Math Problem Statement

In cribbage, each player is dealt 6 cards from a standard deck of cards. In how many ways could a hand contain a) at least two queens

Solution

To solve the problem of determining how many hands of 6 cards in cribbage contain at least two queens, we can break it down as follows:


Step 1: Total number of hands

The total number of ways to deal a hand of 6 cards from a 52-card deck is given by: (526)=52×51×50×49×48×476×5×4×3×2×1=20,358,520\binom{52}{6} = \frac{52 \times 51 \times 50 \times 49 \times 48 \times 47}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 20,358,520


Step 2: Find the complement ("no queens" and "exactly one queen")

Let’s use the complement rule:

  • "At least two queens" is the same as not having 0 queens or 1 queen.

Step 3: Calculate hands with 0 queens

  • There are 4 queens in a deck of 52 cards, so the number of cards excluding queens is 48.
  • The number of hands with 0 queens is given by selecting all 6 cards from the 48 non-queen cards: (486)=48×47×46×45×44×436×5×4×3×2×1=12,271,512\binom{48}{6} = \frac{48 \times 47 \times 46 \times 45 \times 44 \times 43}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 12,271,512

Step 4: Calculate hands with exactly 1 queen

  • We must select 1 queen out of the 4 available:
    (41)=4\binom{4}{1} = 4

  • Then we must select the remaining 5 cards from the 48 non-queen cards: (485)=48×47×46×45×445×4×3×2×1=1,712,304\binom{48}{5} = \frac{48 \times 47 \times 46 \times 45 \times 44}{5 \times 4 \times 3 \times 2 \times 1} = 1,712,304

  • Therefore, the number of hands with exactly 1 queen is: (41)×(485)=4×1,712,304=6,849,216\binom{4}{1} \times \binom{48}{5} = 4 \times 1,712,304 = 6,849,216


Step 5: Apply the complement rule

The number of hands with at least two queens is given by: Total hands(Hands with 0 queens+Hands with exactly 1 queen)\text{Total hands} - (\text{Hands with 0 queens} + \text{Hands with exactly 1 queen}) =20,358,520(12,271,512+6,849,216)=20,358,52019,120,728=1,237,792= 20,358,520 - (12,271,512 + 6,849,216) = 20,358,520 - 19,120,728 = 1,237,792


Conclusion:

The number of hands containing at least two queens is: 1,237,792\boxed{1,237,792}


Would you like further details on this or another explanation?


Here are some related questions:

  1. What is the probability of getting at least two queens in a cribbage hand?
  2. How many hands contain exactly three queens?
  3. How many hands contain all four queens?
  4. What is the probability of having no queens in a 6-card hand?
  5. How does this change if we deal a 5-card hand instead?

Tip: The complement rule is a powerful technique for calculating probabilities or counting cases without directly enumerating complex conditions.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Probability
Complement Rule

Formulas

\binom{n}{k} = \frac{n!}{k!(n-k)!}

Theorems

Complement Rule
Combinatorics for Card Selection

Suitable Grade Level

Grades 9-12