What are the odds in blackjack of a dealer having 20 when a only a face card is showing?

Question

Please take into account the number of decks being used. I would think that the more decks, the less likely and probabliity will be close in the short term, but still. I am trying to better understand my odds when I have a sum of bust cards (12-16) and the dealer has a face card!?! I think this is a great question and I would love it if a math wiz could help. I have let the dealer bust on occasion, when I should be hitting (according to the card) and had luck. The problem is wondering how often probability would guide me to try this technique.

Answer 1

The number of decks doesn't alter the odds of a face card being the dealer's hole card.  there are 4 pictures out of 13 cards, and if you take A, 7,8 and 9 into account that's 8 cards out of 13 which complete the dealers hand without the dealer being able to draw again.

so the odds of a realer making a hand on a picture are 1.75-1,

and 3.25-1 of hitting a picture alone.

Answer 2

In order for the dealer to have 20, his hole card must be a card valued at 10. (Since you asked about him "having 20" instead of "ending up with 20" I am assuming you are not including the cases where the dealer draws additional cards and ends up with 20.)

There are 16 cards valued at 10 per deck, and 52 cards per deck. You know one of those cards is dealer's up card. that makes the odds for N decks

(16N - 1) / (52N -1)

for 1 deck that is 15/51 = 0.294 = 29.4%

for 2 decks that is 31/103 = 0.301 = 30.1%

for 3 decks that is 47/155 = 0.303 = 30.3%

for 4 decks that is 63/207 = 0.304 = 30.4%

...

for 8 decks that is 127/415 = 0.306 = 30.6%

Answer 3

In a 6D game

Standing on stiffs (12-16) against a dealer's 10-value card (face) has an expectancy of 54% loss. Hitting those stiffs has an expectancy of 38% loss.

Similar figures apply to a 1D game.

I have used the card counters' bible (Wong) as my reference, but you can work it out roughly in your head. Consider this: the dealer is showing a 10-value card against your tiny hand. The dealer's 2nd card will be one of the following, A K Q J 10 9 8 7 6 5 4 3 2. Eight of those card will instantly beat you. The 5 that don't beat you 6 5 4 3 2 require the dealer to draw again. The dealer is likely to win 2 and lose 3 of those 5 secondary draws. To summarise: Of the 13 cards that the dealer has as a 2nd card, 10 will turn into winners and 3 become losers. Net result: 7 losses in those 13 cards = 7/13 = 53.8%

So, here, we are not talking about a winning situation, but what to do to minimise losses after having drawn bad cards. You should always take a card.

If you are counting cards and have 16 v 10 and the deck is slightly depleted in small cards, the loss expectancies get worse, but not equally, and in this case stand on 16 is the better play.

My BJ and card counting days go back 36 years.The only time I have ever left cards for the dealer to bust was when a particular dealer used to draw the next card out of the shoe too quickly and accidently disclose it. I won a lot off her but, unfortunately, she gave up the job.

I think it is crazy trying to guess what the dealer has as a hole-card and then trying to guess that the next random card out of the shoe wont help your pathetic hand but will hurt the dealer. Forget it! and play by the book!

Answer 4

If he's showing 1 of 16 cards possible,  his next card will also be 1 of 15 cards possible, minus his first card. the odds are implicit if there are multiple decks, assuming you are unable to count all the possible cards that have been discarded or in use.

so, it ends up being 16/52 x 15/51, if there are NO other cards in play and it is simple the dealer by himself. the probability will vary according to how the hands are distributed (if you see 6 face cards with stands, then the likelihood the dealer has a face card would probability variably go down). variance is probably an issue here.

4/13 x 5/17 = .090 = 9% of the time