One of these charts seems totally wrong

[Cplcam]

The chart at:
http://wizardofodds.com/blackjack/appendix9-6dh17r4.html

Shows for example: when the dealer has an Ace up and you had an 2,Ace and if you didn’t hit you would expect to lose 59.7220% which means on the other hand the dealer would have to be busting 40.278% with an Ace up (59.7220 + 40.278 = 100% no pushes in that hand)

But if you look at the chart produced at:

http://www.qfit.com/CVDPC.htm

It shows the dealer with an Ace up will bust only about 14% of the time (wither it’s one deck or 6 deck).

So there is a HUGE difference between the dealer having an Ace up and busting with one chart at 14% and the other chart showing 40%

Lets use that 6 deck shoe chart / DB any 1st 2 cards / Dealer hits soft 17 / as the example

Which chart is correct? What is the correct percentage?

 

[shadroch]

Without consulting any charts, I believe a dealer will bust approx. 17% of the time with a first card Ace.
I'm sure someone will come around and copy someone else's work, add all sort of pretty colors but say pretty much the same thing.

 

[QFIT]

These are very different tables. Mike's tables exclude BJs. You can tell because the line for dealer A and player AT has an advantage of 1.5. Also, his chart does NOT say you will lose 60% of hands. It says you have an EV of -.60. Quite different. If you lost 60%, that means you would win 40% for an EV of -.20 not -.60.

 

[Cplcam]

When the chart says if you stand with your >Ace / two< against the dealers ace you can expect to lose -0.597220 of every dollar bet which is the same as saying you would lose 59.7220% of your bet. So the flip side to that hand would be the dealer busts 40.278% with >Ace / two< against the dealers ace (59.7220% + 40.278% = 100%)

 

[Kasi]

When the chart says if you stand with your >Ace / two< against the dealers ace you can expect to lose -0.597220 of every dollar bet which is the same as saying you would lose 59.7220% of your bet. So the flip side to that hand would be the dealer busts 40.278% with >Ace / two< against the dealers ace (59.7220% + 40.278% = 100%)

The -.59 assumes, I think, the dealer has already checked for BJ. It sounds like that what you mean too since you'd never get the chance to stand or hit your A,2.

The dealer will bust 20.13% of the time if you don't count when he has a BJ.
So, roughly, after you play this hand 10 times, 2 times you will win a dollar. 8 times you will lose your dollar (his total will be a 17 thru 21).

So you end up 6 units down after betting 10 units. Hence the -.59 ( say 6) .... EV.

Hope that helps lol.

 

[QFIT]

When the chart says if you stand with your >Ace / two< against the dealers ace you can expect to lose -0.597220 of every dollar bet which is the same as saying you would lose 59.7220% of your bet. So the flip side to that hand would be the dealer busts 40.278% with >Ace / two< against the dealers ace (59.7220% + 40.278% = 100%)

No, losing 60% of every dollar does not mean you lose 40% of those hands. Because you win the other hands and that offsets part of the overall EV. Suppose you lost 50% of hands and won 50% of hands. Does that mean you have an EV of -.5?

 

[KenSmith]

When the chart says if you stand with your >Ace / two< against the dealers ace you can expect to lose -0.597220 of every dollar bet which is the same as saying you would lose 59.7220% of your bet. So the flip side to that hand would be the dealer busts 40.278% with >Ace / two< against the dealers ace (59.7220% + 40.278% = 100%)

Let me try...

Expected loss by standing with (A,2) vs Ace = -0.597220

That means p(Dealer Busts) - p(Dealer does not bust) = -0.597220
We know the two probability numbers must sum to 100%.
Let's add p(DB) + p(DNB) to the first side and 1.00 to the second side.

We get 2 X p(DB) = 0.40278
Divide both sides by 2 and we see that the p(DB) = 0.20139

Yep, the dealer Ace busts 20.13% of the time in 6 Decks H17.
See the charts here:
http://www.blackjackinfo.com/bjtourn-dealercharts.php

 

[Cplcam]

Thanks KenSmith yours was the clear explaination

I see what you mean

Put in the simplest terms:

For every $1 you bet if you were to stand on the A,2 againest the A

In the long run for every $1 bet ~ You win from the Dealer busting .20139 cents (20.139% of the $1) and lose 0.79861 cents (79.861% of the $1) so you take the -0.79861 cents you lost and and the .20139 cents you won:
-0.79861 + .20139 = -0.59722 cents average loss on each $1 bet

 

[Kasi]

Yep, the dealer Ace busts 20.13% of the time in 6 Decks H17.

So you say.

But I said "The dealer will bust 20.13% of the time if you don't count when he has a BJ."

You just say it so much better and clearer than I do :grin:

And, actually, you do :grin: