Problem Statement

How many 6-sided dice with values on each side in the set \(\{1, 2, 3, 4, 5, 6\}\) are there with the property that when rolled twice, for each integer \(2 \leq k \leq 12\), there is positive probability that the sum is exactly \(k\)?

Note:

  • Not every value in the set needs to be used.
  • Two dice are considered indistinguishable if they contain the exact same multiset of face values.
  • Each face appears with equal probability.

Original Problem Link: Click here

Solution

We’re essentially being asked: do we really need all 6 numbers on a die to ensure that every sum from 2 to 12 can occur when rolling it twice?

We’ll figure this out systematically.

Step 1: What is the Goal?

We want all values \(k \in \{2, 3, \dots, 12\}\) to be attainable as the sum of two dice rolls (of the same die).
So we need to ensure that for each \(k\), at least one pair (a, b) with \(a, b \in \text{die faces}\) satisfies \(a + b = k\).

Step 2: Can We Remove Any Face?

Let’s try removing one value at a time from {1, 2, 3, 4, 5, 6}, and check whether every sum from 2 to 12 remains possible:

🔸 Remove 1:

  • Can’t make 2 = 1+1 → Invalid

🔸 Remove 2:

  • Can’t make 3 = 1+2 → Invalid

🔸 Remove 3:

Let’s check if all sums 2 through 12 are still attainable:

  • 2: 1+1
  • 3: 1+2
  • 4: 2+2
  • 5: 1+4
  • 6: 2+4
  • 7: 2+5
  • 8: 4+4
  • 9: 4+5
  • 10: 5+5
  • 11: 5+6
  • 12: 6+6 ✅
    → Valid, even without 3.

🔸 Remove 4:

Check again:

  • 2: 1+1
  • 3: 1+2
  • 4: 1+3
  • 5: 2+3
  • 6: 3+3
  • 7: 1+6
  • 8: 2+6
  • 9: 3+6
  • 10: 5+5
  • 11: 5+6
  • 12: 6+6 ✅
    → Valid, even without 4.

🔸 Remove 5:

  • Can’t make 10 = 5+5 → Invalid

🔸 Remove 6:

  • Can’t make 12 = 6+6 → Invalid

Step 3: Count the Valid Dice

We now count how many distinct multisets of dice satisfy the condition:

1. Original Die:

  • Uses all six values
    → 1 way

2. Dice without 3:

  • Replace 3 with any of the other 5 values → still 6 faces
    → 5 choices

3. Dice without 4:

  • Replace 4 with any of the other 5 values
    → 5 choices

Final Answer: \(\text{Total Valid Dice} = 1 + 5 + 5 = \boxed{11}\)

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