# 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.

• Just F.Y.I. for those of you thinking about doing a computer simulation like me, I am wondering if I can just do a full deck shuffle and then discard any cards up to the first 10 valued card and then just use the rest of the deck to get the 21 or bust. It seems reasonable since it is very unlikely to get all the low cards first (such as A,A,A,A,2,2,2,2...). I imagine it changes the probability some but not a lot since it is very likely a 10 valued card will appear 'early' in the 52 card deck, so we would only be discarding a few cards each hand or I could 'tack' them on the end of the deck. Commented Sep 2, 2017 at 15:32
• That sounds perfectly reasonable. Remember that in a perfect shuffle, the card specifically after the ten of clubs is equally likely to be any of the other 51 cards in the deck (using the original first card if 10 of clubs happened to be last). Another option instead of discarding the first many cards until reaching a 10valued card is to just work with a 51 card deck to begin with where it is missing the ten of clubs. Commented Sep 2, 2017 at 16:05
• Yes, if the first card is not already 10 valued, swap it in as the first card and put the original first card elsewhere in the deck, ideally where we 'yanked' any actual found 10 valued card in the deck. As a "spinoff" of this particular question, it is interesting to analyze which is the best ranked card to maximize the chances of 21 and which minimizes. It seems A (value 11) is the best card and maybe a 10 valued card is the worst (to get 21). However, in real blackjack, a 10 value as the first card is quite good since you can get 21 or 20 fairly easily and those are good hands for a win. Commented Sep 2, 2017 at 16:33
• now... that process does not sound like it will avoid affecting results. Stick with the discarding of the top however many cards, or cycling them to the bottom of the deck, but don't insert them somewhere into the middle of the deck. Alternatively and probably most easily, just follow my suggestion and shuffle a 51 card deck instead. If you don't have access to a method to shuffle a nonstandard deck, then shuffle a standard deck, pretend you dealt yourself a 10 of clubs to begin with, and then if when drawing more cards you accidentally come across a 10 of clubs, ignore it and redraw. Commented Sep 2, 2017 at 16:40
• With how you phrased your most recent comment, it sounded like you would swap the position of the top card with the next available 10value if the top card was not 10valued. This changes the probability then of the second card in the deck (after the swap) being a 10valued from $\frac{16}{52}$ to $\frac{16\cdot 15}{52\cdot 51}$ as the only way the second card was 10valued was if both first and second were 10valued (else if 2nd was not 10valued it wouldn't be affected by the swap and remains not 10valued, or if 2nd was 10valued it would be affected by teh swap if it occurred) Commented Sep 2, 2017 at 16:45

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.