BlackJack Card Counting Probabilities

Question

Note: This is related to this question but I'm going to frame the problem in a completely different way (hopefully a more productive way).

Lets say we have a 2-deck blackjack game, half of the cards have been seen, 52 remain unseen. We know that we have seen 5 more low cards than high cards (low cards = 2/3/4/5/6, high cards = 10/J/Q/K/A, neutral = 7/8/9).

Ultimately I'm trying to determine the probability that the next card dealt will be a high card.

There are 104! possible shuffles of 2 decks of cards, but in this case we don't care about suits, or even the specific card value, we just care about Hi/Lo/Ne(utral)

Taking that into account the number of possible shuffles is 104! / (40!*40!*24!) - which still equals a very huge number (2.5e+46) - too big for me to write code to brute force this problem.

Problem 1 - How many of those possible shuffles results in 5 more low cards than high cards in the first 52 cards?

Problem 2 - What % of those have 17 high cards remaining? 18? 19? etc (there are somewhere between 17-28 high cards remaining unseen)

Answer 1

Letting $f(a,b,c)$ be the number of $52$-card subsets of the $104$-card deck consisting of

• $a$ low cards.$\\[4pt]$
• $b$ neutral cards.$\\[4pt]$
• $c$ high cards.

with cards of the same type (low, neutral, high) regarded as indistinguishable, we get $$f(a,b,c)={\small{\binom{40}{a}\binom{24}{b}\binom{40}{c}}}$$ and the number of such subsets such that $a=c+5$ is $$\sum_{c=12}^{23}f(c+5,47-2c,c)$$ which sums to $$75905105572808453675031567360\approx 7.590510557{\times}10^{\,28}$$ For $17\le x \le 28$, let $p(x)$ be the probability that there are at least $x$ high cards remaining in the $52$ undealt cards, given that there were $5$ more low cards than high cards in the first $52$ cards.

Then we get $$ p(x)=\frac {{\displaystyle{\sum_{c=12}^{40-x}f(c+5,47-2c,c)}}} {{\displaystyle{\sum_{c=12}^{23}f(c+5,47-2c,c)}}} $$ which yields the values \begin{align*} p(17)&=1\\[4pt] p(18)&=\frac{\;81821539442\;}{\,81821552267}\approx \,0.9999998433\\[4pt] p(19)&=\frac{\;1063641316396\;}{\,1063680179471}\approx \,0.9999634636\\[4pt] p(20)&=\frac{\;81678973942\;}{\,81821552267}\approx \,0.9982574478\\[4pt] p(21)&=\frac{\;79610267071\;}{\,81821552267}\approx \, 0.9729742942\\[4pt] p(22)&=\frac{\;19423037681\;}{\,23377586362}\approx \,0.8308401637\\[4pt] p(23)&=\frac{1}{2}\\[4pt] p(24)&=\frac{\;3954548681\;}{\,23377586362}\approx \,0.1691598363\\[4pt] p(25)&=\frac{\;2211285196\;}{\,81821552267}\approx \,0.0270257058\\[4pt] p(26)&=\frac{\;142578325\;}{\,81821552267}\approx \,0.0017425522\\[4pt] p(27)&=\frac{\;38863075\;}{\,1063680179471}\approx \,0.0000365364\\[4pt] p(28)&=\frac{\;12825\;}{\,81821552267}\approx \,0.0000001567\\[4pt] \end{align*}