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"What is the probability that the Dealer ends up with a 17 in BlackJack?"

According to this website, the number is 14.58% https://www.blackjackonline.com/strategy/odds/

For those unfamiliar with the rules of BlackJack for the dealer, the dealer will flip over cards until they total at least 17. If the dealer's hand is less than 17 they flip over another card.

For approaching this problem I first thought you could use the Stars and Bars theorem to your advantage, detailed in this post Rolling dice such that they add up to 13 — is there a more elegant way to solve this type of problem?

And also more details here: https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

Essentially, the goal of the stars and bars theorem is to count the number of tuple sets (order counts) that add up to a specific number $n$ from a choice of $k$ positive integers.

For instance, if you would like to know the number of permutations of summing to the number 3 using two positive integers, the answer would be $2$. Namely $(1,2),(2,1)$. This is done by computing

$(n-1) \choose (k-1)$

Unfortunately in this case I don't know how much use it is. When counting the number of ways to reach 17 one might consider the different possible ways of summing to 17 by using integers (which would be represented by cards) but $k$ is variable because the dealer could reach 17 by getting a $10$ and a $7$ but can also reach $17$ by getting two $5$'s and a $7$.

Does anyone have a suggestion as another approach to use?

David G. Stork
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user699279
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    It is worse than that. We also in this situation are pulling from a single deck without replacement, so the outcomes of one card are dependent on the next card (mitigated somewhat if we add an additional finite number of decks). Furthermore, you have aces counting as either $1$ or as $11$ and you have many more ways of getting $10$ due to the face cards and so $10$'s occur more frequently than other numbers. Taking all of this into account, I do not see a more convenient approach than massive case-work, brute force, or simulation. – JMoravitz Aug 05 '20 at 17:33
  • I'm starting to feel that way as well. I think one way I was able to do it by calculating the ways to reach the value of 3 with 52 cards:

    You can reach it by using 3 (one's): so (4 choose 3)/(52 choose 3)

    • 1 One and 1 Two: (4 choose 1 + 4 choose 1)/(52 choose 2)
    • 1 Three: (4 choose 1)/(52 choose 1)

    But that becomes increasingly more difficult. The other way is using Marchov chains I imagine, but again not exactly elegant

     – user699279 Aug 05 '20 at 17:38
  • "4 choose 1 + 4 choose 1" That should be a multiplication, not an addition. And yes, this is able to be described by markov chains... however if phrased inefficiently it could have upwards of a few million different states (and phrased efficiently, not a whole lot better) and so is not really a feasible approach. – JMoravitz Aug 05 '20 at 17:52
  • Yes, my mistake. This person seem to fair well though: https://blogs.sas.com/content/operations/2016/06/20/computing-an-optimal-blackjack-strategy-with-sasor/ – user699279 Aug 05 '20 at 18:10

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