# Casino Black Jack Probabilities (Non-Blackjack $21$s)

So when I look online for this everything wants to talk about just BlackJacks themselves but I want to find the probability of any player getting $$21$$ after having hitting $$N$$-times when playing vs a casino with $$X$$-deck shoes. For simplicity I only need to know the values with $$n$$ is $$1, 2$$, or $$3$$ and when $$X$$ is $$6, 7$$ or $$8$$. But a generalized formula would be great.

For a baseline we can assume just the player vs dealer but if having more players affects the odds of any player getting this I would be intrigued to know how to account for that. Max players of $$7$$.

If this is easier to do with code instead of formulas that works for me as well.

• More players does not affect this at all. Knowledge of other players' cards however does. This is a highly tedious question with a lot of case work. You can do better by considering an infinite shoe instead and using that as an approximation. To do so, look at the $x^{21}$ coefficient of the expansion of $(x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+4x^{10}+x^{11})^{n+2}/13^{n+2}$ noting that this will be a slight overestimation, not only because of the infinite shoe but also because we allowed aces to appear twice in the above in both respective roles. Sep 1, 2020 at 22:29
• That said, brute force should be feasible... We look at all partitions of $21$ into $n$ parts, all of size $11$ or less, there are only so many. Looking at partitions instead of composition helps reduce the number of cases. We have $49$ different cases for $n=4$. For each case, decide which deck and suits are used for each respective rank (as well as which face card in the case of 10's). For instance for the $11+4+4+2$ case there are $24\times \binom{24}{2}\times 24$ different four-card hands (order not mattering) out of the $\binom{52\times 6}{4}$ total possible hands in this case. Sep 1, 2020 at 22:39