can someone help me with a bit of data

can someone help me with a bit of data

I cant find the post but someone posted that there were about 1,300 hands the dealer can get without busting - does anyone have a copy of them or a source for them. Most of the stats I find on the web are just percentages and statistics from sim software, I need to find the definitive list of hands a dealer can get and optionally the list a player can get and for the bonus a list of bust hands for both the dealer and player. I'd rather not have to write a program to figure it out if someone has it already.

Dealer's 1,300 unbusted-consecutive-hands

gern blanston,

I'm sorry, but I can't help you to find that post.

May I ask you a question? If so, what can you do with the info about the dealer's 1,300 unbusted-consecutive-hands? If you like to try your luck with a $5 double-up-progression (no doubledowns, nor splits, nor insurances, etc), you don't need to know any thing more than 35 hands because of this table:

 

Table of $5 double-up-progression 	 

1	5	 

2	10	 

3	20	 

4	40	 

5	80	 

6	160	 

7	320	 

8	640	 

9	1,280	 

10	2,560	 

11	5,120	 

12	10,240	 

13	20,480	 

14	40,960	 

15	81,920	 

16	163,840 

17	327,680 

18	655,360 

19	1,310,720 

20	2,621,440 

21	5,242,880 

22	10,485,760 

23	20,971,520 

24	41,943,040 

25	83,886,080 

26	167,772,160 

27	335,544,320 

28	671,088,640 

29	1,342,177,280 

30	2,684,354,560 

31	5,368,709,120 

32	10,737,418,240	 

33	21,474,836,480	 

34	42,949,672,960	 

35	85,899,345,920	

Never mind about the dealer's 1,300 unbusted-consecutive-hands!
You would lose $86 billions if the dealer wins 35 hands in the row!
Not even Einstein can beat the dealer's unpredictable winning streak, lol.

The post said nothing about "consecutive" (nt)

Amazing how brutal a progression can be

You would lose $86 billions if the dealer wins 35 hands in the row!

Actually you lose that much on just the 35th hand alone.

OOOPS - WHAT I MEANT WAS....

I am looking for the list of possible hands the dealer can get that give him a total between 17 and 21 - like a/10 and 7/10, and 4/5/9 like that - all the hand combos/card combos the dealer can potentially get.

Ditto. Total losses would be 172 billions. Add up 1st thru 35th loss. (nt)

These answers are irrelevant

It did not occur to me in the least that the poster was asking a question about progression systems. The actual question is a tough one, but very interesting. I do hope someone answers it.

Just for you or for "just me"

O.F., just for you, I kick around with these numbers,
& come up with these tables for 21 & 17 quickly:

						 

HandTot	Card 1	Card 2	Card 3	Card 4	Card 5	Card 6 

21	1	2	3	4	5	6 

21	2	3	4	5	6	1 

21	3	4	5	6	3	 

21	4	5	6	6		 

21	5	6	7	3		 

21	6	7	8			 

21	7	8	6			 

21	8	9	4			 

21	9	10	2			 

21	10	11				 

21	11	10				 

 

 

HandTot	Card 1	Card 2	Card 3	Card 4	Card 5	Card 6 

17	1	2	3	4	5	2 

17	2	3	4	4	4	 

17	3	4	5	5		 

17	4	5	6	2		 

17	5	6	6			 

17	6	7	4			 

17	7	8	2			 

17	8	9				 

17	9	8				 

17	10	7				 

17	11	6				 

						

You or (the rest of stone-casters, tongue-in-cheeks) can finish up the above by doing the tables for 18, 19, & 20.
BTW, there are more combos for the above 17 & 21 if you want to do the tedious tables as if they were the tidetables.

Not even close

For example, you can also get 21 via:

9,2,A,A,A,A,6

9,2,A,2,A,A,5

9,2,A,3,6

etc etc etc etc etc

I don't see a whole lot of hands that should be there.

Good try.

Ditto

Let me give another try...

											 

Say a singledeck game...											 

HandTot	Card 1	Card 2	Card 3	Card 4	Card 5	Card 6	Card 7	Card 8	Card 9	Card 10	 

21	1	1	1	1	2	2	2	2	3	3	 

21	1	1	1	1	2	2	2	2	4	5	 

21	1	1	1	1	2	2	2	2	5	4	 

21	1	1	1	1	2	2	2	2	6	3	 

21	1	1	1	1	2	2	2	2	7	2	 

22	1	1	1	1	2	2	2	2	8	2	 

21	1	1	1	1	2	2	2	2	9		 

21	1	1	1	1	2	2	2	3	4	4	 

21	1	1	1	1	2	2	2	3	5	3	 

21	1	1	1	1	2	2	2	3	6	2	 

21	1	1	1	1	2	2	2	3	7	1	Cheated! (5 aces) 

Etc� Wow,  in order to finish up the tables for all the combos,  

we could be here until the 2nd coming of Christ!											 

											

I can help but....

I would not be willing without some sort of compensation. At this point I have the methodology to output dealer's hands to a file, but could do more. I don't know if it would useful to you. I'll just post a sample that shows most of the elements:

 

Dealer hand = 14 

 

 1  2  3  4  5  6  7  8  9  T    Total  Cards  Count    Elem 

 

 0, 0, 0, 0, 0, 0, 1, 0, 0, 0        7      1      1       1 

 1, 0, 0, 0, 0, 1, 0, 0, 0, 0        7      2      1       2 

 0, 0, 0, 0, 0, 1, 0, 0, 0, 0        6      1      1       3 

 0, 1, 0, 0, 1, 0, 0, 0, 0, 0        7      2      1       4 

 2, 0, 0, 0, 1, 0, 0, 0, 0, 0        7      3      1       5 

 1, 0, 0, 0, 1, 0, 0, 0, 0, 0        6      2      1       6 

 0, 0, 0, 0, 1, 0, 0, 0, 0, 0        5      1      1       7 

 0, 1, 0, 1, 0, 0, 0, 0, 0, 0        6      2      1       8 

 2, 0, 0, 1, 0, 0, 0, 0, 0, 0        6      3      1       9 

 1, 0, 0, 1, 0, 0, 0, 0, 0, 0        5      2      1      10 

 0, 0, 0, 1, 0, 0, 0, 0, 0, 0        4      1      1      11 

 0, 1, 1, 0, 0, 0, 0, 0, 0, 0        5      2      1      12 

 2, 0, 1, 0, 0, 0, 0, 0, 0, 0        5      3      1      13 

 1, 0, 1, 0, 0, 0, 0, 0, 0, 0        4      2      1      14 

 0, 0, 1, 0, 0, 0, 0, 0, 0, 0        3      1      1      15 

 0, 2, 0, 0, 0, 0, 0, 0, 0, 0        4      2      1      16 

 2, 1, 0, 0, 0, 0, 0, 0, 0, 0        4      3      1      17 

 1, 1, 0, 0, 0, 0, 0, 0, 0, 0        3      2      2      18 

 3, 0, 0, 0, 0, 0, 0, 0, 0, 0        3      3      1      19 

.
Above is all of the possibilities if dealer starts with a total of 14. Except for element 18, there is only one way to draw each combination. For example, look at element 10 which consists of a draw of 1 ace and 1 four. Since dealer's starting total is 14, he must draw the ace first, so the count for this combo is 1. Element 18 consists of 1 ace and 1 two. Here dealer's final total can be achieved 2 ways; ace first or two first, so the count for this combo is 2.

A complete set of files would be for dealer hands 7,8,9,10,11,12,13,14,15,16 and 6,5,4,3,2,1,0 (2 files each for S17 and H17.) The total number of possible dealer hands (dealer hand=0) is 1677 for S17 and 1740 for H17.

Any way, if you think it may be something you are interested in, let me know.

kaycee

forgot to add....

A complete set of files would also include for S17 dealer hands of soft 2,3,4,5,6 and for H17 dealer hands of soft 2,3,4,5,6,7.

kaycee