KindiakMath

Basic Blackjack

In this set of posts, we strategise over the mechanics behind a game of Blackjack.

Definition 1. A poker card deck comprises of 52 cards. Each card is uniquely characterised by two pieces of information: the value and the suit.

  • The value is an element from the set {A, 2, 3, …, 10, J, Q, K}.
  • The suit is chosen from the set {Clubs, Diamonds, Hearts, Spades}.

For each value, there exists 4 cards, one for each suit. Conversely, for each suit, there exists 13 cards, one for each value. Cards whose value belong to {J, Q, K} are called picture cards. The symbols J, Q, K denote Jack, Queen, and King respectively.

At the start of each round, two cards are dealt to the player without replacement.

Problem 1. Calculate the probability that the player’s hand contains two cards with the same value. We call this a double.

(Click for Solution)

Solution. Given each value A, 2, 3, …, 10, J, Q, K, the probability of a double is given by

\frac 4{52} \times \frac 3{51} = \frac 1{13} \times \frac 1{17} .

Since there are 13 total values, the required probability is

\mathbb P(\text{double}) = 13 \times \frac 1{13} \times \frac 1{17} = \boxed{ \textstyle \frac 1{17} }.

In Blackjack, the goal is to achieve the highest total score subject to a maximum total of 21. More precisely, for each card with value v \neq \mathrm A, define its score s(v) as follows:

s(v) = \begin{cases} v & \text{if}\quad v \in \{2,3,\dots, 10\}, \\ 10 & \text{if}\quad v \in \{\mathrm J, \mathrm Q, \mathrm K\}. \end{cases}

Given the initial hand (v_1, v_2) with values v_1, v_2, define its total score by

s(v_1,v_2) := s(v_1) + s(v_2).

Problem 2. Evaluate the total score of the hand (K, 6).

(Click for Solution)

Solution. By definition, s(\mathrm K, 6) = s(\mathrm K) + s(6) = 10 + 6 = \boxed{ 16 }.

For hands involving A (called Ace), define s(A) := 1. Furthermore,

s(\mathrm A, v) = s(\mathrm A) + s(v) + \begin{cases}10& \text{if}\quad s(\mathrm A) + s(v)\leq 11, \\ 0& \text{if}\quad s(\mathrm A) + s(v) > 11.\end{cases}

Regard score-summing as commutative, so that s(v, A) = s(A, v).

Problem 3. Evaluate the total score of the hands (A, A), (A, 2), (A, K) respectively.

(Click for Solution)

Solution. In all hands, s(\mathrm A) + s(v) \leq 11. Therefore,

\begin{aligned} s(\mathrm A, \mathrm A) &= s(\mathrm A) + s(\mathrm A) + 10 = 1 + 1 + 10 = 12, \\ s(\mathrm A, 2) &= s(\mathrm A) + s(2) + 10 = 1 + 2 + 10 = 13, \\ s(\mathrm A, \mathrm K) &= s(\mathrm A) + s(\mathrm K) + 10 = 1 + 10 + 10 = 21.\end{aligned}

Definition 2. A player obtains a Blackjack if s(v_1, v_2) = 21.

Problem 4. Calculate the probability that a player obtains a Blackjack.

(Click for Solution)

Solution. We observe that a player with a hand (V_1, V_2) has a Blackjack if and only if at least one of the cards is an Ace, and the other has score 10. Therefore,

\begin{aligned} \mathbb P(\text{Blackjack}) &= \mathbb P(V_1 \in \{\mathrm A\}) \cdot \mathbb P(V_2 \in \{10, \mathrm J, \mathrm Q, \mathrm K\}) \\ & \phantom{==} + \mathbb P(V_1 \in \{10, \mathrm J, \mathrm Q, \mathrm K\})\cdot \mathbb P(V_2 \in \{ \mathrm A \}) \\ &= \textstyle \frac 4{52} \cdot \frac{16}{51} + \frac {16}{52} \cdot \frac{4}{51} \\ &= \textstyle \frac{16}{663} \\ &\approx \boxed{ 2.41\% }. \end{aligned}

If a player does not get a Blackjack, he gets to choose to add one more card or stand. Given a hand (v_1,\dots, v_n), define its total score by

s(v_1,\dots, v_n) := s(v_1) + \cdots + s(v_n).

If a player’s total score exceeds 21, we say that the player busts and loses immediately.

Problem 5. A player has an initial hand (K, 6). If the player chooses to hit and obtain a card w, what is the probability that he busts with the hand (K, 6, w)?

(Click for Solution)

Solution. Since there are no Aces, the player’s current score is 16 by Problem 2. Furthermore, s(A) = 1 with no extra additions by 10. Therefore,

\begin{aligned} \mathbb P(\text{bust with }(\mathrm K, 6, w)) &= \mathbb P(w \in \{6,\dots, 10, \mathrm J, \mathrm Q, \mathrm K\}) \\ &= \mathbb P(w \in \{6, \mathrm K\}) + \mathbb P(w \in \{7,8,9, 10, \mathrm J, \mathrm Q\}) \\ &= \textstyle 2 \times \frac {3}{50} + 6 \times \frac{4}{50} \\ &= \textstyle \frac 35 = \boxed{ 60\%}. \end{aligned}

If any hand has an Ace, define

s(v_1,\dots, v_n) = s(v_1) + \cdots + s(v_n) + \begin{cases}10& \text{if}\quad s(v_1) + \cdots + s(v_n)\leq 11, \\ 0& \text{if}\quad s(v_1) + \cdots + s(v_n) > 11.\end{cases}

A non-Blackjack hand that has an Ace and s(v_1) + \cdots + s(v_n)\leq 11 is called a soft total.

Problem 6. Given an initial soft total (A, v), the player hits and obtains a card with value W. Calculate the probability that s(A, v, W) = 21.

(Click for Solution)

Solution. Since the total is soft, it is not a Blackjack, so that v \in \{\mathrm A, 2, \dots, 9\}. Regarding the second ace A as 1, for each v, 10 - v \in \{1,2,\dots, 9\}.

\begin{aligned} \mathbb P(s(\mathrm A, v, W) = 21) &= \mathbb P(s(W) = 10-v) \\ &= \mathbb P(W = 10-v). \end{aligned}

If v \notin \{5, 9\}, then 10-v \notin \{1,v\}. Hence,

\begin{aligned} \mathbb P(s(\mathrm A, v, W) = 21) &= \mathbb P(W = 10-v). = \textstyle \frac 4{50} = \boxed{ \textstyle \frac 2{25} }.\end{aligned}

If v \in \{5, 9\}, then 10-v \in \{1, v\}. Hence,

\begin{aligned} \mathbb P(s(\mathrm A, v, W) = 21) &= \mathbb P(W = 10-v). = \boxed{ \textstyle \frac 3{50} }.\end{aligned}

Each game is played against the dealer. The dealer will have one card exposed and the other card concealed. After the player no longer hits, the dealer will keep hitting until its score total exceeds 16, after which, the dealer is no longer allowed to add any more card.

Let p_n denote the probability that the dealer busts given that his initial hand (v_1,v_2) has no Aces and an initial total s(v_1, v_2) = n.

Problem 7. Evaluate p_{16}.

(Click for Solution)

Solution. By definition, the dealer must bust in his next hit W, so

\textstyle p_{16} = \mathbb P(s(W) > 5) = \boxed{ \textstyle \frac 8{13} }.

Remark 1. These problems don’t even account for the reward system in Blackjack, which requires a more detailed analysis of expected values, or before that, the case when the dealer has a soft hand (e.g. Aces in his initial hand).

—Joel Kindiak, 5 Apr 26, 1847H

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