dealer six card auto win

beyondbj

Well-Known Member
#1
sorry , my topci title is wrong , i should have asked the following rule's house edge

the dealer get six card should be a stop , not a win , he will still compare with the players card after he stopped


1. one deck

2. bj 1:1

3. DAS

4 no respilt

5 only 10 and 11 double allowed

6. six card charlies

7. dealer also hit up to six card to stop (even its below 17)

8. dealer have hole card

9. no surrender

10 S17


( there is no counting can do since its an electronic bj , where I dont know when will it shuffle or it actually shuffled each time)
1. one deck

2. bj 1:1

3. DAS

4 no respilt

5 only 10 and 11 double allowed

6. six card charlies

7. dealer also hit up to six card to stop (even its below 17)

8. dealer have hole card

9. no surrender

10 S17


( there is no counting can do since its an electronic bj , where I dont know when will it shuffle or it actually shuffled each time)
 

Dog Hand

Well-Known Member
#3
Tough... I've already done the calcs!

beyondbj,

You asked what the change in the Basic Strategy EV would be for a BJ game in which any unbusted dealer's hand containing 6 cards would be an automatic winner for the dealer. You failed to specify the house rules and conditions, so I arbitrarily chose a SD, H17, D10, NoDAS game for a heads-up B.S. player playing one spot and getting Rule of 1. I also assumes that a player's BJ would automatically win, so the dealer would NOT attempt to draw a 6-card hand to beat him.

Now the change in BS EV consists of two terms: pushes (under "normal" rules) that the player loses, and wins (under "normal" rules) that the player loses. For example, if the player has 17, and the dealer draws to a 6-card 17, then the player loses, instead of pushes. This costs him one bet. Similarly, if the player has 17 and the dealer draws to a 6-card 15, and the next card busts the dealer, then the player loses a hand that he would have won: this costs him two bets.

Below is the output compiled from a half-dozen CVData sims, each of 400-million rounds, using "normal" BJ rules. Column 1 shows the player's hand, ranging from "Stiff" to 21 (non-BJ). These groups correspond to the half-dozen CVData runs... they are grouped according to the player's total we are considering. The next two columns show the dealer's hand: Column 2 is for when the dealer completes her hand on card 6 (so four hit cards); Column 3 is for when more than 6 cards are required. In each case, the dealer's result is shown: either a total of 17-21, or else a "bust". Column 4 shows the total number of hands played for each of the half-dozen sims: it is more than 400-million due to splits. Column 5 shows the total number of times the indicated result was achieved. For example, if we look down to a player's total of 19, we see that, when the player totaled 19:

  1. the dealer drew to a 6-card 19 10,832 times
  2. the dealer drew to a more-than-6-card 19 558 times
  3. the dealer drew to a 6-card hand with a total of 18 or hard 17 19,081 times (not a soft 17, since I assumed the game is H17)
  4. the dealer drew to a more-than-6-card bust 5,146 times.

Ok... for your hypothetical 6-card-autoloser game, the player would have lost every one of these 4 cases. Since under "normal" rules, he would have pushed under cases 1 and 2, each of those losses cost him 1 unit (shown as "-1" in Column 7); and since he would have won under cases 3 and 4, each of those losses cost him 2 units (shown as "-2" in Column 7).

Column 6 is the ratio of Column 5 to Column 4, expressed as percentage of outcomes. Column 8 is the product of Columns 6 and 7, and shows the change in BS EV for each of these cases. Since they are all independent, we sum Column 8 to find the answer:

The BS EV of the game is decreased by just over 0.07%.

Hope this helps!

Dog Hand

HTML:
Player	Dealer	Dealer		Hands	"Wins"	Percent	Result	Product
	6-card	>6 cards					
Stiff		bust	407,469,536	16,872	0.00414% -2	-0.00828%

17	17		407,470,484	9,385	0.00230% -1	-0.00230%
		17			451	0.00011% -1	-0.00011%
		bust			5,219	0.00128% -2	-0.00256%

18	18		407,475,224	9,401	0.00231% -1	-0.00231%
		18			470	0.00012% -1	-0.00012%
	<18				8,879	0.00218% -2	-0.00436%
		bust			5,007	0.00123% -2	-0.00246%

19	19		407,480,408	10,832	0.00266% -1	-0.00266%
		19			558	0.00014% -1	-0.00014%
	<19				19,081	0.00468% -2	-0.00937%
		bust			5,146	0.00126% -2	-0.00253%

20	20		407,471,208	16,670	0.00409% -1	-0.00409%
		20			884	0.00022% -1	-0.00022%
	<20				43,009	0.01056% -2	-0.02111%
		bust			7,378	0.00181% -2	-0.00362%

21 noBJ	21		407,471,420	2,478	0.00061% -1	-0.00061%
		21			137	0.00003% -1	-0.00003%
	<21				8,495	0.00208% -2	-0.00417%
		bust			992	0.00024% -2	-0.00049%

					%Tie	0.01258% DelEV	-0.07152%
					%Win	0.02947%
 

nightspirit

Well-Known Member
#4
Dog Hand said:
beyondbj,

You asked what the change in the Basic Strategy EV would be for a BJ game in which any unbusted dealer's hand containing 6 cards would be an automatic winner for the dealer. You failed to specify the house rules and conditions, so I arbitrarily chose a SD, H17, D10, NoDAS game for a heads-up B.S. player playing one spot and getting Rule of 1. I also assumes that a player's BJ would automatically win, so the dealer would NOT attempt to draw a 6-card hand to beat him.

Now the change in BS EV consists of two terms: pushes (under "normal" rules) that the player loses, and wins (under "normal" rules) that the player loses. For example, if the player has 17, and the dealer draws to a 6-card 17, then the player loses, instead of pushes. This costs him one bet. Similarly, if the player has 17 and the dealer draws to a 6-card 15, and the next card busts the dealer, then the player loses a hand that he would have won: this costs him two bets.

Below is the output compiled from a half-dozen CVData sims, each of 400-million rounds, using "normal" BJ rules. Column 1 shows the player's hand, ranging from "Stiff" to 21 (non-BJ). These groups correspond to the half-dozen CVData runs... they are grouped according to the player's total we are considering. The next two columns show the dealer's hand: Column 2 is for when the dealer completes her hand on card 6 (so four hit cards); Column 3 is for when more than 6 cards are required. In each case, the dealer's result is shown: either a total of 17-21, or else a "bust". Column 4 shows the total number of hands played for each of the half-dozen sims: it is more than 400-million due to splits. Column 5 shows the total number of times the indicated result was achieved. For example, if we look down to a player's total of 19, we see that, when the player totaled 19:

  1. the dealer drew to a 6-card 19 10,832 times
  2. the dealer drew to a more-than-6-card 19 558 times
  3. the dealer drew to a 6-card hand with a total of 18 or hard 17 19,081 times (not a soft 17, since I assumed the game is H17)
  4. the dealer drew to a more-than-6-card bust 5,146 times.

Ok... for your hypothetical 6-card-autoloser game, the player would have lost every one of these 4 cases. Since under "normal" rules, he would have pushed under cases 1 and 2, each of those losses cost him 1 unit (shown as "-1" in Column 7); and since he would have won under cases 3 and 4, each of those losses cost him 2 units (shown as "-2" in Column 7).

Column 6 is the ratio of Column 5 to Column 4, expressed as percentage of outcomes. Column 8 is the product of Columns 6 and 7, and shows the change in BS EV for each of these cases. Since they are all independent, we sum Column 8 to find the answer:

The BS EV of the game is decreased by just over 0.07%.

Hope this helps!

Dog Hand

HTML:
Player	Dealer	Dealer		Hands	"Wins"	Percent	Result	Product
	6-card	>6 cards					
Stiff		bust	407,469,536	16,872	0.00414% -2	-0.00828%

17	17		407,470,484	9,385	0.00230% -1	-0.00230%
		17			451	0.00011% -1	-0.00011%
		bust			5,219	0.00128% -2	-0.00256%

18	18		407,475,224	9,401	0.00231% -1	-0.00231%
		18			470	0.00012% -1	-0.00012%
	<18				8,879	0.00218% -2	-0.00436%
		bust			5,007	0.00123% -2	-0.00246%

19	19		407,480,408	10,832	0.00266% -1	-0.00266%
		19			558	0.00014% -1	-0.00014%
	<19				19,081	0.00468% -2	-0.00937%
		bust			5,146	0.00126% -2	-0.00253%

20	20		407,471,208	16,670	0.00409% -1	-0.00409%
		20			884	0.00022% -1	-0.00022%
	<20				43,009	0.01056% -2	-0.02111%
		bust			7,378	0.00181% -2	-0.00362%

21 noBJ	21		407,471,420	2,478	0.00061% -1	-0.00061%
		21			137	0.00003% -1	-0.00003%
	<21				8,495	0.00208% -2	-0.00417%
		bust			992	0.00024% -2	-0.00049%

					%Tie	0.01258% DelEV	-0.07152%
					%Win	0.02947%
Dog Hand, I admire the effort you put into your answers for such posts across all boards since a long time. :) There is always something to learn from it. Thank you and keep it coming ;)
 

k_c

Well-Known Member
#5
Dog Hand said:
beyondbj,

You asked what the change in the Basic Strategy EV would be for a BJ game in which any unbusted dealer's hand containing 6 cards would be an automatic winner for the dealer. You failed to specify the house rules and conditions, so I arbitrarily chose a SD, H17, D10, NoDAS game for a heads-up B.S. player playing one spot and getting Rule of 1. I also assumes that a player's BJ would automatically win, so the dealer would NOT attempt to draw a 6-card hand to beat him.

Now the change in BS EV consists of two terms: pushes (under "normal" rules) that the player loses, and wins (under "normal" rules) that the player loses. For example, if the player has 17, and the dealer draws to a 6-card 17, then the player loses, instead of pushes. This costs him one bet. Similarly, if the player has 17 and the dealer draws to a 6-card 15, and the next card busts the dealer, then the player loses a hand that he would have won: this costs him two bets.

Below is the output compiled from a half-dozen CVData sims, each of 400-million rounds, using "normal" BJ rules. Column 1 shows the player's hand, ranging from "Stiff" to 21 (non-BJ). These groups correspond to the half-dozen CVData runs... they are grouped according to the player's total we are considering. The next two columns show the dealer's hand: Column 2 is for when the dealer completes her hand on card 6 (so four hit cards); Column 3 is for when more than 6 cards are required. In each case, the dealer's result is shown: either a total of 17-21, or else a "bust". Column 4 shows the total number of hands played for each of the half-dozen sims: it is more than 400-million due to splits. Column 5 shows the total number of times the indicated result was achieved. For example, if we look down to a player's total of 19, we see that, when the player totaled 19:

  1. the dealer drew to a 6-card 19 10,832 times
  2. the dealer drew to a more-than-6-card 19 558 times
  3. the dealer drew to a 6-card hand with a total of 18 or hard 17 19,081 times (not a soft 17, since I assumed the game is H17)
  4. the dealer drew to a more-than-6-card bust 5,146 times.

Ok... for your hypothetical 6-card-autoloser game, the player would have lost every one of these 4 cases. Since under "normal" rules, he would have pushed under cases 1 and 2, each of those losses cost him 1 unit (shown as "-1" in Column 7); and since he would have won under cases 3 and 4, each of those losses cost him 2 units (shown as "-2" in Column 7).

Column 6 is the ratio of Column 5 to Column 4, expressed as percentage of outcomes. Column 8 is the product of Columns 6 and 7, and shows the change in BS EV for each of these cases. Since they are all independent, we sum Column 8 to find the answer:

The BS EV of the game is decreased by just over 0.07%.

Hope this helps!

Dog Hand

HTML:
Player	Dealer	Dealer		Hands	"Wins"	Percent	Result	Product
	6-card	>6 cards					
Stiff		bust	407,469,536	16,872	0.00414% -2	-0.00828%

17	17		407,470,484	9,385	0.00230% -1	-0.00230%
		17			451	0.00011% -1	-0.00011%
		bust			5,219	0.00128% -2	-0.00256%

18	18		407,475,224	9,401	0.00231% -1	-0.00231%
		18			470	0.00012% -1	-0.00012%
	<18				8,879	0.00218% -2	-0.00436%
		bust			5,007	0.00123% -2	-0.00246%

19	19		407,480,408	10,832	0.00266% -1	-0.00266%
		19			558	0.00014% -1	-0.00014%
	<19				19,081	0.00468% -2	-0.00937%
		bust			5,146	0.00126% -2	-0.00253%

20	20		407,471,208	16,670	0.00409% -1	-0.00409%
		20			884	0.00022% -1	-0.00022%
	<20				43,009	0.01056% -2	-0.02111%
		bust			7,378	0.00181% -2	-0.00362%

21 noBJ	21		407,471,420	2,478	0.00061% -1	-0.00061%
		21			137	0.00003% -1	-0.00003%
	<21				8,495	0.00208% -2	-0.00417%
		bust			992	0.00024% -2	-0.00049%

					%Tie	0.01258% DelEV	-0.07152%
					%Win	0.02947%
I thought the question meant that if dealer's hand consisted of 6 cards then player automatically wins. Another way of stating would be if dealer hasn't made a hand of 17-21 in 5 cards then (non-busted) player wins.

I could be wrong though.
 
#6
For player: Player six cards automatically win.

For Dealer: If dealer hit to 6 card, it will stop. If the six cards total is 14, then compares this 14 to the player's hand...the higher one win.
 

Sucker

Well-Known Member
#8
I don't need to run sims to be able to tell you that the house edge is going to be right around 3%. You're probably better off playing just about any other slot machine in the house except this one.
 
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