blackorange
blackorange
Joined: Sep 17, 2010
  • Threads: 8
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September 28th, 2010 at 8:32:28 AM permalink
What is the odds/% of a dealer standing on 6 full shuffled decks? 17 18 19 20 21?

Imagine: one dealer, no players. Play to stand. Odds. Please.

cheers

J
PapaChubby
PapaChubby
Joined: Mar 29, 2010
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September 28th, 2010 at 10:34:25 AM permalink
Based on the Wizard's blackjack appendix 2B, I think the answer is 71.8% if the dealer stands on soft 17, and 71.4% if the dealer hits a soft 17.
blackorange
blackorange
Joined: Sep 17, 2010
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October 3rd, 2010 at 8:12:04 AM permalink
Thanks- interersting result. I would have thought that there are 5 standing stations, and 11 values. My initial thought would be that only 5 in 13 would 'get in the pond', the other 6 values throwing long or short.

What would the result be for instance if there was only one's to 11, and only one 10. Would it then rough out to be only 5/13 landing in the pond?
PapaChubby
PapaChubby
Joined: Mar 29, 2010
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October 3rd, 2010 at 4:58:33 PM permalink
No. Your 5/13 hypothesis assumes a uniform distribution, in which each outcome is equally likely. Like rolling a die, and the probability of rolling a 6 is one in six. However, the blackjack result will be arrived at by adding multiple cards together. The result is a more complicated distribution function. Like rolling two dice in craps, where 7 is a much more likely result than 2 or 12. Off the top of my head, I'd estimate the probability of standing in your game to be somewhere around 55-60%.

The only time your 5/13 probability would be true is if your had was already valued at 16, and you were drawing an additional card to get into the "stand zone". And now you're really only talking about 5/10 as well, 'cause you removed all the face cards.

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