Probability of One Player Drawing Cards Labeled 1 and 2 from a Deck of 24 Cards
In this article, we will explore the probability of a single player drawing the cards labeled 1 and 2 from a deck of 24 cards when four players each draw five cards at random. We'll break down the problem into manageable parts and provide a detailed step-by-step solution using combinatorial mathematics.
Introduction to the Problem
The problem at hand involves a deck of 24 uniquely numbered cards, from 1 to 24. Four players, each drawing five cards randomly, we want to determine the probability that one of these players will draw the specific cards labeled 1 and 2.
Step-by-Step Solution
Total Number of Ways to Choose 5 Cards from 24
First, we need to determine the total number of ways to choose 5 cards from a deck of 24. This can be represented using the binomial coefficient (binom{24}{5}), which calculates all possible combinations.
The formula for the binomial coefficient is:
[binom{n}{k} frac{n!}{k!(n-k)!}]For our specific case:
[binom{24}{5} frac{24!}{5!(24-5)!} frac{24 times 23 times 22 times 21 times 20}{5 times 4 times 3 times 2 times 1} 42504]Ways to Choose Cards with 1 and 2
Next, let's consider the scenario where a specific player is chosen to draw the cards labeled 1 and 2. After selecting these two cards, we need to randomly select 3 more cards from the remaining 22 cards.
The number of ways to choose 3 cards from the remaining 22 cards is given by:
[binom{22}{3} frac{22!}{3!(22-3)!} frac{22 times 21 times 20}{3 times 2 times 1} 1540]Calculating the Probability
The probability that one specific player draws the cards labeled 1 and 2 and then draws 3 additional cards is the ratio of the number of favorable outcomes to the total number of outcomes.
The probability is given by:
[P(text{Player draws 1 and 2}) frac{binom{22}{3}}{binom{24}{5}} frac{1540}{42504}]Simplifying this ratio, we get:
[P(text{Player draws 1 and 2}) approx 0.0362 text{ or } 3.62%]Alternative Methodology
Another approach to solving this problem involves considering the sequence in which cards are drawn. We can calculate the probability by considering any specific player drawing the first two cards as 1 and 2 and then considering the positions of these cards within the five draws.
Probability Using Sequence Consideration
The probability that a specific player draws the cards labeled 1 and 2 as the first two cards is (frac{1}{24} times frac{1}{23}).
Since the 1 could be drawn at any one of the 5 positions and the 2 at any one of the remaining 4 positions, the total probability for that player to have both a 1 and a 2 is:
[P(text{Player draws 1 and 2}) 20 times frac{1}{24} times frac{1}{23} frac{20}{552} frac{5}{138} approx 0.0362 text{ or } 3.62%]Alternatively, we calculate the probability that the 1 is drawn at all, which is (frac{20}{24}), and then the probability that the same player draws the 2 in one of the remaining 4 positions, given the 1 is not in that position, which is (frac{4}{23}).
Multiplying these probabilities gives the same result:
[P(text{Player draws 1 and 2}) frac{20}{24} times frac{4}{23} frac{5}{69} approx 0.0362 text{ or } 3.62%]Conclusion
We concluded that the probability of one player drawing the cards labeled 1 and 2 from a deck of 24 cards when four players each draw five cards is approximately (frac{5}{69}) or around 3.62%. Both methods provided the same result, validating the accuracy of our calculations.