# 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.

• If you have seen only 7,8,9's the count hasn't changed, but the odds do change, so it's an incomplete and somewhat primitive system. Commented Jun 16, 2018 at 9:15
• The count does not tell us how many ten-valued cards remain (not to speak of the number of remaining cards , we also would have to know). So, I don't think the probability that a ten-valued card appears can actually be calculated. Commented Jun 16, 2018 at 9:20
• I agree it is primitive, but the odds in blackjack are so close it only takes a small advantage to push them into the players favor. The count is definitely incomplete information, but having the information that there have been 5 more low cards seen than high cards (and an unknown number of 7-9's), seems like information that I could use to adjust the probability calculations. Commented Jun 16, 2018 at 9:21
• @Peter you can't calculate the exact number of winning cards left, but you can calculate the distribution of that number given the information you have, which gives you an answer. In the same way if you know nothing you can't calculate how many winning cards are left, but it is still meaningful to say the probability of a winning card is $16/52$. Commented Jun 19, 2018 at 9:45
• @EspeciallyLime But we still have the problem of not knowing how many cards remain in the deck. And this number of cards , of course , influences the probability that the next card is ten-valued. Commented Jun 19, 2018 at 17:25

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 $$\left(\frac{4}{5}\right)\left(\frac{45}{104}\right)=\frac{9}{26}\approx\,0.3461538462$$

• I think it should be $47-2c$ rather than $47-c$. Also, OP asked about the probability of a T, J, Q or K, but a high card could be an A. Commented Jun 19, 2018 at 9:42
• Thanks. A few fixes are needed. I'll bring it back soon. Commented Jun 19, 2018 at 9:51
• Now fixed, I think. Commented Jun 19, 2018 at 10:25