This is my second try, but I think is a MUCH better argument than what I had before. So I deleted the more complicated answer I tried to give earlier.
The total # of possible deals to the two players is: $$\binom{52}{2}\binom{50}{2}=1,624,350$$
(i.e., choose two cards for the first player and then two for the second.)
Then, the # of ways to deal a blackjack to both players would be: $$\binom{4}{1}\binom{16}{1}\binom{3}{1}\binom{15}{1}=2880$$
(i.e, choose which ace to give to the first player, then which 10,J,Q,K for the first player, then which ace for the second player, and then which 10,J,Q,K for the second player.)
Also, the # of ways that exactly one player gets a blackjack would be: $$\binom{2}{1}\binom{4}{1}\binom{16}{1}\times(\binom{3}{2}+\binom{3}{1}\binom{32}{1}+\binom{47}{2})=151,040$$
(i.e., choose which player to give the blackjack, then choose which ace to give them, and then which 10,J,Q,K to give them. Then for the other player we either give them two aces from the 3 that are left, give them one ace from the 3 that are left and 1 of the 32 2-9 cards, or give them two non-aces from the 47 non-aces that are left.)
So, the probability of at least one player getting a blackjack is: $$\frac{2880+151,040}{1,624,350}$$
Hence the probability of neither player getting a blackjack is: $$1- \frac{2880+151,040}{1,624,350} \approx 90.5\%$$
The number of ways, in a deck of $n$ cards, to win this game (out of the $n!$ total ways for the deck to have been shuffled) is given in OEIS as sequence A144188.
Playing with only 13 cards your chances of winning are about 5.246%.
Playing with half a deck, yet using suit order to break ties, your chances are down to 0.095%. What I find amazing is that this is much less than the square of the chances for a 13-card deck (which we would expect because it involves 25 guesses and the 13-card deck involves only 12) and is even 1.6 times less than the square of the winning chances with a 14 card deck and is less than the chances of winning two consecutive games, with a 15 card deck and a 14 card deck.
For large $n$ the chances of winning drop off by about a factor of 0.737 for each additional guess needed.
Best Answer
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
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 $$ \left(\frac{4}{5}\right)\left(\frac{45}{104}\right)=\frac{9}{26}\approx\,0.3461538462 $$