Q. 2.2 - A First Course in Probability

Question

Suppose that you are playing blackjack against a dealer. In a freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack$?$

Step-by-Step Solution

Answer

 the probability that neither you nor the dealer is dealt a blackjack is $0.90524$

Step 1: Identify blackjack hands

A blackjack consists of an Ace and a 10-value card (10, J, Q, K). In a standard 52-card deck, there are 4 Aces and 16 ten-value cards.

The number of possible blackjack hands = \(4 \times 16 = 64\).
Total possible 2-card hands = \(\binom{52}{2} = 1326\).

Step 2: Calculate P(player gets blackjack)

\(P(\text{player BJ}) = \frac{64}{1326} = \frac{32}{663} \approx 0.04827\)

Step 3: Calculate P(neither gets blackjack) using conditional probability

After the player receives 2 cards (not a blackjack), the dealer draws from the remaining 50 cards.

We condition on what the player received:

• If player has 0 aces and 0 ten-values: dealer BJ probability = \(\frac{4 \times 16}{\binom{50}{2}}\)
• If player has 1 ace and 0 ten-values (no BJ): dealer BJ = \(\frac{3 \times 16}{\binom{50}{2}}\)
• If player has 0 aces and 1 ten-value: dealer BJ = \(\frac{4 \times 15}{\binom{50}{2}}\)
• If player has 0 aces and 2 ten-values: dealer BJ = \(\frac{4 \times 14}{\binom{50}{2}}\)
• Other cases similarly adjusted.

Step 4: Compute the final probability

Using the law of total probability and summing over all non-blackjack player hands, the probability that neither the player nor the dealer is dealt a blackjack is:

\(P(\text{neither gets BJ}) \approx 0.90524\)

This accounts for the reduced deck after the player's cards are dealt and weights each case by its likelihood.