God-Blessed Blackjack

Whether either side has blackjack is determined immediately from the initial two cards. If at least one side has blackjack, the answer is directly one of $0$, $1.5R$, and $-R$ according to the rules.

For the remaining normal game, divide all rewards by $R$: a win is $+1$, a draw is $0$, and a loss is $-1$. Multiply the final expected value by $R$.

Number the $N$ remaining cards from $0$ to $N-1$. Let bitmask $P$ be the set of cards additionally drawn by jtw7913 and bitmask $D$ the set additionally drawn by the dealer. During jtw7913's turn, $D=0$. After jtw7913 Stays, $P$ is fixed and only $D$ grows.

Every card is in exactly one of three states: still in the deck, drawn by jtw7913, or drawn by the dealer. Hence there are at most $3^N$ possible $(P,D)$ states.

Precompute for every mask $M$:

• $\mathrm{playerScore}[M]$: the score of jtw7913's initial two cards plus the cards in $M$.
• $\mathrm{dealerScore}[M]$: the score of the dealer's initial two cards plus the cards in $M$.

To evaluate a hand, first count every ace as 1 and let the sum be $s$. If there is at least one ace and $s+10\leq21$, count one ace as 11 and use $s+10$; otherwise use $s$.

Let $G(P,D)$ be the normalized expected result after jtw7913 has already Stayed, when his additional-card set is $P$ and the dealer's is $D$. If the dealer score exceeds 21, then $G(P,D)=1$. If the dealer score is at least 17 or the deck is empty, compare the final scores and return one of $1,0,-1$.

Otherwise the dealer must Hit. If $U$ is the set of remaining cards,

$$G(P,D)=\frac{1}{|U|}\sum_{i\in U}G(P,D\cup\{i\}).$$

For memoization, encode each card by one ternary digit: 0 if it remains in the deck, 1 if it belongs to $P$, and 2 if it belongs to $D$. Thus all dealer states fit in one array of size $3^N$.

Let $F(P)$ be the maximum normalized expected result when jtw7913 has additional-card set $P$ and may still choose an action. The value of Staying now is

$$S(P)=G(P,0).$$

If the current score is 21 or the deck is empty, jtw7913 automatically Stays, so $F(P)=S(P)$. Otherwise he may also Hit. With remaining-card set $U$,

$$H(P)=\frac{1}{|U|}\sum_{i\in U}h_i,$$

where $h_i=-1$ if drawing card $i$ busts, and otherwise $h_i=F(P\cup\{i\})$. Therefore

$$F(P)=\max(S(P),H(P)).$$

The answer for a normal game is $F(0)\cdot R$.

Score preprocessing takes $O(N2^N)$ time. There are at most $3^N$ dealer states, and each examines at most $N$ cards. The total time complexity is $O(N3^N)$ and the space complexity is $O(3^N)$.

Solution written by GPT5.6