A die is thrown three times. The chance that the highest number shown on the die is 4 is p, then the value of `[1//p]` is where [.] represents greatest integer function is _________.

Explanation

To solve the problem, we need to find the probability \( p \) that the highest number shown on a die thrown three times is 4. We will break this down into cases based on how many times the number 4 appears in the throws.

### Step 1: Understanding the Conditions
The highest number can be 1, 2, 3, or 4. Since we want the highest number to be exactly 4, all three throws must show numbers from the set {1, 2, 3, 4}.

### Step 2: Counting the Total Outcomes
When a die is thrown three times, the total number of outcomes is:
\[ 6^3 = 216 \]

### Step 3: Case 1 - Exactly 1 occurrence of 4
If 4 appears exactly once, the other two throws must show numbers from {1, 2, 3}. The number of ways to choose which throw shows 4 is \( \binom{3}{1} = 3 \). The other two throws can each be any of the three numbers (1, 2, or 3), so:
\[ \text{Number of outcomes} = 3 \times 3 \times 3 = 27 \]
Thus, the probability for this case is:
\[ P(\text{exactly 1 four}) = \frac{3 \times 27}{216} = \frac{81}{216} = \frac{1}{8} \]

### Step 4: Case 2 - Exactly 2 occurrences of 4
If 4 appears exactly twice, the third throw must show a number from {1, 2, 3}. The number of ways to choose which two throws show 4 is \( \binom{3}{2} = 3 \). The remaining throw can be any of the three numbers:
\[ \text{Number of outcomes} = 3 \times 3 = 9 \]
Thus, the probability for this case is:
\[ P(\text{exactly 2 fours}) = \frac{3 \times 9}{216} = \frac{27}{216} = \frac{1}{8} \]

### Step 5: Case 3 - Exactly 3 occurrences of 4
If all three throws show 4, there is only one outcome:
\[ \text{Number of outcomes} = 1 \]
Thus, the probability for this case is:
\[ P(\text{exactly 3 fours}) = \frac{1}{216} \]

### Step 6: Total Probability
Now, we sum the probabilities from all three cases:
\[ p = P(\text{exactly 1 four}) + P(\text{exactly 2 fours}) + P(\text{exactly 3 fours}) = \frac{1}{8} + \frac{1}{24} + \frac{1}{216} \]

### Step 7: Finding a Common Denominator
The least common multiple of 8, 24, and 216 is 216. We convert each fraction:
\[ \frac{1}{8} = \frac{27}{216}, \quad \frac{1}{24} = \frac{9}{216}, \quad \frac{1}{216} = \frac{1}{216} \]
Now, summing these gives:
\[ p = \frac{27 + 9 + 1}{216} = \frac{37}{216} \]

### Step 8: Finding \( \left\lfloor \frac{1}{p} \right\rfloor \)
Now, we need to find \( \frac{1}{p} \):
\[ \frac{1}{p} = \frac{216}{37} \approx 5.8378 \]
Thus, the greatest integer function gives:
\[ \left\lfloor \frac{1}{p} \right\rfloor = 5 \]

### Final Answer
The value of \( \left\lfloor \frac{1}{p} \right\rfloor \) is \( \boxed{5} \).