Find the probability of being dealt an "aces over fives" full house (three aces and two fives). (Round your answer to six decimal places).

To determine the probability of being dealt an "aces over fives" full house in a standard 52-card deck, we need to calculate the total number of ways to get this specific hand and then divide it by the total number of possible 5-card hands.

### Step 1: Calculate the Total Number of Possible 5-Card Hands

The total number of possible 5-card hands from a 52-card deck is given by the combination formula:

\[ \binom{52}{5} = \frac{52!}{5!(52-5)!} \]

Calculating this:

\[ \binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960 \]

### Step 2: Calculate the Number of Ways to Get "Aces Over Fives"

An "aces over fives" full house consists of three aces and two fives. 

- The number of ways to choose 3 aces from the 4 available aces is:

\[ \binom{4}{3} = 4 \]

- The number of ways to choose 2 fives from the 4 available fives is:

\[ \binom{4}{2} = 6 \]

Therefore, the number of ways to get this specific hand is:

\[ 4 \times 6 = 24 \]

### Step 3: Calculate the Probability

The probability \( P \) of being dealt an "aces over fives" full house is the number of favorable outcomes divided by the total number of possible 5-card hands:

\[ P = \frac{24}{2,598,960} \]

Let's calculate this fraction to six decimal places:

\[ P \approx \frac{24}{2,598,960} \approx 0.000009 \]

Thus, the probability of being dealt an "aces over fives" full house, rounded to six decimal places, is:

\[ 0.000009 \]

Do you want details or have any questions?

Here are 8 related questions that expand on this information:
1. What is the probability of being dealt a full house in general from a 52-card deck?
2. How do you calculate the probability of being dealt four of a kind?
3. What is the probability of being dealt a straight flush?
4. How does the probability of being dealt a specific full house compare to being dealt a flush?
5. What is the probability of being dealt two pairs?
6. Can you explain the difference between calculating combinations and permutations in the context of card hands?
7. How would the probability change if using a non-standard deck, such as one with jokers?
8. What is the probability of being dealt a hand with no pairs or special combinations (a high card hand)?

**Tip:** When calculating probabilities involving combinations, ensure you clearly define the specific outcomes of interest and correctly use the combination formula to count the number of favorable outcomes.

Learn how to calculate the probability of being dealt an "aces over fives" full house in a standard 52-card deck. This problem involves combinatorics and probability theory, specifically using the combination formula to count favorable outcomes. Suitable for students in Grades 10-12.

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