If a dealer starts with a two card 12...

[Infinite Monkey]

... is he more likely to bust or to make a hand?

If a dealer starts with a two card soft 17 -- is she more likely to bust or to make a hand?

Anyone know?

 

[jack.jackson]

... is he more likely to bust or to make a hand?

If a dealer starts with a two card soft 17 -- is she more likely to bust or to make a hand?

Anyone know?

Im just spitballin here, but this should be pretty close!

1.Dealer will make a pat hand 7/13 times when he draws to hard 12.

2. Dealer will make a pat hand 10/13 when he draws to S17.

What I can say for sure is there both greater than 50%

 

[Infinite Monkey]

Sorry for saying so, I don't think that's right.

It seems to me that 9 out of 13 times the dealer won't bust on the next card and out of those 9 times, 5 cards will give him a pat hand, not 7. Four will give him another stiff (A, 2, 3, 4 --> 13, 14, 15, and 16). And for those four stiff hands, they each have a set of cards that will bust them, make them pat, or create of them another stiff (except for 16, which can't be a stiff any more).

And then the resulting stiff hands face more cards, and so on.

So the question is more complicated.

I suppose I could write up all potential occurrences by hand and count up the total number of situations the dealer busts vs. the total number of situations in which the dealer makes a hand and then I'll know. I just thought some smart person out there might know how to do the math more easily.

I'd also love to know the answer for a soft 17 (A6). As well as for the dealers first two cards when they add up to 3 (a 2 and an A), 4, 5, 6, and 7.

My guess is that, like for 14, 15, 16, the 4, 5 and 6 will make the dealer more likely to bust. Not sure about the 7 or the 2 or the 3 though.

Anyone?

 

[Canceler]

I just thought some smart person out there might know how to do the math more easily.

This is entirely possible. People around here know how to do some amazing things.

While we're waiting, though...

You realize that your questions are just "curiosity" questions, right? Knowing the answers won't help you play blackjack any better, because, except for holecarders, once you know which two cards the dealer has, it's too late to do anything about it.

 

[Kasi]

Well, if it's an S17 game, I think I know the answer :grin:

I think I know a dealer will bust more often in an H17 game than a S17 game. Yet H17 benefits the house. Go figure.

As to the rest, I ain't nearly smart enuf.

Although I'd love to know the frequencies that you're asking about.

Or, even why you're asking about the chances of a dealer busting on his 2-card totals.

 

[bj bob]

Well, if it's an S17 game, I think I know the answer :grin:

I think I know a dealer will bust more often in an H17 game than a S17 game. Yet H17 benefits the house. Go figure.

QUOTE]
I do believe that in S-17, although the dealer does bust more often on one hand; however , on the other hand ,the hands in which he does not bust end up stronger , as in the player's BS of always hitting on A-6 across the board ,which outweighs the bust ratio overall.
Hope you guys can sift through all of my subordinate clauses today since English hasn't been my primary language since the Fiesta Friday night.

 

[Geoff Hall]

Infinite Deck Calculation

I've just used an infinite deck (as it makes for easier probabilities ) and it's actually quite close to 50/50 unless I've made a mistake with my calculations.

I just did the dealer starting with 12 as this was much easier than the soft 17.

The ways that a dealer can bust if on 12 are (brackets are the cards that can produce the 'bust'):-

10
A (9,10) , A,A (8,9,10) , A,A,A (7,8,9,10) , A,A,A,A (6,7,8,9,10)
A,2 (7,8,9,10) , A,A,2 (6,7,8,9,10) , A,2,A (6,7,8,9,10) , A,3 (6,7,8,9,10)
2 (8,9,10) , 2,A (7,8,9,10) , 2,2 (6,7,8,9,10) , 2,A,A (6,7,8,9,10)
3 (7,8,9,10) 3,A (6,7,8,9,10)
4 (6,7,8,9,10)

= (assuming infinite deck) :-

4/13 + 1/13.5/13 + 1/13.1/13.6/13 + ... + 1/13.8/13 (I'm not writing them all out)

If I get time I'll check my calculations but I make the probability of a dealer busting with 12 as 48.6%.
This goes against my initial thoughts as I assumed it would be higher than 50%.
As I said, I'll check through my work but it's getting a little late now.

 

[SilentBob420BMFJ]

I've just used an infinite deck (as it makes for easier probabilities ) and it's actually quite close to 50/50 unless I've made a mistake with my calculations.

I just did the dealer starting with 12 as this was much easier than the soft 17.

The ways that a dealer can bust if on 12 are (brackets are the cards that can produce the 'bust'):-

10
A (9,10) , A,A (8,9,10) , A,A,A (7,8,9,10) , A,A,A,A (6,7,8,9,10)
A,2 (7,8,9,10) , A,A,2 (6,7,8,9,10) , A,2,A (6,7,8,9,10) , A,3 (6,7,8,9,10)
2 (8,9,10) , 2,A (7,8,9,10) , 2,2 (6,7,8,9,10) , 2,A,A (6,7,8,9,10)
3 (7,8,9,10) 3,A (6,7,8,9,10)
4 (6,7,8,9,10)

= (assuming infinite deck) :-

4/13 + 1/13.5/13 + 1/13.1/13.6/13 + ... + 1/13.8/13 (I'm not writing them all out)

If I get time I'll check my calculations but I make the probability of a dealer busting with 12 as 48.6%.
This goes against my initial thoughts as I assumed it would be higher than 50%.
As I said, I'll check through my work but it's getting a little late now.

but are you counting the fact that the dealer may go from A6 to A6A, A62, A63, or A64? those would all count as not busting, so you have to figure that in, so 4 cards can give the dealer a pat hand, so 4/13 must be figured in there somehow, i dont feel like thinking about it tho, im going to bed

 

[Geoff Hall]

I only did the calculation for Hard 12.

I agree, Soft 17 would be far more laborious but is easy to see that there is a far greater chance that the dealer will make a hand rather than 'bust' as there are 8 out of 13 cards that the dealer can get to make a hand with 1 shot.

 

[k_c]

... is he more likely to bust or to make a hand?

If a dealer starts with a two card soft 17 -- is she more likely to bust or to make a hand?

Anyone know?

Dealer's probabilities on a hard 12 or soft 17 (assuming soft 17 is hit) will depend upon the composition of the shoe from which cards are drawn.

The probabilities for an infinite deck where ranks of 1-9 each make up 1/13 of the deck and tens 4/13 the probabilities are:

drawing to hard 12:
17) 0.10346545719957 
18) 0.10346545719957 
19) 0.10346545719957
20) 0.10346545719957
21) 0.10346545719957
bust) 0.482672714002149

drawing to soft 17:
17) 0.342193569291845 
18) 0.111424338522614 
19) 0.111424338522614 
20) 0.111424338522614 
21) 0.111424338522614
bust) 0.212109076617699

Sorry about all the decimal places. I just pasted the results without rounding. If you give me a finite shoe comp, I can give results for that.

k_c

 

[Infinite Monkey]

Wow, Thank you all these efforts to answer my question. I really appreciate it. K_C, you say you can do this for various deck compositions? Can you do it for single deck, double deck and six decks? If I'm asking too much of you or your time, feel free to punt the request and thanks again for the info.

~ I.M.

 

[k_c]

Wow, Thank you all these efforts to answer my question. I really appreciate it. K_C, you say you can do this for various deck compositions? Can you do it for single deck, double deck and six decks? If I'm asking too much of you or your time, feel free to punt the request and thanks again for the info.

~ I.M.

Basically, any comp is fine as long there are enough cards in the shoe to finish the round.

k_c