What is probability of getting 21 if the first card dealt is always a 10 valued card? I was curious to know what the probability of getting 21 is if using a standard, well shuffled 52 card deck, cards are dealt (without replacement for each hand) for a 1 player game but assuming the first card dealt is always a 10 valued card (such as 10, J, Q, or K).  Note that 21 can be had with as little as 1 additional card (an A) or as many as 7 additional cards (2,A,A,A,A,2,3 in any order except A first).  At the end of each hand (21 or bust), the cards are then returned to the deck and reshuffled well and the next hand is played.
I am curious to know if by always getting an initial "10 card", if that makes the probability of getting 21 higher than if some random card is the first card (such as a 7).
I scanned other posts for blackjack but none of them seemed to talk about an initial 10 valued card as a precondition.
I plan on running a computer simulation for this eventually but am interested in the math approach too.
 A: Probably not worth trying to calculate this without computer help.
If you denote by $X_n$ the sum at time $n$ and you just add one card at a time then you have a probability transition matrix given by:
$P(X_{n+1}=X_n +k|X_n) = 1/13$ for $k=1,...,9$, $P(X_{n+1}=X_n+10|X_n)=4/13$
and the rest zero (here we do not take into account that probabilities changes in time, which makes the problem quite a lot harder).
Let $Q_{ij}= P(X_{n+1}=j|X_n=i)$ denote the matrix obtained for $0\leq i,j\leq 20$ and add to that matrix $Q_{21,21}=1$ which corresponds to stopping if you reach 21. Q is not really a probability matrix since all probabilities above 21 are simply lost.
Let $S$ be the event that we end up with 21.
Starting out with the vector $E_0=(1,0,0,...,0)\in {\Bbb R}^{21}$ we get:
$$P(S|X_0=E_0)=\lim_{n\rightarrow \infty} E Q^n = (0,...,0,0,0.1398)$$
implying that there is probability 0.1398 to end up with 21 starting from nothing. More generally, starting out with the vector $E_{j}=(0,...,1,...,0)$ (a 1 in the j'th column) we get (I don't guarantee that my matrix is error-free though):
$$\begin{array}
   {ccccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
     0.1398 & 0.2129 & 0.1559 & 0.1519 & 0.1476 & 0.1432 & 0.1386 & 0.1340 & 0.1293 & 0.1247 & 0.1200 
 \end{array} $$
Note that starting with an ace increase the probability quite a lot (reasonable as after that getting two 10's is fairly likely). Starting with 10 is, however, less favorable than starting from scratch.
