problem 4 (Deck of Cards, 10 points). A standard deck of cards has 52 cards. Each card has a rank
and a suit. There are 13 ranks: A (Ace), 2, 3, 4, 5, 6, 7, 8, 9, 10, J (Jack), Q (Queen), K (King).
There are 4 suits: clubs (♣), diamonds (♢), hearts (♡), and spades (♠). Diamonds and hearts are
red, and clubs and spades are black.
Four cards are drawn randomly one-by-one without replacement. What is the probability that:(d) You receive cards that can be arranged into consecutive ranks (e.g. 4, 6, 7, 5), including those
that already consecutive (e.g. 4, 5, 6, 7)?

### Problem Breakdown:
We need to find the probability of drawing four cards from a deck that can be arranged into consecutive ranks. A standard deck of cards has 52 cards, each with 13 possible ranks and 4 possible suits. Drawing without replacement means the sequence of drawing affects the outcome.

### Key Points:
1. **Consecutive Ranks**: Four cards should have distinct ranks that can be ordered consecutively. This includes ranks like {4, 5, 6, 7} or {9, 8, 7, 6}.
2. **Ranks Available**: In total, there are 13 ranks, so for a group of four cards to be consecutive, the lowest rank in the group can only go up to 10 (e.g., {10, J, Q, K}). Starting at rank 10 is the highest point because going beyond this would exceed four ranks.
   
### Step-by-Step Approach:

#### Total Possible Outcomes:
The total number of ways to draw 4 cards from a deck of 52 cards is given by the combination:
\[
\binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270,725
\]

#### Successful Outcomes:
##### 1. **Choosing 4 Consecutive Ranks**:
- First, choose the starting rank of the consecutive sequence. The possible starting ranks are from Ace (A) to 10, giving us 10 options for the starting point (since the sequence must have exactly 4 consecutive ranks).
  
##### 2. **Assigning Suits**:
- For each of the 4 ranks in the chosen sequence, we have 4 suits available. Thus, for any specific sequence of ranks, the number of ways to assign suits is:
\[
4^4 = 256
\]
This accounts for the fact that each rank can come from any of the four suits.

##### 3. **Total Successful Outcomes**:
- There are 10 possible sequences (starting from ranks 1 to 10), and for each sequence, there are 256 ways to assign suits. Therefore, the total number of favorable outcomes is:
\[
10 \times 256 = 2,560
\]

#### Final Probability:
The probability is the ratio of successful outcomes to the total possible outcomes:
\[
P = \frac{2,560}{270,725} \approx 0.00946
\]
This is approximately a 0.946% chance.

### Final Answer:
The probability that four randomly drawn cards can be arranged into consecutive ranks is approximately **0.00946** or **0.946%**.

---

Would you like more details or have any questions on this?

Here are five related questions for deeper understanding:
1. How would the probability change if we were allowed to draw with replacement?
2. What is the probability of drawing four cards that are all the same rank?
3. What is the probability of drawing exactly two red cards and two black cards in four draws?
4. How does the probability change if we are only concerned with cards from a single suit?
5. Can you calculate the probability of drawing four cards with no consecutive ranks?

**Tip**: Always make sure to consider both the rank and suit when calculating probabilities in a deck of cards.

Four cards are drawn randomly one-by-one without replacement. What is the probability that you receive cards that can be arranged into consecutive ranks (e.g. 4, 6, 7, 5), including those that are already consecutive (e.g. 4, 5, 6, 7)?

Discover how to calculate the probability of drawing four cards that can be arranged into consecutive ranks from a standard 52-card deck. This problem uses combinatorics and probability theory, exploring key concepts like the binomial coefficient and counting principles. Suitable for students in Grades 10-12.

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