In this problem, we're dealing with a modified deck of cards where all the face cards (Jack, Queen, and King) have been removed. Let's break down the scenario:

1. Understanding the Modified Deck:

• A standard deck has 52 cards. After removing the face cards (Jack, Queen, and King), each suit (hearts, diamonds, spades, clubs) will have 10 cards each: Ace, 2 through 10.
• This means we are left with a total of $40$ cards in the deck.

2. The Objective:

• You are to draw 4 cards in total: the first two drawn simultaneously without replacement and the next two also drawn simultaneously without replacement.
• The goal is:
  • The first two cards should add up to 15.
  • The next two cards should add up to 17.

3. Possible Pairs that Sum to 15:

Let’s first find all pairs of values in the modified deck that sum up to 15.

• $10 + 5 = 15$
• $9 + 6 = 15$
• $8 + 7 = 15$

So the possible pairs that sum to 15 are $(10, 5), (9, 6), (8, 7)$.

4. Possible Pairs that Sum to 17:

Now, let's find all pairs that sum to 17.

• $10 + 7 = 17$
• $9 + 8 = 17$

So the possible pairs that sum to 17 are $(10, 7), (9, 8)$.

5. Total Possible Ways to Draw Cards:

Since there are 40 cards, the number of ways to choose 4 cards out of these 40 is calculated as: $\binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4 \times 3 \times 2 \times 1} = 91,390$ So there are 91,390 possible ways to draw any 4 cards from the deck.

6. Favorable Outcomes:

We now need to calculate the number of favorable outcomes where the first two cards sum to 15 and the next two sum to 17.

Case 1: First Pair Adds to 15

The valid pairs for a sum of 15 are $(10, 5), (9, 6), (8, 7)$. Each pair can be chosen in $4 \times 4 = 16$ ways (since there are 4 suits for each number). So, there are: $3 \times 16 = 48 \text{ ways to choose two cards that sum to 15}.$

Case 2: Second Pair Adds to 17

The valid pairs for a sum of 17 are $(10, 7), (9, 8)$. Each pair can be chosen in $4 \times 4 = 16$ ways. So, there are: $2 \times 16 = 32 \text{ ways to choose two cards that sum to 17}.$

7. Total Number of Favorable Outcomes:

The number of favorable outcomes is the product of the favorable outcomes for the two draws: $48 \times 32 = 1,536.$

8. Probability Calculation:

Finally, the probability is the ratio of favorable outcomes to total outcomes: $P = \frac{1,536}{91,390} \approx 0.0168.$

Thus, the probability that the first two cards sum to 15 and the next two sum to 17 is approximately $0.0168$ or $1.68\%$.

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Would you like further explanation on any part of the solution?

Here are some related questions for further exploration:

1. What would be the probability if we also include face cards?
2. How would the probability change if we drew cards one at a time instead of simultaneously?
3. What are the odds of drawing a pair that sums to a specific number in one draw?
4. How does the probability change if the deck is reshuffled after each draw?
5. What would the probability be if we draw 5 cards instead of 4?

Tip: Understanding combinations ($\binom{n}{r}$) and how to apply them is crucial for solving problems involving selections and probabilities in card games.