence. Always wonderful to see one of the giants in BJ research come around for a visit.

Don