Curious for both regular shoe BJ as well as Spanish.
The major change needed from basic strategy (after "Hit 20 with 6 or more cards") appears to be, "Never split Aces." A soft 12 is more likely to lead to a seven-card 21 than what you might end up with from your split.
For example, there is a casino dealing spanish 21, where the first 7 card 21 is worth $500 with a $3 minimum bet plus 21 cents of -EV per required side betting. Given what you said above, I'm ballparking that the true odds (in Spanish 21) with all rational deviations would be about 1 in 500, but I'm not sure what it would cost me in EV on the base bet.
If I am understanding DonGuy's post, for a 6 deck, strategy change of hitting any 6 card 20 and never splitting Aces it would have a probability of 0.00055556. This is way less than a dealer's 7 card 21 being 0.000029251 or 1 / 34,186 which makes sense that it would be less but that much less?
So is the probability of a 6 deck 7 card 21 0.00055556 or if not can someone help me figure out the probability of an 8 deck 7 card 21?
Thank you
Quote: tooncestdcI've seen the odds of a dealer BJ as about .000024.<snip>link to original post
tooncestdc,
WHAT???? Where did you see the above-quoted figure?
The dealer averages a BJ about once in 21 rounds: off the top of a 6D shoe, the dealer's BJ probability is 2*96*24/(312*311) = 0.047489...
Dog Hand
Thanks
if the bonus was paid on either hand and say you couldn't/wouldn't double after split, then my estimate was you'd split 10(789) 9832(4 6789) 6 (4 678) 654(456789) A(6789). Also I can imagine some close doubling might be a bad idea since the cost of not being able to hit 3-card 12+ is quite high. Also not splitting 3s vs low cards is a fairly close decision, so splitting 11 might make it worth it.
Similarly a quick look vs 10 you stand 2-card 17+ 2-card soft 20, whereas vs 6 you do stand 2-card 13+ 3-card 15+ 4-card 17+ and 2-card soft 20. I think you probably hit any 3-card soft total.
FFS STOP JACKIINNG AROUND...what's the promo and where? (-:Quote: Viper21I appreciate the look at strategy changes. What about if you just went for it every single hand ignoring the main bet win/loss? So basically hit until you either bust or reach a hard 21 like what DonGuy posted. What would be the probability of reaching a 7 card 21.
link to original post
Quote: Viper21I appreciate the look at strategy changes. What about if you just went for it every single hand ignoring the main bet win/loss? So basically hit until you either bust or reach a hard 21 like what DonGuy posted. What would be the probability of reaching a 7 card 21.
link to original post
To calculate this probability, we could ignore the dealer's card and simply look at all the ways that seven cards with possible values of 1-11 can be combined to add up to 21, and then calculate the probability of each combination and the number of distinct permutations of each combination (to represent the different sequences in which each set of seven cards can be dealt.)
I don't think this would be very hard.
Quote: tooncestdcI've seen the odds of a dealer BJ as about .000024. And I've seen returns of 7-card 21 bonuses that paid 3-1, 5-1, etc. But let's say there was a promotion that paid 50-1 or 100-1 on a 7-card 21. How high could the chances of drawing a 7-card 21 get as you care less and less about winning your basic bet? What are common strategy changes in addition to obvious ones like hitting all 6-card 20 or less.
Curious for both regular shoe BJ as well as Spanish.
link to original post
Give me the exact rules of the promotion and who is running it and I'd be happy to provide a full strategy.
5 cards = 1 in 67
6 cards = 1 in 280
7 cards = 1 in 1594
This was just a 10-second simulation. I'll let it run a few hours to see how it changes things.
Quote: WizardI'm in a good mood today and will go a little ways with this, not knowing the details. Based on a quick simulation, there are probabilities of reaching 5 to 7 cards, with a goal of getting to 21 or busting:
5 cards = 1 in 67
6 cards = 1 in 280
7 cards = 1 in 1594
This was just a 10-second simulation. I'll let it run a few hours to see how it changes things.
link to original post
Make sure your simulations are treating an Ace as one and not eleven so you can continue to draw more cards. If you have a six card 21 using an ace as eleven obviously you can dar another card with a good chance of getting 500-1.
No of cards | BUST | 21 | |
3 | .382 339 553 937 | .078 288 575 330 | 1 in 12.773 |
4 | .306 711 949 862 | .042 015 335 597 | 1 in 23.801 |
5 | .128 728 524 373 | .015 068 961 709 | 1 in 66.362 |
6 | .034 456 304 362 | .003 681 935 622 | 1 in 271.596 |
7 | .006 737 944 102 | .000 681 960 340 | 1 in 1 466.361 |
8 | .001 034 753 232 | .000 099 150 367 | 1 in 10 085.691 |
9 | .000 127 998 499 | .000 011 927 013 | 1 in 83 843.289 |
10 | .000 015 125 655 | < Hands not resolved |
Guessing that 1 hand in 12 with 10+cards makes 21, this makes the total about .000 794 298 191 or 1 hand in 1259 being a 7-card+ 21.
Quote: DRichMake sure your simulations are treating an Ace as one and not eleven so you can continue to draw more cards. If you have a six card 21 using an ace as eleven obviously you can dar another card with a good chance of getting 500-1.
link to original post
I am. I even have the player hitting a blackjack, because he only has 11. The simulation absolutely always hits to 21, even with a 10-10, where there is no hope of a 5+ card 21.
Here are the results of a longer simulation. The "cards" shows the number of cards in the 21-point hand.
Cards | Count | Probability | Inverse |
---|---|---|---|
Bust | 52,104,124,978 | 0.859949 | 1.16 |
3 | 4,759,037,984 | 0.078545 | 12.73 |
4 | 2,557,594,660 | 0.042212 | 23.69 |
5 | 908,819,311 | 0.015000 | 66.67 |
6 | 216,326,234 | 0.003570 | 280.09 |
7 | 38,049,196 | 0.000628 | 1,592 |
8 | 5,220,188 | 0.000086 | 11,607 |
9 | 572,119 | 0.000009 | 105,904 |
10 | 50,292 | 0.000001 | 1,204,760 |
11 | 3,487 | 0.000000 | 17,375,910 |
12 | 192 | 0.000000 | 315,571,868 |
13 | 14 | 0.000000 | 4,327,842,761 |
Total | 60,589,798,655 | 1.000000 |
p.s. My simulation is based on six decks.
195,707,534,756,955,275,874 / 247,064,529,073,450,392,704,413
or 1 / 1262.417.
This assumes (a) you hit on all hands, (b) you do not double or split, and (c) if you are dealt a blackjack, you are allowed to treat it as an 11 and hit.
Quote: ThatDonGuyUsing an infinite deck, I get an exact probability (which matches simulation) of getting 21 with 7 or more cards as:
195,707,534,756,955,275,874 / 247,064,529,073,450,392,704,413
or 1 / 1262.417.
This assumes (a) you hit on all hands, (b) you do not double or split, and (c) if you are dealt a blackjack, you are allowed to treat it as an 11 and hit.
link to original post
I would assume that splitting bigger cards would be correct.
Quote: DRichQuote: ThatDonGuyUsing an infinite deck, I get an exact probability (which matches simulation) of getting 21 with 7 or more cards as:
195,707,534,756,955,275,874 / 247,064,529,073,450,392,704,413
or 1 / 1262.417.
This assumes (a) you hit on all hands, (b) you do not double or split, and (c) if you are dealt a blackjack, you are allowed to treat it as an 11 and hit.
link to original post
I would assume that splitting bigger cards would be correct.
link to original post
DRich,
Good point! To have ANY chance of a 7-card 21, you would have to split 9's and all 10's.
8-8 has the chance of drawing 5 consecutive Aces, but I would guess that splitting gives a better chance of reaching the goal.
7-7 is probably also a split: do you have a better chance of drawing 5 cards totaling 7, or two chances at 6 cards totaling 14?
On the low end, I would guess that splitting Aces, deuces, 3's, and 4's would be incorrect.
5's and 6's are tough to guess.
Dog Hand