Probability of Drawing Face Cards with Replacement in a Standard Deck

Introduction

When dealing with probability, one common scenario involves drawing cards from a standard deck to determine the likelihood of certain outcomes. A question that often arises is: what is the probability of drawing two face cards with replacement? In this article, we will explore how to calculate this probability, addressing common scenarios and providing a clear explanation.

Understanding Basic Concepts

A standard deck of playing cards consists of 52 cards, divided into four suits (hearts, diamonds, clubs, and spades). Each suit has 13 cards, including the face cards: jack, queen, and king. Therefore, there are a total of 12 face cards in a deck.

Probability of Drawing a Face Card

The probability of drawing a face card on the first draw is:

[ P(text{A}) frac{12}{52} frac{3}{13} ]

Since the card is replaced, the deck still contains 52 cards for the second draw. Therefore, the probability of drawing a face card on the second draw is the same:

[ P(text{B}) frac{12}{52} frac{3}{13} ]

Calculating the Probability of Both Cards Being Face Cards

Since the draws are independent (due to replacement), the probability of both cards being face cards can be calculated as the product of the individual probabilities:

[ P(text{A and B}) P(text{A}) times P(text{B}) frac{3}{13} times frac{3}{13} frac{9}{169} approx 0.053254 ]

Alternative Method: Complementary Probability

Another way to find the probability of drawing at least one face card in two draws is by calculating the probability of the complementary event (neither card is a face card) and subtracting it from 1.

The probability of drawing a card that is not a face card on the first draw is:

[ P(text{not-A}) frac{40}{52} frac{10}{13} ]

Since the card is replaced, the probability of drawing a non-face card on the second draw is the same:

[ P(text{not-B}) frac{40}{52} frac{10}{13} ]

Therefore, the probability of both cards not being face cards is:

[ P(text{not-A and not-B}) P(text{not-A}) times P(text{not-B}) frac{10}{13} times frac{10}{13} frac{100}{169} approx 0.591715976 ]

Thus, the probability of drawing at least one face card is:

[ P(text{at least one face card}) 1 - P(text{not-A and not-B}) 1 - frac{100}{169} frac{69}{169} approx 0.4083 ]

Conclusion

In summary, the probability of drawing two face cards with replacement is approximately 0.053254, or 9/169. This calculation is fundamental in understanding basic probability and can be applied to various similar scenarios in card games and other probability-based studies.

To explore more detailed and complex probability questions, you can check out my Quora Profile for more insights and discussions.