BlackJack Card Probability when Counting Cards In a single deck blackjack game - if you're not counting cards - the probability that the next card will be a 10/J/Q/K is 16/52.
I'm trying to figure out how to adjust the probabilities when you are counting cards.  For those that might not be familiar, a common card counting system (HiLo) works by keeping a running "count", when you see a 2-6 you add 1 to the count, when you see a T/J/Q/K/A you subtract 1 from the count.  When the count is positive it means the odds are better for the player (the remaining deck(s) is richer in high cards vs low cards).
Let say that it's a 2 deck game, and half the cards have already been dealt - 52 cards remain.  The count is +5, that means there have been 5 more low cards seen so far than high cards.  What is the probability that the next card dealt will be a 10/J/Q/K? It's gotta be more than 16/52, because we know the deck is richer in high cards based on the count.  I just don't know how to model/calculate it.

As in my answer to your later question

$\qquad$BlackJack Card Counting Probabilities

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


*

*$a$ low cards$\;(2,3,4,5,6)$.$\\[4pt]$

*$b$ neutral cards$\;(7,8,8)$.$\\[4pt]$

*$c$ high cards$(10,\text{J},\text{Q},\text{K},\text{A})$.


with cards of the same type (low, neutral, high) regarded as indistinguishable, we get
$$f(a,b,c)={\small{\binom{40}{a}\binom{24}{b}\binom{40}{c}}}$$

Given that there were $5$ more low cards than high cards in the first $52$ dealt cards, the probability that the next card to be dealt card is a high card $(10,\text{J},\text{Q},\text{K},\text{A})$ is
$$
\frac
{{\displaystyle{\sum_{c=12}^{23}f(c+5,47-2c,c)(40-c)}}}
{52{\displaystyle{\sum_{c=12}^{23}f(c+5,47-2c,c)}}}
=\frac{45}{104}\approx\, 0.4326923077
$$
and the probability that it's a $10$-type card $(10,\text{J},\text{Q},\text{K})$ is approximately 
$$
\left(\frac{4}{5}\right)\left(\frac{45}{104}\right)=\frac{9}{26}\approx\,0.3461538462
$$
